CHAPTER-1

Number TheoryNumber is a fundamental concept of mathematics. The first conceptions of numbers were acquired by man in remote antiquity. It began with the counting of people, animals and various articles. The process of counting produced the numbers one, two, three, etc. In ancient times numbers were denoted by straight-line strokes. This notation was found inconvenient to write large numbers. Therefore symbols were invented for large numbers. These symbols are known as numerals. The most popular and widely used number system is the Hindu-Arabic system that consists of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Addition (+), Multiplication (), Subtraction (-), Division (), Involution (raising a number to a power) and Evolution (finding the root of a number) are the six fundamental arithmetic operations. Addition and subtraction, multiplication and division and involution and evolution are pair wise inverse operations.

Natural Number

The No’s 1, 2, 3, . . . . . are called natural numbers. They are also known as counting no positive integers. The set of natural no’s is usually denoted by N i.e. N={1, 2, 3,…}

Note: 1. 1 is the least natural no.2. There is no greatest natural no.

Whole Numbers

The No. 0, 1, 2, 3,. . . . . are called whole No’s. They are also known as non-negative integers. The set of whole no is usually denoted by W.

W= {0, 1, 2, 3,…..}

Note: ‘0’ is he only whole no which is not a natural no.

Integers

The natural no’s together with their negative counterparts and zero called integers. The set of integers is denoted by I or Z.

Z = {…….-2, -1, 0, 1, 2,…..}

Note: 1. The set Z = {-1, -2, -3,…..} is called the set of negative integers.

2. ‘0’ is neither a positive nor a negative integer.

Rational Numbers

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A number which can be expressed in the form P/Q, where P is an integer and Q is a natural number is called a rational number. The set of rational numbers is denoted by Q.

Example: 2, -2, -5.5, 5.5 and 10.8752 etc are all rational numbers.

Note: Every integer is as rational number, for eg if x is an integer, if can be written as x/1

Irrational Numbers

A number which cannot the expressed in form P/Q, P is an integer and Q is a natural number is called irrational number. An irrational number is also known as incommensurable quantity. The set of irrational numbers is usually noted by Q’.

Examples: 2=1.414 ………, 3=1.732 ……, 5=2.236 …., are all irrational numbers.

Note:

1. The ratio of the circumference to diameter of a circle is an irrational number and is denoted by . Its value is 3.14159….. The rational number 22/7 very often taken as an approximated value of . In fact a better approximation of is 355/113.

2. The sum 1+1/1+1/12+1/123+….. is an irrational number and is denoted by ‘e’ and its value is 2.718….

Real Numbers

The rational numbers and irrational numbers are together called real numbers. The set of real numbers is denoted by R.

The German mathematician Richard Dedekind established that “Every real number can be represented by a point on a line called the real number line and conversely every point on the real number line represents a real number”

Concepts of Decimals

Terminating Decimals

A real number is called a terminating decimal if the number of digits succeeding the decimal part is finite.

Example: 1, 2, 12, 34 and 123.456 are all terminating decimals.

Non-terminating recurring decimals

A real number is called a non-terminating recurring decimal if a certain number of continuous digits repeat infinitely after the decimal point. It is also known as a non-terminating periodic decimal.Examples: 0.333…….and 5.23467467….are non-terminating recurring decimals. They are

denoted by 0.3 and 5.23467 respectively. Note: Every non-terminating non-recurring decimal is an irrational number.

Concepts of FractionsA common fraction is a part of units or several equal parts of units. The numbers which indicated how many parts a unit is divided into is called the denominator of the fraction and the number

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indicating how many parts are taken is the numerator of the fraction. 4/7. 2/3, 5/1, are examples of common fractions.

Proper Fractions

A proper fraction is one in which the numerator is less than the denominator.

Example: 4/7, 1/3, 1/7, 1/5 etc. A proper fraction is always less than 1.

Improper Fractions

An improper fraction is one in which the numerator is greater than or equal to the denominator.

Example: 5/4. 3/2, 9/5, 7/3, etc. Improper fractions are a form of mixed fractions. An improper fraction is always greater than or equal to1.

Mixed Fraction

Fraction consisting of whole numbers and fraction are mixed fractions.

Example: 5½, 3¾ etc.

To change a mixed number into a fraction:

a. Multiply the whole number by the denominator of the fractional part.

b. Add the result to the numerator of the fractional part.

c. Divide the result by denominator of the fractional part.

Example: 5½

Step 1 52=10

Step 2 10+1=11

Step 3 11/2 Result

Complex Fractions

Fractions whose numerator and denominators are fractions are known as complex fractions.

Decimal Fractions

A set of digits after a decimal is known as a decimal fraction. For example 0.464, 0.5, 0.33, 0.49 etc. The numbers are known as decimal fraction because they can be converted to proper fractions by dividing and multiplying them by 10n (where n is the number of digits after the decimal point)

Arithmetical Operations With Fractions

Multiplying Fractions

In order to multiply fractions, multiply their numerators and divide the result by the product of their denominators.

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Examples:

1. a/b c/d = ac/bd 2. a/b c/d e/f = ace/bdf

Dividing Fractions

In order to divide two fractions, multiply one fraction by the inverted second fraction

Example: a/b d/c = a/b c/d

Equivalent Fractions

Two fractions are equivalent if on reducing both the fraction to their lowest terms they are found to be equal.

Example: 4/5 and 40/50 are equivalent fractions because 40/50 = 4/5 on being reduced to its lowest terms.

Note: The value of a fraction is not changed if the numerator and denominator are multiplied by the same number.

Example: 4/5= 4*3/5*3= 12/15

This is known as reducing the fraction to higher terms. The value of the fraction is not changed if the numerator ad denominator are both multiplied (or both divided by the same number)

Even and Odd Numbers:An integer is called an even number if it is a multiple of 2, otherwise it is called an odd number.

Thus the set of the even integers is {0, +2, +4…..} and the set of odd integers is {+1, +3, +5, ……}

Note:

1. The sum, difference and product of two even numbers is again an even number. For example 8+4=12 is even, 16-12=4 is even 64=24 is even.

2. The sum and difference of two odd numbers is an even number. For example 3+5=8 is even and 21-7=14 is even.

Prime NumbersThe integers P > 1 is called a prime number if and only if the integer has got exactly two distinct factors.

Example: 2, 3, 5, 7, 11, ………… are all prime numbers.

Note: 1. Around 2,500 years age. Euclid established that shore are an infinite number of primes.

2. The largest prime number known till date is 230,21,377-1 and has 9,09,526 digits. It was discovered very recently day a 19-year old student called Roland Clarkson from the California State University.

3. 2 is the only even prime number, for any other even number will have 2 as a factor.

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Composite NumbersAn integer C > 1 is called a composite number, if and only if it has at least three or more factors including 1 and the integer itself.

Example: 4, 9, 10……..are all composite numbers.

Twin PrimesPrime numbers differing by 2 are called twin primes.

Example: 3 and 5, 5, and 7, 11 and 13 are all twin primes.

Prime TripletA set of three prime numbers, differing by 2 in called a prime triplet.

Example: The only set of prime triplets is {3, 5, 7}.

Perfect Numbers

A number is known as a perfect number it the sum of all the possible divisors of the number (excluding the number itself) is equal to the value of the number. For example, take integer 6.

The possible divisors of 6 are 1, 2 and 3 and 1 + 2+ 3 = 6. Similarly for 28, 496, 8128 etc., the sum of the possible divisors equal the number itself.

Complex NumbersThere is no real number which when squared given a negative number. In higher mathematical calculations often we come across such situations where in we need to find the roots of negative numbers. The set of real numbers is inadequate in handling such calculations. In order that these calculations are facilitated, imaginary numbers were defined. Imaginary numbers are root of negative numbers. The number I = -1 was defined with the property that is i2 =-1.

The sum of the real number and an imaginary number constitutes a complex number. For example 2 + -5 can also be denoted an 2 i5 (where I = square root of –1).

FactorsAny natural number will have at least two factors, they are 1 and the number itself.

Only the number 1 has a single factor i.e. itself consider for example:

i. 2 Factors of this number are 1, 2.ii. 4 Factors are 1, 2, 4iii. 6 Factors are 1, 2, 3, 6iv. 28 Factors are 1, 2, 4, 7, 14, 28

MultipleWhen a given a number ‘X’ is multiplied by an integer (Y) we get the multiple of that number ‘X’. It like adding the value X, Y times say 12. It is multiple of 3, 4, also 2, 6

3 4 = 3+3+3+3=12

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X Y

2 6 =2+2+2+2+2+2=12 X Y

Highest Common Factor (HCF)

It is the highest common factor to two or more given numbers. It is also known as greatest or highest common factor and is denoted as either HCF or GCF

Example: (a) 5 is the HCF of 10, 25(b) 1 is the HCF / GCF of 7, 9, 2.

Note: The HCF of any to prime numbers is always ‘One’.

Method to find other HCF / GCF of given numbers.

i. Factorization method :

Express each number is product of primes. Now take the product of common factors, this is HCF.

Example: 30, 24, 36, 810

30 =23524 =222336 =2233810 = 333325

HCF = 2 3 = 6

ii. By division method:

Suppose two number are given, Divide the greater number by lesser number, divide the lesser by the reminder, Divide the first reminder by the second reminder and so on till there is no reminder. The last divisor is the HCF required.

Example:

a. Find HCF of 2002, 182

By Division Method:

182) 2002 (11 2002 0000

HCF = 182;

b. 189, 1197

189) 1197 (6 1134

63) 189 (3

6

189 000

HCF = 63

The HCF of 189, 1197 is a factor of 1197 – (189 6) =63

Again the common factor of 63, 189 is 63 which gives a reminder of zero after dividing it with 189 The HCF of these two numbers is 63.

Note: In case of more than two numbers, say three nos. choose any two of them and find their HCF. The HCF of these two and the third gives HCF of the third numbers.

Least Common Multiple (LCM):

Least common multiple of two or more numbers is the smallest (least) number that is exactly divisible by each of them. Eg: 24

a. The number 24 is the LCM of 2, 3, 4, 6, 8, 12…….

b. 45, is the LCM of =3, 5, 9, 15

Method to find out LCM of a given number:

i. By method of factorization:

Resolve each one of the given numbers into prime factors, them their LCM is the product of highest power of all factors, that occur in these numbers.

Ex: LCM of 40, 81, 256, 111

40 = 4 10 = 2225 = 23 5

81 = 34

111 = 3 37

LCM = 23 34 51 37 = 119880

ii. By Formula:

Product of numbers = HCF LCM

LCM= Product of Numbers/HCF

Ex. LCM 21, 54, is LCM= 21 54/3 = 378

iii. Using factors:

LCM of 15, 24, 64, 100

LCM = 222222355= 26 3 52

= 4800

Important Rule

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Let A and B be two numbers, then

Product of A and B = HCF(A, B) LCM (A, B)

HCF and LCM of fractions

1. HCL of fractions = HCF of Numerators/LCM of Denominators

2. LCM of fractions = LCM of Numerators/HCF of Denominators

Example: Find the HCF and LCM of 2/7 = 4/24

i. HCF = 2/74 = 1/140 = 1/14ii. LCM = 4/7

Concepts of Divisibility of NumbersWe will categories the Divisibility criteria in three broad types:

i. For Numbers 2 to 11ii. Using the concept of seed for bigger prime numbers (not discussed in this booklet as it is not required for Campus Placements)iii. Using co-prime concept for composite numbers(not discussed in this booklet as it is not required for Campus Placements)

Divisibility for Numbers from 2 to 11

i. Divisibility by 2: A number is said to be divisible by two if its units digit is ‘even’ or ‘zero’.

E.g.: 24, 36, 100, 10, 2 etc.

ii. Divisibility by 3: A number is said to be divisible by three, if the sum of the digits is divisible by 3.

E.g.: 27=2+7=9 is divisible by ‘3’ 27 is div. by 3339=3+3+9=15 is divisible by ‘3’ 339 is div. by 3

The method is known as “digit sum”, where in we add up the individual digits of a given umber and divide the result to check for divisibility.

iii. Divisibility by 4: If the last two digits of a given number is divisible by ‘4’ or the last two digits of the given number is ‘00’, then the number is divisible by 4.

E.g.: 256, 5120, 1336

iv. Divisibility by 5: A number is said to be divisible by if the units digits of the given number is 5 or ‘0’

E.g.: 225, 125, 100 etc

v. Divisibility by 6: A number is said to be divisible by 6 if it is divisible by 2 and 3 both. This will be further explained in the co prime concept for composite numbers.

E.g.: 12, 216, 78 etc.

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vi. Divisibility for 7: Test for divisibility of 7.

See if the number is a large number i.e. more than 6 digits then, Do the following.

E.g.: Consider this numbers: 8235437

The above number is divisible by 7 if the sum of the product of the digits of the number from left to right with 1, -2, -3, -1, 2, 3…….Successively us divisible by 7 or is equal to ‘0’.

= 8x(1) + 2x(-2) + 3x(1-3) + 5x(-1) + 4x2 + 42 +33+ +71

= 8 – 4 – 9 - 5 + 8 + 9 + 7

= 23 – 9

= 14 is divisible by ‘7’

We would discuss this again in the seed concepts of divisibility.

vii. Divisibility of Eight: A number is said to be divisible by eight if its, last three digits are divisible by eight or if the last three digits are ‘000’.

E.g.: 2512, 14256, 17200

viii. Divisibility by Nine: A number is said to be divisible by ‘9’ is the sum of the digits is divisible by ‘9’. I.e., we are again taking into account the digits um

E.g.: 43731 = 4 + 3 +7 +3 +1 = 18 is div. By 9.

ix. Divisibility by ten (10): A number is said to be divisible by 10 if it last digits is ‘0’

E.g.: 1000, 500, 100, etc.

x. Divisibility for 11: A number is said to be divisible by 11 if the difference between the sum of the digits in odd position and even position is zero or eleven or a multiple of eleven.

E.g.: 1331 = 1+3 = 43 +1 = 4 4 – 4 = 0, is divisible by 11.

9174 = 9+ 7 = 16 sum of digits of even position1+4 = 5 sum of digits of odd position 16 – 5 = 11 is divisible by 11.

Square of a numberWhen a number is multiplied by itself the number obtained is called the square of that number.

E.g.: Square of 5 = 5 5 = 25;

Square of 13 = 13 13 = 169;

Square of 9 = 9 9 = 81; etc

Perfect Square:

The square of a natural number is called perfect square.

Some properties of square number:

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i. Square of a number cannot end with odd number of zeros.

ii. Square of a number cannot end with 2, 3, 7, 8.

iii. Square of 1, 5, 6 always end with the digits 1, 5, 6 respectively.

iv. Square of an even number is always even.

v. Square of an odd numbers is always odd.

vi. Every square number is a multiple of 3, or exceeds a multiple of 3 by unity.

vii. Every square number is a multiple of 4, or exceeds a multiple of 4 by unity.

viii. Square of a negative number is always positive.

ix. To find the square of a number ending in 5, multiply the ten’s digits by next higher integer and annex 25.

E.g.: 1052 = 10 (ten’s digits) 11 (next higher integer) = 11025; 25 is annexed to get the Ans.

xi. If a square ends in 9, the proceeding digit is even.

Cube of Number

Cube a number is obtained by multiplying the number twice by itself: * *

i. Cube of a number can end in any integer from 0 to 9.

ii. Cube of a negative number will always have negative sign in the result.

Square & Cube roots

1. The number of digits in the square root of a ‘n’ digit number is

a. n+1/2 when n is odd

b. n/2 when n is even

2. The square root of any number is either positive or negative.

3. The cube root of any positive number is always positive.

Square root of a number

1. Prime factorization method

Find square root 3600 = 3600 22223355 = 2235 = 60

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2. Division method: Find the square root of 2116. Do the following steps

Square Root of a Perfect Square:Step 1: The integers are grouped in pair (2 at a time)

From the RHS of number i.e. from the units digits

We have two groups 21, 16

Step 2: The integer whose square is less than or equal total first group is written on the top and the left of the number.

Step 3: The Square of the integer is subtracted from the first group

Step 4: Now take the next pair of numbers. We get 516 as divided.

Step 5: Double the divisor (which make it 8) and add a number (in this case 6) which multiplied by 8 (in this care 86) gives a numbers less than or equal to 516. 86 6 = 516; But (here 6) in the quotient. There is no reminder. So square of 2116 is 46.

Square Root of not a Perfect Square:

a. To find the square root of 51

First find the nearest square to 51 i.e. 49. Subtract 49 from 51 = 2

Now, 49 is the reminder of 7. Now, divide the reminder 2 by two times 7, (27) to get

= 2/27 = 1/7 = 0.1412.

Add this to 7, and you have 7.14 as the square root of 51.

Step 1: 51 - 49=2

Step 2: 2/14 = 1/7 =0.14

Step 3: 7 + 0.14 = 7.14 square root

b. To find square root of 80

The nearest square number is 81.

Subtract 80 from 81. = 1

Now 81 is square of 9.

Divide the reminder 1 by two times 9.

= 1/29=1/18 = 0.055;

Subtract this from 9 = 8.95 as the square root of 80

Step 1: 80 = 81 –1

Step 2: -1/18 =-0.05

Step 3: 9 – 0.05 = 8.95

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Memorize the following

i. Square from 1 to 60, look in for some pattern.

ii. Square roots from 1 to 50; look in for some patterns

iii. Cube from 1 to 30

iv. Cube roots from 1 to 30

v. Reciprocals from 1 to 30;

Concepts of Last digit

For any no. the digit in units place is called as last digit.

Observe some important facts in the last digits and memorize till the fourth power.

1. Even numbers:

21 = 02 41 = 04 61 = 06 81 = 0822 = 04 42 = 16 62 = 36 82 = 6423 = 08 43 = 64 63 = 216 83 = 51224 = 16 44 = 256 64 = 1296 84 = 409625 = 32 45 = 1024 65 = 7776 85 = 32768

* Observe the fifth power of the above number will have the same last digit as the first power.

2. Odd numbers:

31 = 03 51 = 05 71 = 07 91 = 0932 = 09 52 = 25 72 = 49 92 = 8133 = 27 53 = 125 73 = 434 93 = 72934 = 81 54 = 625 74 = 2401 94 = 656135 = 243 55 =3125 75 = 16807 95 = 59049

* Observe that again the fifth power of the above given odd number will have the same digit as the first power.

* See that fourth power of even number is ending with ‘6’ multiplied by any even number (say x) when multiplied with this number ‘6’ will have the same last digit of (x).

Example: 6 12 = 7216 4 = 646 18 = 108

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* Also observe that the fourth power of every odd number ends in 1; when any given number is multiplied with this fourth power, the product will have the same last digit as the no. used as multiplier.

81 7 = 56781 6 = 486

* Observe that for numbers ending in 0, 1, as 5 the last digit is 0, 1, 6 and 5 respectively.

Some examples of last digit concept

1. Say the number is of the form xy.

* Observe x: Is the last digit odd or even? If it is even, any power of that should end in even only. If last digit of x is odd, any power of this should end in odd only.

* Break up ‘y’ as a multiple of 4 + odd /even depending on the example (case) i.e express y = 4g + k (k can be odd or even and will be definitely less than 4)

* Now x to power of 4g + odd /even will have the same last digit as power of last digit of ‘x k if x is even.

Example: What is the last digit in the following?

1. 2898

Step 1: x = 28 even last digit should be even.

Step 2: y = 93 = 4 23 + 1 = 92 + 1 4g + k (odd)

Step 3: 284g+k = 8k = 81 = 8 Where z = 23

The last digit is eight = 8.

2. 2334

Step 1: x = 23 =Odd Last digit should be odd

Step 2: y = 34 = 4 8 + 2 = 32 + 2 | 4g + k (even)

Step 3: 234g+k = 3k = 32 = 9

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2. Binomial expansion: How last digit is obtained?

(a+b)n =nC0an + nC1an-2b2 + ----- + nC2an-2ba + nCn-1a1bn-1 + nCn-1a1bn-1 + nCna0bn

Remember this expansion will always have (n+1) terms.

= a[1st n terms] + bn where M is the polynomial expansion Remember this for applicationslast term

In general

(a + b)n = aM + bn

(a – b)n = aM + (-b)a

Lets consider the same example as given before

1. (28)93 = (20+8)93 = 20 M + 8 93

= 8(92+1) = 8(4g+1) = 84z 81, where g=23

This will always end in ‘6’

Using the fact that any even number when multiplied when multiplied with six will give the same last digit we have, 6 81 = 8. Since we are only concerned about the last digit is 8.

2. Lets take an easy example:

27 = (0+2)7 = O M +27 where z=1= 24x1+3 = 24xg+3 = 24g 23

= 16g 8 = ---6 8= ----8

3. Lets see for an odd no.: What is the last digit

(23)34 = (20 +3)34 = 20 M + 334

= 332+2 = 348+2 = 34g+2 = ---132 (Since 34g will always end in 1)=32 = 9

The last digit is 9

4. Lets do 1 more:

(43)75 = (40+3)75 = 40 M + 375

=372+3 = 34xg+3 = ---133 = -----7

The last digit is 7.

Remember:

Even4x will always end in the digit: ‘6’Odd4x will always end in the digit: ‘1’

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Computations using Vedic MethodologyComputations form the most important aspect of problem solving in tests that disallow the use of

computation devices. All of us, over a period of time would have become conversant with small numbers

and hence computations involving small numbers are quick and accurate. As we have not developed the

same type of instincts with large numbers and fractions, it takes us more time to compute with these

numbers. Vedic Mathematics takes this fact into account and reduces large computations into small

numbers with the help of formulae so that computations can be quicker and more accurate. For example,

the usual procedure in schools and colleges in case of multiplication is to cram up multiplication tables up

to 20 and 30. But according to Vedic Mathematics, the multiplication tables are not required above 5. With

the help of simple sutras any multiplication work can be performed. We shall examine some of the simple

sutras and their application in computations.

Multiplication

Using Bases – 10, 100, etc.

Suppose we have to multiply 99 by 97.

(a) We should take, as base for our calculations, that power of 10 which is nearest to the number

to be multiplied. In this case 102 or 100 is that power.

(b) Put the numbers 99 and 97 above and below on the left hand side.

99 -197 -3

96 3

(c) Subtract each of them from the base (100) and write down the remainders (01 and 03) on the

right hand side with a connecting minus sign (-) between them, to show that the numbers to

be multiplied are both less than 100.

(d) The product will have two parts, one on the left side and one on the right. A vertical dividing

line may be drawn for the purpose of demarcation of the two parts.

(e) Now the left hand side digits (of the answer) can be arrived at in one of 4 ways.

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(i) Subtract the base 100 from the sum of the given numbers (99 and 97) i.e., 196, and put

196-100 i.e., 96 as the left hand side part of the answer. 99 + 97 - 100 = 96

(ii) Subtract the sum of the two deficiencies (01 + 03 = 04) from the base (100) and you get

the same answer (96) again. 100 - (01 + 03) = 96

(iii) Cross subtract deficiency 3 on the second row from the original number 99 in the first row.

And you get (99 - 3) i.e., 99 - 03 = 96 or

(iv) Cross subtract in the converse way i.e., 1 from 97). And you get 96 again as the left-hand

side of the portion of the required answer. 97 - 1 = 96.

(f) Now vertically multiply the two deficiencies (01 and 03). The product is 03. And this is the

right hand side portion of the answer.

(g) Thus 99 x 97 = 9603

This method holds well in all cases and is therefore capable of infinite application.

Solved Examples

1. 9 x 9 (Base: 10) 2. 999 x 996 (Base: 100)

3. 8 x 8 (Base: 10) 4. 9 x 6 (Base: 10)

8 -2   9 -18 -2   6 -4

6 4   5 4

The algebraic explanation for this is (x - a) (x - b) = x(x - a - b) + ab, where x takes the base value,

a and b are deficiencies.

In cases where the multiplication of deficit digits yields a product consisting of more than one digit,

the second digit must be carried forward to the left-hand side.

For example, 7 x 6

7 -3

6 -4

3 +

1

2

4 2

Here 3 x 4 yields 12 and 1 is carried forward to the left hand side and is added to 3 to get the product

as 42.

9 -1   999 -19 -1   996 -4

8 1   995 4

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Further Examples

1. 8 x 5 2. 93 x 95

8 -2 93 -7

5 -5 95 -5

3 0 -2 * -5 = 10 88 35

1 (Carry)

4 0

3. 89 x 95 4. 87 x 87

89 -11 87 -13

95 -5 87 -13

84 55 74 169

74+1 69

75 69

As you must have observed, we have till now considered only examples in which the multiplicands are

both on the lower side of the base. If both the multiplicands are on the higher side of the base, the

story remains the same.

Solved Examples

1. Multiply 101 with 103

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101 + 1

103 + 3

104 / 03 = 10403

Do not forget to put an extra 0 on the right hand side, since in this case our base 100 has 2 zeros.

We must have 2 digits on the right hand side.

2. Multiply 109 with 121

109 + 09

121 + 21

130 / 189 = 130 + 1 / 89 = 13189

The algebraic explanation for this is : (x+a) (x+b)= x(x+a+b) + ab

3. Multiply 1021 with 1034

1021 + 21

1034 + 34

1055 / 714 =1055714

(As base is 1000, there should be 3 digits on the right side. No carry-over is required)

4. Multiply 10134 with 10111

10134 + 134

10111 + 111

10245 /

We multiply 134 x 111 as

134 + 34

111 + 11

145 / 374 = 14874

10134 + 134

10111 + 111

10245 / 14874 = 102464874

The method is applicable for numbers as large as possible.

5. 1001 x 101

This one is not so tricky as it looks, we choose our base as 100 (remember that the differences for

both the numbers have to be calculated from the same base)

1001 + 901

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101 + 1

1002 / 901 = 1002 + 9 / 01 = 101101

Further down the line, we will see how to multiply numbers choosing different bases for them.

6. 102 x 97

How should we go about multiplying these? One of the numbers is above the base (100) while the

other one is below. Before proceeding with solving such kind of problems, we will have to learn

the concept of vinculum in Vedic Mathematics.

Concept of Vinculum

We will come across a lot of numbers with Vinculum above them in Vedic Mathematics, e.g. 102 , 10

.

The bar is the Vinculum. How do we convert between the normal numbers and Vinculum numbers,

and actually first of all, what does the Vinculum mean?

We will first see how the numbers with Vinculum above them are converted to normal numbers, can

you then figure out what does it mean?

The 2 shlokas which deals with vinculum state, “All from 9 and last from 10” and “One less from the

previous digit”.

So suppose we have to convert 210 to a normal decimal number. Then we proceed as follows:

1. “All from 9 and last from 10”: All the digits under the vinculum should be subtracted from 9,

except for the last one, which should be subtracted from 10.

So in this case, 9 should be subtracted from 9 and 1 from 10. Thus we get the last 2 digits of our

answer as: 09

2. “One less from the previous digit”: One should be subtracted from the digit preceding the

vinculum.

Since in this case it is 0, we borrow 1 from the digit to the left of it (as we do in normal mathematics)

to arrive at the first part of our answer: 209

Thus, answer is: 20909

Solved Examples

1.

Step 1 yields 88

Step 2 yields 09

So our answer is 988.

2.

Step 1 yields 54

19

Step 2 yields 22

So our answer is 2254.

Further Examples

1. 2.

Step 1 yields 1 Step 1 yields 66

Step 2 yields 0 Step 2 yields 1

Answer is 01 Answer is 166

Just try to see whether you have understood the concept of vinculum by trying out the following

examples:

1. 2. 3. 4.

Answers are:

1. 433 2. 9980001 3. 2611 4. 2257

Do you see a pattern?

So does that tell you what actually does a vinculum signify?

Let’s get back to the problem of multiplying 102 with 97. The method is still essentially what it was,

except that a vinculum comes into play.

102 + 2

97 – 3

99 /

We cross-subtract (or cross-add) as we used to do earlier for the left hand side of our product. Since for

the right side, we are multiplying + 2 with – 3, we should put – 6, our vinculum is exactly that, as you

can see from the “patterns” above.

Our method of converting vinculum numbers to normal numbers is essentially to carry out the

subtraction. Now 99 = 9894 and that is our answer.

Solved Examples

1. 103 97 2. 111 91 3. 1024 981

20

103 + 03 111 + 11 1024 + 24

97 – 03 91 - 9 981 - 19

100 / = 9991 102 / = 10101 1005/ = 1004544

4. 121 x 81 5. 9 x 11 6. 976 x 1041

121 + 21 9 – 1 976 - 24

81 – 19 11 + 1 1041 + 41

102 / 10 / = 99 1017 / = 1016016

= 102 – 3 /

= 99 / = 9801

7. 46 x 51

In this case, we can choose our base as 10 or 100, but the difference with the base in both cases will

be very large. Wouldn’t it be better if we could choose 50 as our base? We certainly can and this

brings us to our new extension of this method of multiplication.

Using Sub-Bases

1. First of all, choose the normal base, the sub-base will be calculated from that. Let’s choose, in this

case, our base to be 10.

2. Choose the sub-base, in this case, it is 50. Calculate the differences from the sub-base.

3. Multiply the number as we used to do earlier, the only extension being that the left part of the

answer has to be multiplied by . The right hand side of the answer should have

number of digits equal to number of digits in base (not sub-base), and it will be obtained by

multiplying the differences.

Thus:

46 – 4

51 + 1

47 / ( = 5, hence multiply by 5)

= 235 / = 2346

Note that we can also use 100 as our base, in which case:

46 – 4

51 + 1

21

47 / ( = ½)

= 23 / (Take on the right hand side, it becomes 50)

= 23 / 46 (since 50 – 04 = 46)

= 2346.

Solved Examples

1. 241 x 263

Choose 1000 as the base and 250 as the sub-base.

241 – 9 (Calculate the differences from the sub-base)

263 + 13

254 /

= 63 / (take on the right-hand side, it becomes 500)

= 63 / 383 (500 – 117 = 383)

= 63383

2. 313 x 283

It will be better to take 100 as the base and 300 as the sub-base in this case, instead of taking 250

as the base.

313 + 13

283 – 17

296 /

3

= 888 /

= 886 /

(As, right hand side can have only two digits, take on the left side, so it becomes 888 – 2 = 886)

22

= 88579

3. 486 x 486

Take 1000 as the base and 500 as the sub-base.

486 – 14

486 – 14

472 / 196

= 236 / 196

= 236196

4. 1024 x 1031

Take base as 1000 sub-base is also 1000.

1024 + 24

1031 + 31

1055 / 744 = 1055744

Thus we see that this method encompasses the previously encountered methods.

How did we multiply 24 and 31 (the differences) in the previous example? One method is to use the

normal mathematics, another one is to use Vedic Mathematics (take 10 as base and 30 as sub-base)

but there is a third much easier method.

24 x 31 = 24 x 30 + 24 x 1 = 720 + 24 = 744

The basic law of number theory that we use here: a (b + c) = a b + a c is, as you will

remember, called the distribution of multiplication over addition. Used properly, it is a much more

powerful tool for some kind of multiplications than both the conventional multiplication and Vedic

Mathematics.

Example:

98 x 34 = 98 x (30 + 4) = 98 x 30 + 98 x 4

Now 98 x 30 = 98 x 3 x 10, so we find 98 x 3 and then put a 0.

How do we calculate 98 x 3? One way is to multiply. Other is to realize that:

98 x 3 = (100 – 2) x 3 = 100 x 3 – 2 x 3 = 300 – 6 = 294

Similarly, 98 x 4 = 400 – 2 x 4 = 392.

23

Or, since we know from the first step that 98 x 3 = 294, add 98 to it to arrive at 98 x 4.

How do we add 98 to 294? One method is to add. Other is to add 100 to 294 and then subtract 2 to

arrive at our answer 392.

Hence,

98 x 34 = 2940 + 392 = 2940 + 400 – 8 = 3332

The calculations are always much faster and accurate if we try to find some bases (multiplies of 10)

near our numbers.

We have seen till now how to multiply 2 numbers both of which are closer to some base (or sub-base).

Can we multiply two numbers one of which is closer to one base and another to another base?

Yes we can. Let’s see how we can do that with an example.

Example:

Multiply 34 with 87.

Our base for 34 is 10, and for 87 too is 10.

For 34, our sub-base is 30, for 87 our sub-base is 90.

87 – 3

34 + 4

The left hand side of the answer is calculated as:

1. Let n1 and n2 be the two numbers; d1 and d2 their differences, b1 and b2 the bases and r1 and r2 the

ratio of the sub-bases with bases.

Then left hand side = n1 x r2 + d2 x r1 = n2 x r1 + d1 x r2

Here n1 = 87, n2 = 34, d1 = - 3, d2 = 4, r1 = 9, r2 = 3, b1 = b2 = 10

(n1 x r2 + d2 x r1) = 87 x 3 + 4 x 9 = 297

(n2 x r1 + d1 x r2) = 34 x 9 – 3 x 3 = 297

2. Right hand side will have same number of digits as the base

Thus,

87 – 3

34 + 4

297 /

= 296 / = 2958

24

Example:

Multiply 897 with 21.

Here the two bases two are different. For 897, it is 100; for 21 it is 50. Let the ratio of the bases be b,

and the ratio of the sub-bases to the respective bases be r1 and r2. Then the left side will be calculated

as:

n1 x r2 x denominator of b + d1 x r1 x numerator of b

= n2 x r1 x numerator of b + d1 x r2 x denominator of b

Thus in this case:

Sub-base for 897 = 900 r1 = 9

Sub-base for 21 = 20 r2 = 2

And, b = = 10

Left side = 897 x 2 x 1 + 1 x 9 x 10 = 1794 + 90 = 1884

Left side = 21 x 9 x 10 – 3 x 2 x 1 = 1890 – 6 = 1884

The right side will be obtained by multiplying the differences, and will contain the minimum number of

digits between the two bases.

Thus,

897 – 3

21 + 1

1884 / = 18837

Example:

2013 x 521

1. For 2013, the base is 1000 and sub-base is 2000.

2. For 521, let’s choose the base as 100 and sub-base 500. (We could have chosen 1000 as the base

too)

r1 = 2 r2 = 5 b = = 10

2013 + 13

521 + 21

/ 273

Left hand side = 2013 x 5 + 21 x 2 x 10 = 10065 + 420 = 10485

It’s also equal to = 521 x 10 x 2 + 13 x 5 = 10420 + 65 = 10485

25

Right hand side should have 2 digits.

Thus:

2013 + 13

521 + 21

10485 / 273 = 1048773

Solved Examples

1. 789 x 341

Choose base as 100 for both. Right hand side will have 2 digits.

r

789 – 11

341 + 41

341 x 8 – 11 x 3 /

= 2728 – 33 / = 2695/ = 2691 = 269049

2. 10179 x 449

Base is 10000 for 10179, and 1000 for 449 . b = 10

r

10179 + 179 1

449 – 51

10179 x - 5 /

= 5089 - 510 /

= 4579 /

= 4579 - 9 / (Right hand side can have only 2 digits)

= 4570 / = 4570 / 500 – 129 = 4570371

26

What exactly does ½ mean? It means to carry over 5 or 50 or 5000 as the case may be to the right-

side, the number of digits determined by the number of digits that should be on the right-side.

3. 12441 x 12441

12441 + 2441

12441 + 2441

14882 /

r

Now, 2441 – 59 ¼ Base is 10000, sub-base 2500

2441 - 59 ¼

2441 x - 59 x / 3481 = 2382 x / 3481 = 595 / 3481 = 5958481

14882 / 5958481 = 14882 + 595 / 8481 = 154778481

(To calculate 14882 + 595, do 14882 + 600 – 5)

4. 331 x 623

Choose base as 100 for both.

r

331 + 31 3

623 + 23 6

331 x 6 + 23 x 3 / 713 = 2055 / 713 = 206213

Note:

31 x 23 = 23 x 30 + 23 = 690 + 23 = 713.

You will be able to do all this mentally after some time if you practice enough, and there won’t be

any need of writing all the steps.

Similarly, 331 x 6 = 300 x 6 + 30 x 6 + 1 x 6 = 1986

1986 + 69 = 1986 + 70 – 1 = 1986 + 100 – 30 – 1 = 2055

It is not so long winded as it seems, it all becomes very quick with very little but continuous effort.

5. 7723 x 1541

Choose 1000 as base for each

r

7723 – 277 8

1541 + 541 1

27

7723 + 541 x 8 /

= 7723 + 4328 /

= 12051 /

r

277 – 23 3 (Choose base as 100 for both)

541 + 41 5

277 x 5 + 41 x 3 /

= 1385 + 123 /

= 1508 /

= 1499 / = 149857

Thus, our final answer is 12051/ = 12051 – 149 / = 11902 / = 11901143

Though we see that we are able to arrive at the answers, we see that it becomes more and more

complex as the deviations from the base become higher and higher. In such cases, the method below

becomes much more useful and economical.

Cross Multiplication Method

The method discussed earlier is convenient only when the two numbers being multiplied are close to

each other. In cases when the difference is large, the method takes time. There is another method

which can be used for such multiplication.

Suppose we need to multiply 12 by 21.

(a) Multiply the left-hand most digit 1 of the multiplicand vertically by the left-hand most digit 2 of the

multiplier, and set this down as the left hand most part of the answer.

(b) Then multiply 1 and 1 and 2 and 2 cross-wise, add the two to get 5 as the sum and set this down

as the middle part of the answer.

(c) Multiply the right hand most digit 2 of the multiplicand vertically by the right hand most digit 1 of

the multiplier and set this down as the right hand most part of the answer.

12

21

1 x 2; 1 x 1 + 2 x 2; 2 x 1 = 252

Examples

1. 43

23

4 x 2; 4 x 3 + 3 x 2; 3 x 3 = 9; 18; 9 = 8 + 1; 8 ; 9 = 989

28

2. 17

31

1 x 3; 1 x 1 + 7 x 3; 7 x 1

3 : 22 : 7 = 3 + 2; 2 ; 7 = 527

3. 15

24

1 x 2 : 1 x 4 + 5 x 2 : 5 x 4

2 : 14 : 20 = 2 + 1 : 4 + 2 : 0 = 360

The algebraic explanation for the method is (ax + b) (cx + d) = acx2 + (ad + bc) x + bd

In the first example of 12 21, x = 10, a = 1, b = 2, c = 2 and d = 1.

The same method can be extended for multiplying higher numbers also.

4. 123 x 31

(Add a 0 before 31 and then carry out calculations as before.)

123

031

3/ 7 / 11 / 3 = 3813

3 x 1 = 3, 2 x 1 + 3 x 3 = 11, 1 x 1 + 3 x 0 + 2 x 3 = 7, 1 x 3 + 2 x 0 = 3, 1 x 0 = 1

5. 12441 x 12441

12441

12441

1/4/12/ 24 / 34 / 36 / 24 / 8 / 1 = 154778481

We see that we can arrive at our answer in one-line by this method and the power of this method

becomes all the more visible as the multiplicands become more and more bigger.

Did you notice another thing about this method? We can calculate the answer both from left to right as

well as right to left.

6. 12

23

2 /

2 x 1 = 2, 3 x 1 + 2 x 2 = 7 gives the second digit of our answer, and 3 x 2 = 6 the last digit.

Thus 12 x 23 = 276.

Just try all our examples from the bases and sub-bases section using this method of get a hang of it.

29

Squaring of Numbers

612 =

We can calculate the square of a number (a+1)2 if we know the square of a, as (a+1)2 = a2 + 2a + 1

Thus, 612 = 3600 + 121 = 3721

Similarly, 792 = 6400 – 159 = 6241

Besides, we can use all the other methods for multiplying we have learned so far.

This introduction to Vedic Mathematics will enable you to cut down your basic computation time

significantly. We will explore more such techniques in the next module. Let us now examine a few

other approaches which will help you carry out regular computations faster.

Digit Sum Method

Let us examine this method with the help of an example.

Example: What is the value of 28981 x 129?

1. 3798549 2. 3738549 3. 3738547 4. 37358459

The conventional approach would be to multiply the two numbers and refer to the correct answer

choice. This obviously takes time.

The simpler approach would be to use the method of estimation. Instead of choosing the correct

answer out of the four, we can eliminate three wrong answers and arrive at the correct answer. Let us

now examine the approaches to eliminating the wrong answers.

28981 has 1 in the unit place and 129 has 9 in its unit place. When we multiply 28981 and 129 the unit

place of the product should be 9 x 1 = 9. Answer choice (3) has a number ending with 7 and hence

cannot be the answer. Therefore eliminate answer choice (3). 28981 is close to and less than 29000

and similarly 129 is close to and less than 130. 29000 x 130 gives 3770000. The product of 28981 and

129 should therefore be less than 3770000. But answer choice (1) has a number greater than 3770000

and hence can be eliminated. Now we are left with only two answer choices, (2) and (4) and we can

use the digit sum method to find the correct answer.

According to the digit sum method, the digit sum of the product of digit sums of two numbers must be

equal to the digit sum of the product of two numbers. Digit sum of 28981 is 2 + 8 + 9 + 8 + 1 = 28 =

2 + two numbers is 1 x 3 = 3. As per the digit sum rule, the digit sum of the product of 28981 and 129

must also be equal to 3. Consider answer choice (4). Digit sum of 3735845 is 3 + 7 + 3 + 5 + 8 + 4 +

5 + 9 = 8. Hence (4) cannot be the correct answer choice. The only choice left now is (2) and therefore

the correct answer must be (1). (The digit sum of 3738549 is 3 + 7 + 3 + 8 + 5 + 4 + 9 = 3).

Fractions

1. A fraction always has a numerator and a denominator. It can be defined in the form of , where p

and q are integers.

30

2. If the denominators of two fractions are equal, then the fraction which has a higher numerator will

be higher in value.

3. If the numerators of two fractions are equal than the fraction which has a lower value of

denominator will be higher in value.

Example:

Which of the following has the highest value?

1. 2. 3. 4.

The conventional method would be to find the decimal value of the fractions and then compare. This

method would be tedious and time consuming. You will be quicker if you convert all fractions to the

same value of denominator or numerator. For example, in this case, it would be convenient for us to

convert all fractions into fractions with a numerator value of 1. Therefore, can be written as

and so on. Then it becomes easier to compare them. Converting all fractions with numerator value of 1

we get the following:

1. 2. 3. 4.

Since all fractions have the same numerator value, the highest fraction will be the one which will have

the lowest denominator value.

In this case, answer choice (1) has the lowest denominator value and hence will be the highest

fraction.

Example:

Which of the following is the lowest?

1. 2. 3. 1.5 4.

The conventional approach followed here would be to find the value of each of the fractions. This

would be tedious and time consuming. Since three of the answer choices have numbers with square

root it would be better to convert all the numbers under the square root. can be written as

or . Similarly 1.5 can be written as or and as . The answer

choices would become

1. 2. 3. 4.

Obviously the lowest is and hence the answer is (4).

31

Factorials

A Factorial, n!, is defined as an integer, F = n (n – 1) (n – 2) … 2 1

Example:

What is the highest power of 3 that can divide factorial 10 perfectly?

1. 3 2. 4 3. 7 4. None of these

Factorial of a number ‘n’ is defined as the product of all numbers from 1 to n. Therefore factorial 10 is

the product of all numbers from 1 to 10. In order to find the highest power of 3 that can divide factorial

10, we must determine the numbers in 1 to 10 that are divisible by 3. 3, 6 and 9 are the three

numbers that are divisible by 3. 3 and 6 are divisible by 31 while 9 is divisible by 32.

Therefore the highest power of 3 that can divide factorial 10 is 4 and the answer is (2).

Worked Examples1. If 97 is multiplied by a certain number, it is increased by 7566. Find the number.

1. 87 2. 79 3. 75 4. 81

Solution:

Let the number be x, 97x = 97 + 7566, 97x = 7663, x = = 79

2. The area of a square park is 53361 m2. Find the length of wire required for the fence.

1. 924 2. 797 3. 533 4. 764

Solution:

Side of the square park =

Side of the square = 231

Length of the wire = 231 x 4 = 924 m

3. A general wishing to draw up his 17429 men in the form of a solid square found that he had five

extra men. Find the number of men in the first row.

1. 120 2. 264 3. 178 4. 132

Solution:

No. of men in the form of a square = 17429 - 5 = 17424

No. of men in the front row = = 132 men.

32

4. Find the number whose 7th part multiplied by its 11th part gives 1232.

1. 306 2. 308 3. 309 4. 301

Solution:

Let the number be x

= 1232 x2 = 1232 x 7 x 11

x2 = 94864 x = = 308

5. Which one of the following is the smallest?

1. 2. 3. 4.

Solution:

, , ,

These fractions are now . The smallest fraction is .

6. A number is divisible by 9 if the sum of its digits is divisible by 9. Which of the following is divisible

by 45?

1. 63345 2. 72365 3. 72144 4. 98145

Solution:

For a number to be divisible by 45, it should be divisible by 9 and 5. Choice (3) is not divisible by 5

and can be eliminated.

The sum of the digits in (a), (b) and (4) are 21, 23, 27 respectively. Only 27 is divisible by 9.

Therefore, choice (4) is the correct choice.

7. (235)2 equals

1. 56225 2. 55225 3. 55275 4. 58225

Solution:

[TIP: A square of a number ending in 5 can be obtained by multiplying the ten’s digit by the next

higher integer and annexing 25.]

23 x 24 = 552

2352 = 55225. Answer is (2)

8. Which of the following is a perfect square?

1. 26579 2. 25779 3. 26569 4. 26659

Solution:

33

If a number n ends in an odd digit, then its square can be calculated as follows:

Let number n = an an-1 ………… a1 (where a1, a2 ………… an are the digits)

Then, n x n = anan-1………….a2a1

anan-1………….a2a1

The last digit = a1 x a1 = a12

The second last digit = a2 x a1 + a2 x a1 = 2a1a2 + carry over from a12

If a1 is odd, then carry over for a12 is always even.

2a1a2 + carry over of a12 is always, even if a1 is odd.

Thus, the second last digit of the square of an odd number is always even.

Hence (3).

9. 75 = ?

1. 14843 2. 15941 3. 16807 4. 16813

Solution:

75 = 72 x 72 x 7

Now, 72 = 49

72 x 72 will end in the last digit of 9 x 9 = 81, i.e., 1

72 x 72 x 7 will end in the last digit of 1 x 7 = 7. Hence (3).

10. Can you determine what does 3343 end in?

3343 ends in 7.

[TIP: For even numbers ending in 0, 2 and 8, the second last digit of the square is even.]

11. The LCM of two numbers is 312 and the HCF is 4. If one of the numbers is 24, find the other

number.

1. 62 2. 64 3. 52 4. 48

Solution:

The product of two numbers is equal to the product of the HCF and LCM of the numbers.

312 4 = 24 x

x = = 52

Answer is (3)

12. Which one of the following numbers are divisible by 11?

1. 9,163,627 2. 4,176,737 3. 8,142,672 4. 7,264,916

Solution:

34

The number 11 divides, only those numbers whose sum of digits occupying odd positions is either

equal to the sum of digits occupying even positions or differs from it by a number which is divisible

by 11.

9 + 6 + 6 + 7 = 28; 1 + 3 + 2 = 6

28 - 6 = 22 11 = 2

Answer is (1)

13. Which of the following divide 1781 - 581?

1. 11 2. 6 3. 10 4. 4

Solution:

1781= (172)40 x 17

= ((A number ending in 9)2)20 x 17 (As 172 will end in the last digit of 72, which is 9)

= (A number ending in 1) 20 x 71

= A number ending in 1 x 17

= A number ending in 7

581 will end in 5.

Thus the difference will end in (7-5) = 2

Since any multiple of 11 ends with 1, while any multiple of 10 ends in 0, (a) and (c) are ruled out.

Now, a81 – b81 = (a27)3 - (b27)3 = (a27 - b27) (a 2x27 + a27 b27 + b 2x27)

(As x3 – y3) = (x – y) (x2 + xy + y2), here x = a27 and y = b27

We can see that similarly a27 - b27 will give us a multiple of the form a9 – b9, which in turn will give a

multiple a3 – b3, which in turn will give a multiple a – b.

Here, a = 17, b = 5. a – b = 12

Since both 6 and 4 divide 12, they will divide 1781 - 581.

Hence both (2) and (4) divide 1781 - 581

14. The remainder when 784 is divided by 342 is:

1. 0 2. 1 3. 49 4. 341

Solution:

There should be some relation between the numbers involved, otherwise the question wouldn’t

have been asked in the first place.

There exists such a pattern: 73 = 343, thus 784 = (343)26

So we have to find the remainder of

35

Let x = 342, then this becomes

Now (x+1)2 = x2 + 2x + 1, (x+1)3 = x3+3x2+3x+1 and so on.

The point to note is that x occurs in each term of the expansion except for the last one, which is

always 1. Thus, on dividing by x, the remainder will be 1, one x of each of the remaining terms will

cancel out with the x in denominator.

Hence (2).

Practice Exercise 11. What least number must be subtracted from 4564 to get a number exactly divisible by 35?

1. 14 2. 15 3. 16 4. 17

2. What least number must be added to 5364 to get a number exactly divisible by 80?

1. 14 2. 76 3. 16 4. 17

3. Find the number which is nearest to 2344 & exactly divisible by 34.

1. 2346 2. 2345 3. 1234 4. 4321

4. Find the number which is nearest to 6443 and exactly divisible by 42.

1.2346 2. 6425 3. 1234 4. 4321

5. Find the number nearest to 5030, which is exactly divisible by 64.

1. 2346 2. 6425 3. 5056 4. 4321

6. What is the unit digit of 734 + 691?

1. 6 2. 5 3. 3 4. None of these

7. 1924 x 436 = ?

1. 838864 2. 828864 3. 838874 4. None of these

8. What is the ten’s digit of 34046?

1. 8 2. 2 3. 4 4. None of these

9. The LCM of two numbers is 3528. If you divide the LCM by 126, you will get the HCF of the two

numbers. If one of the numbers is 504, what is the other number?

1. 49 2. 245 3. 196 4. None of these

10. 1075 x 92 = ?

36

1. 99870 2. 98900 3. 97950 4. None of these

11. Let x be a natural number. Then (x-1) x (x+1) is divisible by: (All x –1, x, x+1 > 0)

1. 3 2. 6 3. 12 4. (1) and (2)

12. What is the last digit in the given 12441 x 12441

1. 3 2. 6 3. 12 4. 1

13. What is the last digit in the given 123 x 31

1.2 2. 3 3. 12 4. (1) and (2)

14. To find the HCF of 12, 9, 24

1. 3 2. 6 3. 12 4. 1

15. To find the HCF of 2002, and 182

1. 233 2. 623 3. 182 4. 13

Practice Exercise 21. The area of a square is 124 sq. cm. The length of each side should be

1. 11.126 2. 11.234 3. 11.016 4. 11.136

2. A wholesale carpet dealer sold 1839 carpets for Rs.8,632.10 each. How much money did he get?

1. 1,58,5,431.9 2. 1,58,74,431.9 3. 1,78,71,431.9 4. 1,58,73,431.7

3. Which is the largest of the following?

1. 2. 3. 4.

4. Which of the following is the smallest?

1. 2. 3. 3 4. 10

5. What is the highest power of 2 which can divide 16!?

1. 10 2. 8 3. 13 4. 15

6. The product of 39463.7 and 4597.3 is

1. 181426568.01 2. 181426468.01 3. 1814264680.1 4. 1814265680.1

7. Find the smallest fraction amongst

1. 2. 3. 4.

8. 48632 x 224 equals

1. 10994525 2. 10893568 3. 11325448 4. 12093658

9. 842.3 x 19.7 equals

37

1. 16593.31 2. 16462.31 3. 17092.02 4. 15262.41

10. 497.27 x 5.4 equals

1. 2686.438 2. 2590.348 3. 2685.258 4. 2728.532

11. The LCM of two numbers is 312 and the HCF is 4. If one of the numbers is 24, find the other

number.

1. 62 2. 64 3. 52 4. 48

12. Which one of the following numbers are divisible by 11?

1. 9,163,627 2. 4,176,737 3. 8,142,672 4. 7,264,916

13. Which of the following divide 1781 - 581?

1. 11 2. 6 3. 10 4. 4

14. The remainder when 784 is divided by 342 is:

1. 0 2. 1 3. 49 4. 341

15. Let a, b, c be distinct digits. Consider a two digit number ‘ab’ and a three digit number ‘ccb’, both

defined under the usual decimal number system. If (ab)2 = ccb and ccb > 300 then the value of b

is:

1. 1 2. 0 3. 5 4. 6

Answer Keys for Practice Exercise 1

1 1 6 2 11 42 2 7 1 12 43 1 8 4 13 24 2 9 3 14 15 3 10 2 15 3

Answer Keys for Practice Exercise 2

1 4 6 2 11 32 2 7 3 12 13 1 8 2 13 24 2 9 1 14 25 4 10 3 15 1

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CHAPTER-2

Ratio and ProportionRatio

Ratio is the relation which one quantity has to another of the same kind. Ratio is the comparison of two quantities.

Example: 2 kg : 3 kg, 5 men : 10 men.

Note that in both the examples the quantities we have compared have the same unit.

The ratio of two quantities l and m denoted by l : m (read as l is to m). Here l and m are called the terms of the ratio where l is called the antecedent and m is called the consequent.

Example: In the ratio 32 : 51, 32 is called the antecedent and 51 is called as the consequent.

Important points to remember1. The ratio of 2 quantities x and y denoted by x : y (read as x is to y the first term x is called the

antecedent and second term y is called the consequent.)

2. In order to find the ratio of two quantities of the same kind they must first be expressed in terms of a common unit.

For instance, the ratio of Rs. 3 and 45 Ps. is 300 : 45.

Or 20 : 3

3. Two or more ratios may be compared by reducing their equivalent fractions to a common denominator.

For instance to compare 5 : 2 and as

and as

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Since , we get 5 : 2 > 1 : 3

4. A ratio is a pure number and has no units.

5. Comparison of Ratios

a. l : m is called the ratio of greater inequality, if l > m.

b. l : m is called the ratio of lesser inequality, if l < m.

c. l : m is called the ratio of same inequality, if l = m.

Examples: 1. 3 : 2 is a ratio of greater inequality since 3 > 2

2. 1 : 6 is a ratio of lesser inequality since 1 < 6

3. 5 : 5 is a ratio of equal inequality since 5 = 5

Two ratios can l : m and n : p can be compared. The different comparisons are as given below:

a. If lp > mn then l : m > n : p

b. If lp < mn then l : m < n : p

c. If lp = mn then l : m = n : p

Examples:

1. Compare the ratios 6 : 5 and 3 : 2.

Clearly (6 x 2 = 12) < (5 x 3 = 15). Hence 6 : 5 > 3 : 2

2. Compare the ratios 4 : 7 and 7 : 2.

Clearly (4 x 2 = 8) < (7 x 7 = 49). Hence 4 : 7 < 7 : 2

3. Compare the ratios 9 : 2 and 4 : 7.

Clearly (9 x 7 = 63) > (4 x 2 = 8). Hence 9 : 2 > 4 : 7

6. Commensurable Quantities

Two quantities are said to be commensurable, if their ratio can be expressed as the ratio of two integers.

Example:

Show that and are commensurable quantities.

Consider : = : = 78 : 26 = 6 : 2. This is the ratio of 2 integers.

Hence and are commensurable numbers.

7. Some Important Results on Inequality

a. If the same positive number is added to both the terms of ratio of greater inequality then the ratio is diminished.

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Example : is ratio of greater inequality.

A positive number 4 is added to both the terms of ratio i.e.,

Since , the result is proved.

b. If the same positive number is added to both the terms of ratio of less inequality then the ratio is increased.

Example : is ratio of lesser inequality.

If 3 is added to both the terms of the ratio

Since , the result is proved.

8. Compound Ratio

If a : b and c : d are two ratios then the ratio ac : bd is called their compound ratio.

For example the compound ratio of 2 : 5 and 3 : 1 is 6 : 5.

9. Important terms and definitions on Ratio

i) Duplicate ratio of a : b is a2 : b2

ii) Triplicate ratio of a : b is a3 : b3.

iii) Sub duplicate ratio of a : b is a½ : b½.

iv) Compounded ratio of a : b and c : d is ac : bd.

v) Duplicate ratio of a : b is a2 : b2.

vi) Sub triplicate ratio of a : b is a1/3 : b1/3.

vii) Reciprocal ratio of a : b is :

ProportionLet us consider two ratios a : b and c : d. When we compare these two ratios i.e., when we compare ad and bc, if ad = bc then we say the ratios are in proportion.

Example:

Consider two ratios 2 : 3 and 4 : 6.

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a = 2, b = 3, c = 4, d = 6.

ad = 2 x 6 = 12 and bc = 3 x 4 = 12.

Since ad = bc the two ratios are in proportion and is denoted by a : b :: c : d and we read it as a is to b is to c is to d.

When a : b :: c : d then a and d are called extremes and b and c are called means.

The product of the means is equal to the product of the extremes.

Important points to remember

1. Mean Proportion

In the proportion a : b :: b : c, the term b is called the mean proportion.

Since the product of the means is equal to product of the extremes, we get

Or .

2. Third Proportional

In the proportion a : b :: b : c, the extreme c is called the third proportional.

3. Fourth Proportional

In the proportion a : b :: c : d, the extreme d is called the fourth proportion.

4. Direct Variation

One quantity A is said to vary directly as another B, when the two quantities depend upon each other in such a manner that if B changes, A changes in the same ratio. Ratio of A and B is a constant always.

Example: If a train moving at a uniform rate travels 40 miles in 60 min, it will travel 20 miles in 30 min, 80 miles in 120 min. This is expressed by telling the distance is proportional to time.

5. The symbol is used to denote variation

A B is read as A varies as B.

6. The ratio of A and B is always a constant.

A B A = constant B

i.e., = constant.

Hence, if A1, A2 are distinct values of A and B1, B2 are distinct values of B, and if

A1 B1 and A2 B2

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then, =

Example: The number of men working is directly proportional to units of work completed. More men, more work can be completed.

7. Directly Proportional Graph

If, X Y

Then, X = KY

where, K CONSTANT

The graph illustrates this relationship.

8. Inversely or Indirectly Proportional

One quantity A is said to vary inversely as another quantity B, when A varies directly as the reciprocal of B.

i.e., A A = AB = K, where K = constant.

Hence, if A1, A2 are distinct values of A and B1, B2 are distinct values of B, and if

A1 and A2

then, =

The graph illustrates this relationship.

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Worked Examples1. Two numbers are in the ratio 5 : 6. If 8 is added to each they are in the ratio 9 : 10. Find the

numbers.

Solution :

Since, the two numbers are in the ratio 5 : 6,

let us assume that the numbers are 5x & 6x.

If 8 is added to each number, we get 5x + 8 & 6x + 8 respectively.

The new ratio = .

But according to the condition given, this ratio is 9 : 10

Equating the two ratios we get

10(5x + 8) = 9(6x + 8) 50x + 80 = 54x + 72

4x = 8 x = 2

first number = 5 x 2 = 10 & second number = 6 x 2 = 12

2. Two numbers are in the ratio 3 : 5. If 6 is subtracted from each they are in the ratio 5 : 9. Find the numbers.

Solution:

Since, the two numbers are in the ratio 3 : 5,

we can assume that the numbers are 3x & 5x.

Subtracting 6 from each of them, they are in the ratio 5 : 9.

i.e., 9(3x - 6) = 5(5x - 6) 27x – 54 = 25x – 30 2x = 24 x = 12

The first number is 3 x 12 = 36 & the second number is 5 x 12 = 60

3. What should be subtracted from each term in the ratio 3 : 5 so that it becomes 5 : 9 ?

Solution:

Let x be subtracted from each term then

9(3 - x) = 5(5 - x) 27 - 9x = 25 - 5x 27 - 25 = 9x - 5x

2 = 4x x =

Thus should be subtracted from each term so as to change the ratio from 3 : 5 to 5 : 9.

4. What should be added to each term in the ratio 7 : 9 so that it becomes 5 : 6 ?

Solution :

Let x be added to each term then

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6(7 +x) = 5(9 +x) 42 + 6x = 45 + 5x

6x – 5x = 45 – 42 x = 3

5. The prices of two sites are in the ratio 6 : 11, 6 months later when the price of the first site was raised by Rs. 600 and the price of the second site was raised by Rs. 800, the ratio of their prices become 3 : 5. Find the original prices of the sites.

Solution :

Let the original prices of two sites be 6x & 11x respectively.

6 months later, the price of first site will be 6x + 600 & the price of second site will be 11x800.

It is given that, six months later, the ratio of two sites will be 3 : 5. Thus we have

5(6x + 600) = 3(11x + 800)

30x +3000 = 33x + 2400

600 = 3x x = 200

Thus the original price of the first site is 6 x 200 = 1,200

and the original price of the second site is 11 x 200 = 2,200

6. In a 1,200 meters race, A beat B by 200 meters and C beat B by 400 meters. If A’s speed is 400 mts/min, find C’s speed.

Solution :

Distance covered by A is 1,200 mts in the given time

Distance covered by B is 1,200 - 200 = 1000 mts in the same time

Similarly,

C’s speed = A’s speed = x 400 = 500

7. There are two clerks A and B drawing the same salary. A saves 15% of his salary while B spends 80%. Find the ratio of the savings made by the former to that made by the latter.

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Solution :

Let us assume that A and B have Rs. 100 as salary.

Then A saves Rs. 15 and B saves 100 - 80 = Rs. 20

The ratio of their savings is 15 : 20 i.e. 3 : 4.

8. If x : y : z = 4 : -3 : 2 & 2x +4y -3z = 20, find the values of x, y, z.

Solution :

Let = k (say)

x = 4k, y = -3k & z = 2k and 2x + 4y - 3z = 20

8k -12k - 6k = 20

-10k = 20 k = - 2

Hence,

x = 4(-2) = - 8

y = (-3)(-2) = 6

z = 2(-2) = - 4

9. Three numbers are in the ratio 4 : 5 : 7. The sum of their cubes is equal to 4256. Find the numbers.

Solution :

Let the first number be 4x, then the second number will be 5x & the third number will be 7x.

Further it is given that, the sum of their cubes is equal to 4256.

(4x)3 + (5x)3 + (7x)3 = 4256 64x3 + 125 x3 + 343 x3 = 4256

532 x3 = 4256 x3 = = 8 x = 2

Thus: first number is 4x = 4 x 2 = 8

second number is 5x = 5 x 2 = 10 and

third number is 7x = 7 x 2 = 14

10. Two lecturers A and B whose salaries are Rs. 7,500 and Rs. 14,500 received an increase in their salaries. If their salaries were increased in the same ratio, what increase did B receive when A’s salary reached Rs. 9,000?

Solution:

A’s old salary = Rs. 7,500

B’s old salary = Rs. 14,500

A’ s new salary = Rs. 9,000

Thus increase in A’s salary = Rs. 9,000 - Rs. 7,500 = Rs. 1,500

Thus ratio of increase = .

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Let us assume the ratio increase of B’ salary as .

Then = (because the increase of A & B are in the same ratio)

Thus x : 14,500 :: 1,500 : 7500

7500x = 14,500 x 1,500

x = = 2,900

Thus B’s salary was increased by Rs.2,900.

Practice Exercise 1

1. Two numbers are in the ratio of 5 : 8. If 9 is added to each they will be in the ratio 8 : 11. The original number in the numerator is

1. 24 2. 15 3. 8 4. 5

2. The ages of a man and his son are in the ratio of 3 : 1. Fifteen years hence the ratio will be 2 : 1. How old is the father now?

1. 45 years 2. 30 years 3. 40 years 4. None of these

3. The ratio of the areas of two squares is 9 : 1. The ratio of their perimeters will be

1. 1 : 2 2. 3 : 1 3. 3 : 2 4. None of these

4. X and Y have their annual incomes in the ratio of 8 : 5 and their annual expenditure is in the ratio 5 : 3. If they save Rs.1400 and Rs.1000 per year respectively, find their annual incomes.

1. Rs.19200, Rs.13000 2. Rs.6400, Rs.4000

3. Rs.12000, Rs.7500 4. Rs.7200, Rs.4500

5. Find three numbers in the ratio of 3 : 2 : 5 such that the sum of their squares is equal to 1862.

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1. 21, 14, 35 2. 11, 14, 25 3. 20, 24, 30 4. 30, 34, 37

6. Two numbers are in the ratio 7 : 1. The difference between them is 44. What is the sum of these two numbers?

1. -208 2. 3. 4. None of these

7. A college has 53 boys and girls in the ratio 5 : 4. If there is an increase of 20% in the number of girls, what is the new ratio of boys to girls?

1. 25 : 24 2. 235 : 570 3. 196 : 531 4. 4 : 5

8. For what value of x will the ratio (7 + x) : (12 + x) be equal to the ratio 5 : 6?

1. 20 2. 13 3. 22 4. 18

9. The ratio between the numerator and denominator of a number is 3 : 7. If the difference between the denominator and numerator is 28, what is the sum of the numerator and denominator?

1. 60 2. 70 3. 65 4. 75

10. The ages of Anil and Sagar are in the ratio of 6 : 5 and the sum of their ages is 44 years. What will be the ratio of their ages after 8 years?

1. 7 : 6 2. 3 : 4 3. 8 : 7 4. 2 : 5

11. Two numbers are in the ratio of 3 : 5. If 9 is subtracted from each, they are in the ratio of 12 : 23. What is the second number?

1. 55 2. 25 3. 40 4. 30

12. A greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 4 leaps of the hound are equal to 5 of the hare. The ratio of distance traversed by the hound and hare area :

1. 24 : 25 2. 25 : 24 3. 5 : 6 4. 6 : 7

13. The ages of A and B are in the ratio 4 : 5; the ages of B and C are in the ratio 3 : 2. The eldest of the three is

1. A 2. B 3. C 4. Can’t say.

14. A contractor undertook to build a road in 100 days. He employed 70 men. After 30 days, he found

that only of the road could be built. How many additional men should be employed to complete

the work in time ?

1. 6 2. 20 3.10 4. None of these

15. The age of a man and his son are now in the ratio of 3 : 1. Fifteen years hence the ratio will be 2 : 1. How old is the father now?

1. 45 years 2. 30 years 3. 40 years 4. None of these

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49

Practice Exercise 21. An alloy consists of two metals iron and copper which are in the ration 1 : 2 and another alloy

contains the same metals in the ratio 2 : 3. How many parts of each alloy must be taken to obtain a third alloy which would contain the same metals in the ratio 5 : 8 ?

1.2 : 3 2. 4 : 5 3. 1 : 1 4. 3 : 10

2. If +b = 2b find a : b.

1. 3 : 1 2. 1 : 3 3. -1 : 3 4. 1 : -3

3. If 9 men and 6 boys can do in 2 days what 5 men & 7 boys can do in 3 days, what is the ratio of work rate of a man to a boy.

1. 3 : 1 2. 2 : 1 3. 3 : 2 4. 4 : 3

4. Rama and Sita started a business with initial investments in the ratio 6 : 5. Their profits were in the ratio 16 : 5. If Rama had invested his money for 8 months, find for how long had Sita invested her money.

1. 4 months 2. 2 months 3. 6 months 4. 3 months

5. 50 lts of wine are mixed with 10% water. How much water must be added to make the mixture contain wine water in the ratio 4 : 1 ?

1.10 lts 2. 7.5 lts 3. 12 lts 4. 8 lts

6. The ratio of the ages of mother and son is 5 : 3. If the sum of their ages is 64 years, find the difference of their ages.

1. 20 years 2. 16 years 3. 25 years 4. 21 years

7. In a country, in a decade, the population increased by 15.9%. If the increase in the urban population was 18% and that in the rural population was 4%. Find the ratio of the urban to rural population at the beginning of the decade.

1. 5 : 1 2. 17 : 3 3. 10 : 3 4. 19 : 5

8. Vessels A and B contain milk and water in the ratio 4 : 5 and 5 : 1 respectively. In what proportion would quantities be taken from the vessels A and B respectively to form a mixture in which milk : water is in the ration 5 : 4.

1. 2 : 5 2. 4 : 3 3. 5 : 2 4. 3 : 4

9. A piece of wire 78 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, the length of the base in cms is

1. 19 cm 2. 18 cm 3. 16 cm 4. 17 cm

10. One man adds 3 liters of water to 12 liters of milk and another man adds 4 liters of water to 10 liters of milk. Compare the amount of milk in the two mixtures.

1. 25 : 26 2. 28 : 25 3. 3 : 2 4. None of these

11. If A : B = 3 : 4 and B : C = 5 : 6 then A : B : C is _____

1. 15 : 20 : 24 2. 2 : 3 : 4 3. 3 : 2 : 7 4. None

12. Two numbers are in the ratio 4 : 5. If each is diminished by 6, they would be in the ratio 3 : 4. Find them.

1. 24, 30 2. 12, 15 3. 16, 19 4. 80, 100

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13. Write 6 : 5.2 as the ratio of integers

1. 5 : 4 2. 6 : 5 3. 8 : 7 4. 13 : 1

14. The incomes of A and B are in the ratio 4 : 3 and their expenditures are in the ratio 2 : 1. If each one saves Rs.1000, their incomes are

1. Rs.2000 and Rs.1500 2. Rs.3000 and Rs.2250

3. Rs.4000 and Rs.3000 4. None of these

15. Divide 27 into two parts so that 5 times the first and 11 times the second are together equal to 195

1. 17 : 10 2. 16 : 11 3. 15 : 12 4. None of these

Answer Keys for Practice Exercise 1

1 2 6 3 11 12 1 7 1 12 23 2 8 4 13 34 2 9 2 14 25 1 10 3 15 1

Answer Keys for Practice Exercise 2

1 3 6 2 11 12 3 7 2 12 23 4 8 3 13 44 2 9 2 14 45 1 10 3 15 1

CHAPTER-3

51

PercentagesConcept of Percentage

The French word ‘Cent’ or the Latin word ‘Centum’ means hundred.

Therefore the word percentage means per hundred or hundredths.

Any fraction with hundred in the denominator is called percentage.

Percent is indicated using the symbol: %

80% is ; 5% is ; 365% is 365

100

A percentage can be represented by a fraction or a decimal.

Important points to remember1. Converting Fractions, Numbers and Decimals to Percentages

5 can also be written as 500%

1.25 can also be written as 125%

3.75 can also be written as 375%

0.575 can also be written as 57.5%

2. A fraction can be converted to percentage by multiplying by 100

100 = 0.8 x100= 80%

Percentage Fraction Decimal

75% 0.75

30% 0.3

65% 0.65

23% 0.23

66% 0.66

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100 = 0.1666 x 100 = 16.66%

3. Finding a percent of a given number.

P% of A =

Example:

a) 10% of 50 = =5

b) 25% of 120= =30

4. Percentage Comparison.

Given x and a, to find x is what% of a

Multiply the ratio by 100 and add % sign, i.e.,

Example:

a) 4 is what percent of 60? 100 = 6.67%

b) 10 is what percent of 100? 100 = 10%

c) 207 is what percent of 100? 100 = 207%

5. Percent Increase and Decrease

In many cases we are interested in knowing by how much fraction of its origin value a quantity has changed. If the value is given in percentage then:

Percentage change =

6. Percentage Error

Error = Actual value - Value taken (AV>VT)

Error = Value taken - Actual value (VT>AV)

If the value of x is taken as y then error is y - x

, where (y – x) is error and x is the actual value

7. Relative Percentage

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In relative percentage, there is percentage comparison between two or more quantities.

Example:

If A’s income is 20% less than B’s, then by what percentage is B’s income more than A?

Solution:

Let B’s income be Rs.100

A’s income would be 20% less than B

Hence, A’s income = 80% of Rs.100 = Rs.80

The percentage by which B’s income would be more than A

= % = 25%

Worked Examples1. Find 5 percent of 400.

Solution:

5% = = 0.05

Hence, the required amount = 0.05 x 400 = 20

2. Find 2% of Rs.18.75 to the nearest paise.

Solution:

2% of Rs.18.75 18.75 = 0.02 x 18.75 = 0.375 = 38 paise.

3. In a school the strength of the students is 260 out of which 65% are boys. Find the number of boys and girls in the school.

Solution:

Total number of students = 260

Given that 65% of the students are boys. Hence, 35% of the students are girls.

No. of girls = 35

100 260 = 91 and, No. of boys = 260 – 91 = 169

There are 169 boys and 91 girls in the school.

4. A piece of cloth 50 meters long shrinks by 0.6%. How much did it shrink?

Solution:

0.6% of 50 meter = 0.3 meters = 30 cm

5. In a spelling test of 80 words, Ravi spelt 80% of the words correctly. How many words did he spell right?

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Solution:

The number of words in the spelling test = 80 words.

Ravi spells 80% words correctly.

No. of words spelt correctly by Ravi = 80

100 80 = 64 words.

6. A man spends 25% of his money and then Rs.75 and then 5% of the remainder. If he had Rs.3291.75 left with him, how much money did he possess initially?

Solution:

Let the money before spending 5% be Rs.100.

So, amount after spending 5% is Rs.(100 - 5) = Rs.95

Actual amount of money after spending 5% = Rs.3291.75

So, actual amount of money before spending 5% = Rs. = Rs.3465

Therefore, amount of money left before spending Rs.75 = Rs.3465 + Rs.75 = Rs.3540

If the original money is Rs.100, then the amount of money left after spending 25% of Rs.75.

Money left after spending 10% Original money

Rs.75 Rs.100

Rs.3540 Rs.4720

Therefore his original sum of money is Rs.4720.

7. Find the percentage error in writing 2.54 as 2.5.

Solution:

Actual value=2.5

Value taken=2.54

Error value=2.54-2.5=0.04

8. The price of rice is raised by 30%, by what percent should a house wife reduce consumption so as to not to increase her total expenditure on rice?

Solution:

Let her initial consumption be 100 kg at a price of Rs.10 per kg.

Hence her total expenditure = 100 10 = Rs.1000

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The new price after increase of 30% is 13 per kg

Let the consumption be X kgs

Then,

13X = 1000

Alternative Method

where X is the increase in percentage.

9. Satish’s income is 20% more than that of Sanjana’s. By what percentage is Sanjana income less than that of Satish?

Solution:

Method 1

Let Sanjana income be Rs.100

Satish’s income is

Sanjana income is less than Satish by

= %

Method 2

, where x is percentage difference

10. If the present population of a town is 20000, which is increasing at r rate of 10% every year what will be population of the town after 2 years?

Solution:

56

(Or Original price - 15% of Rs.40, 000) = 40,000 = Rs.34,000

Practice Exercise 1

1. X’s salary is half that of Y. If X got a 50% increase in his salary while Y got a 25% increase, then what is the percentage increase in the total salary received by both?

1. 50% 2. 25% 3. 33 % 4. 66.66%

2. A man spends 30% of his salary on food and 60% of the balance on clothing. If his total savings per month is Rs.265, what is his total annual income?

1. Rs.11,352 2. Rs.12,815 3. Rs.10,582 4. Rs.12,102

3. A’s income is 10% more than B’s. By what percentage is B’s income less than A’s?

1. 10% 2. 9.09% 3. 10.9% 4. 11%

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4. Kumar earns 25% more while Raman earns 75% more than their colleague Mohan. If the monthly income of Raman is Rs.1750, what is Mohan’s salary as a percentage of the combined salary of the three colleagues?

1. 50% 2. 25% 3. 33 % 4. 40%

5. A mango tree’s height is 30% less than that of a coconut tree. By what percent is the coconut tree taller than the mango tree?

1. 42.8% 2. 40% 3. 28.3% 4. 34.15%

6. What is 20% of 50%?

1. 10% 2. 40% 3. 25% 4. None of these

7. If the population of a town increases by r% every year, then the population after ‘t’ years is

1. 2. 3. 4.

8. When the price of a radio was reduced by 20% the quantity sold increased by 80%. What was the net effect on the sales revenue?

1. 44% increase 2. 44% decrease 3. 20% decrease 4. 34% increase

9. A man spends 40% of his salary on books. Out of this amount, 60% is spent on buying text books, 20% on novels and the remaining on journals. If 80% of the money on journals is spent on buying foreign journals and the remaining on Indian journals, what percentage of his salary is spent on Indian Journals?

1. 2.4% 2. 1.6% 3. 6.4% 4. 3.2%

10. 50% of a% of b is 75% of b% of c. What is c?

1. 0.5a 2. 0.667a 3. 1.25a 4. 1.5a

11. The drivers of 65% of BTS buses are on strike today. 4% of the remaining buses are under repair due to accidents. If the number of buses still operating is 168, what is the total number of BTS buses operating under normal circumstances?

1. 542 2. 500 3. 600 4. 550

12. In a test, a student gets 80% of what he expected. His expected score happens to be 60% more than the class average. What is his score as a percentage of the average score?

1. 115% 2. 160% 3. 144% 4. 128%

13. Two sides of a cuboid are increased by 10% and the third side is decreased by 20%. What is the percentage volume change?

1. No change 2. 3.2% increase 3. 3.2% decrease 4. 2.9% decrease

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14. A block of ice is to be carried through a certain distance. The ice melts by th of its

own weight for every rd of the total distance. What percent of ice remains at the end?

1. 94% 2. 72% 3. 33.33% 4. 66.66%

15. Mohan spends 50% of his monthly allowance on food, 18% on magazines, 10% on writing material and 15% on miscellaneous items. If he saves Rs.840 at the end of the year out of his allowance, what is his monthly allowance?

1. Rs.3000 2. Rs.2200 3. Rs.1000 4. Rs.750

Practice Exercise 21. Star TV is presently connected to 2 million houses in India. If 1000 new connections are

given every day, what is the annual percentage growth in star TV viewer-ship?

1. 18.25% 2. 22.15% 3. 7.07% 4. 11.11%

2. The price of wheat drops by 20%. How many kg can be bought now with the same money that was sufficient to buy 16 kg prior to price change?

1. 18 2. 20 3. 20.5 4. 17

3. In a village, 12% of goats were lost in a flood and 5% of the remainder died due to starvation. If the number of goats remaining is 8,360, what was the original number of goats before the flood?

1. 10,000 2. 8,800 3. 9,400 4. 10,240

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4. Anupama’s income is 15% greater than Sivakami’s. By what percent is Sivakami’s income less than Anupama’s?

1. 13.04% 2. 15% 3. 20% 4. 17.1%

5. A salt solution contains 5% salt and the remaining water. The total volume of the solution is 10 litres. If 2 litres of water evaporate, what percent of the remaining solution is salt?

1. 6 % 2. 5% 3. 7 % 4. 6 %

6. The price of Vadilal ice cream is 50% more than Dollops and that of Walls is 50% more than Vadilal. By what percent is Dollops cheaper than Walls?

1. 50% 2. 66.66% 3. 55.55% 4. 60%

7. The sales of a business in 1977 was 20% less than in 1976. By what percent must the sales in 1978 be increased to bring it to the level of 1976?

1. 20% 2. 25% 3. No increase4. 30%

8. The price of sugar is increased by 20% and its consumption is reduced by 20%. Find out the net increase/decrease in expenditure on sugar.

1. 4% increase 2. 6% decrease 3. 4% decrease 4. No change

9. If the price of each cup of tea served in a restaurant is reduced by 20%, one can buy 3 more cups of tea for Rs.30. What is the price of tea per cup?

1. Rs.2.00 2. Rs.2.25 3. Rs.2.30 4. Rs.2.50

10. A company manufactures electric tube lights which are priced at Rs.25 each. The tube lights consume power @ Re.1 per hour. The company is planning to launch a new type of tube-light which, as per the tests conducted, can cut the electricity bill by 50%. What higher percentage of price can the company fix for the new tube over the old variety if the extra cost is to be recovered in just 10 hours of lighting?

1. 50% 2. 35% 3. 20% 4. 48%

11. A and B are two cinema halls in a city. A charges an entrance fee of Rs.10.80 which includes entertainment tax @ 20% on Basic fee. Basic fee at B is cheaper by 60%, tax being 25%d on basic fee. By what percentage is the entrance fee of A costlier than that of B?

1. 350% 2. 140% 3. 235% 4. 180%

12. A fruit vendor is sanctioned a loan of Rs.5000 from a bank. The loan carries an interest of 40% compounded half-yearly. The vendor intends to recover the entire annual interest within 4 months of operation. By what percent should he load the prices, if he currently has an average monthly income of Rs.4400 and has 100% margin.

1. 25% 2. 5% 3. 15% 4. 30%

13. A rectangle shaped orchard has trees planted in 12 rows and 48 columns. The adjoining land is empty and square in shape, with two sides being half and double of the existing rectangular orchard farm. How many trees can be planted in the square land if the same spacing between trees is followed?

1. 400 2. 900 3. 121 4. 576

60

14. Suresh weighs 65 kg. He has an attack of viral fever and loses 1/5 of his weight. Within a few days he recovers and gains 25% weight. But when the fever relapses after two weeks, he loses 10 kg. By what percentage should he improve his weight in order to reach his normal weight?

1. 20% 2. 18.2% 3. 25% 4. 15%

15. The price of an article is cut by 20%. By what percent should it be increased to sell it at the original price?

1. 15% 2. 20% 3. 25% 4. 30%

Answer Keys for Practice Exercise 1

1 3 6 1 11 22 1 7 2 12 43 2 8 1 13 34 2 9 2 14 15 1 10 2 15 3

Answer Keys for Practice Exercise 2

1 1 6 3 11 22 2 7 2 12 13 1 8 3 13 44 1 9 4 14 25 4 10 3 15 3

CHAPTER-4

Profit, Loss & DiscountImportant points to remember

1. Cost Price (CP) : The price at which a commodity is purchased is called the COST PRICE of the commodity.

2. Selling Price (SP): The price at which a commodity is sold is called the SELLING PRICE of the commodity.

3. The amount that is deducted from the marked price is called Retailer’s Discount.

The difference between the marked price (List price) and the discount becomes the new selling price.

The reduction made on the marked price of the article is called the Discount.

DISCOUNT = Market price – Selling price.

4. The difference of cost price and selling price gives LOSS or PROFIT.

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Profit = Selling price – Cost price

Loss = Cost price – Selling price

Cost price = Selling price – Profit

Selling Price = Cost price + Profit

Cost price = Selling price + Loss

Selling price = Cost price – Loss

If the selling price is equal to the cost price then there is neither Profit nor Loss.

5. When profit or loss is calculated for Rs.100 (or any currency of 100 denominations), we get Profit Percentage and Loss Percentage.

Discount% = 100.

6. To find Selling Price, when Cost Price and Profit /Loss Percentage are given

7. The Selling Price of 2 commodities being the same, if one of them is sold at a Loss of R% and the other sold at a Profit of R%, then there will be an overall LOSS Percentage of

i.e., and, Loss =

Worked Examples1. A man sells two cars for Rs.2,00,000 each. On one he gains 5% and on the other he

loses 5%. Find his gain or loss percent on the whole transaction.

Solution:

Loss % = = .

2. If the selling price of 10 articles is the same as the cost price of 15 articles, find the gain percent.

Solution:

62

Let C.P. of each article be Re.1.

C.P. of 10 articles = Rs.10.

S.P. of 10 articles = C.P. of 15 articles = Rs.15.

Gain = .

3. By selling 100 bags, a shopkeeper gains the selling price of 25 bags. Find his gain percent.

Solution:

(S.P. of 100 bags) - (C.P. of 100 bags) = S.P. of 25 bags

S.P of 75 bags = C.P. of 100 bags

Let C.P. of each bag be Rs.1

then C.P of 75 bag = Rs.75

S.P of 75 bags = Rs.100

Gain = = 33.33%.

4. Santosh sells 2 bags at Rs.200, each he gets a profit of 15% on one bag an a loss of 15% on the other bag. What is overall profit or loss?

Solution:

CP = SP = 200

CP = 200 = Rs.173.91

CP = SP ( ) = 200( ) = 200 = 235.2

Total CP = 235.24 + 173.91 = Rs. 408.25

Total SP = 400

He has loss.

Loss% = = = 2.25%

In the previous problem % profit and % loss was 15%.

Then loss% = = = 2.25%

63

Loss =

Loss =

= = 8.022

But the above is not valid when the cost price of both the articles are same.

5. A dishonest dealer prefers to sell his goods at cost price but uses a weight of 950 gms for 1 kg. Find his gain percent.

Solution:

Gain % = = .

6. By selling an article of Rs.120, one gains Rs.20. Find the gain percent.

Solution:

S. P. = Rs. 120, Gain = Rs. 20

C. P. = Rs. (120 - 20) = Rs. 100

Gain of Rs. 100 is Rs. 20

So, gain % = .

7. By selling an article for Rs.100, a man loses 20% for how much should he sell it to gain 8%?

Solution:

S.P. = Rs.100, loss = 10%

C.P = Rs. = Rs.125

Now C.P = Rs.125, Gain required, = 8%

S.P = Rs. = Rs.135.

8. A man loses Rs.50 by selling some pens at the rate of Rs.3 per piece and gains Rs.100, if he sells them at Rs.4 per piece. Find the number of pieces sold by him.

Solution:

Suppose the no. of all pieces sold is x. Then, the difference of selling prices at the new rates is Rs.150.

Rs.4x – Rs.3x = 150 x = 150

64

Thus the number of pens sold = 150.

9. The percent profit made when an article is sold for Rs.50 is twice that when it is sold for Rs.45. Find the C.P of the article.

Solution:

Let C.P. be Rs.x. Then

 = 

 50 - x = 2(45 - x) x = 40

Thus, C.P. of the article = Rs.40.

10. X spends 80% of his income. His income was increased by 10% and his expenditure was increased by 5%. By what percentage did his savings increase?

Solution:

Let the original income of X = Rs.100

Expenditure = 80% of 100 or Rs.80

His savings = Rs.100 - Rs.80 = Rs.20

X’s new expenditure = (100 + 5)% of 80 = Rs.84

Savings = Rs.110 - Rs.84 = Rs.26

Increase in savings = Rs.26 - Rs.20 = Rs.6

Percentage of increase in savings = = 30%

65

Practice Exercise 11. Find the selling price if the cost price is Rs.150 and the profit is 20%.

1. 120 2. 180 3. 210 4. 100

2. Find the cost price if selling price is Rs.200 and profit is 12%.

1. Rs.175.87 2. Rs.170 3. Rs.178.57 4. Rs.185.77

3. Find the percentage profit if the cost price is Rs.175.00 and selling price is 250.00.

1. 42.8 2. 48.2 3. 48 4. 40.8

4. Find the selling price if cost price is Rs. 1,000 and percentage loss is 10.

1. 1100 2. 1010 3. 1001 4. 900

5. A sells a cycle to B at a profit of 20% and B sells it to C at a profit of 25%. If C pays Rs. 225 for the bicycle, how much did A pay for it?

1. Rs.150 2. Rs.120 3. Rs.125 4. Rs.225

6. A merchant sells two articles for Rs. 150/- each. He sells one at a loss of 10% and the other at a profit of 10%. What is his net profit or loss?

1. Loss of Rs.1.50 2. Gain of Rs.3.03 3. Loss of Rs.3.03 4. No profit, no loss

7. A fruit seller bought some oranges at 4 per rupee and an equal number at 5 per rupee. He then sold the entire quantity at 9 for two rupees. What is his percentage gain or loss?

1. 8 % 2. 2% 3. 1 % 4. No profit, no loss

8. A tape recorder is sold at Rs.935 at a profit of 10%. What would have been the actual profit or loss if it had been sold for Rs.810?

66

1. Rs.55 loss 2. Rs.75 loss 3. Rs.20 gain 4. Rs.40 loss

9. A merchant sells rice and makes a profit of 6%. His cost price increases by 10% and as a result he increases his selling price also by 10%. What percent profit does he earn now?

1. 6% 2. 6.6% 3. 10% 4. None of these

10. Ram buys a Maruti car at a 20% discount and immediately sells it at 20% more than the list price. What will be his profit?

1. 40% 2. 50% 3. 60% 4. 20%

11. The printed sale price of a book is Rs.60. The trader offers two successive discounts of 20% and 30%. The net sale price is subject to a sales tax of 5%. How much does the buyer pay?

1. Rs.35.28 2. Rs.31.50 3. Rs.36.00 4. Rs.32.48

12. John sells two cameras at Rs.1000 each. On one camera he gets a profit of 10%, while on the other he incurs a loss of 10%. What is his overall profit or loss?

1. Gain of Rs.20 2. No profit, no loss 3. Loss of Rs.20.20 4. None of these

13. A retailer buys a dozen pens for Rs. 96. The wholesaler gives him a pen free on the condition that he should sell at Rs. 8 per pen. What will be the retailer’s percentage profit?

1. 8.33% 2. 10% 3. 6.66% 4. 12.25%

14. A cloth merchant sells at a profit of 10%. What is his net profit if he sells goods and bills Rs.1000 on a particular day, given that the measuring tape of the merchant measures 995 mm, instead of 1m?

1. Rs.95.45 2. Rs.82.33 3. Rs.110.25 4. Rs.78.75

15. A trader buys 78 kg of wheat for Rs.892. He sells 40% of this quantity at a loss of 20%. What should be his percentage mark up on the remaining in order that he earns an overall profit of 25%?

1. 40% 2. 55% 3. 28% 4. 45%

67

Practice Exercise 21. Sheela buys eggs at Rs. X per dozen and sells them at 8 times that price per hundred.

What is her percentage profit or loss?

1. 16% profit 2. 4% profit 3. 4% loss 4. 16% loss

2. Ram buys an article at th its value and sells it at 50% more than its value. What is his

percentage profit?

1. 50% 2. 100% 3. 75% 4. 125%

3. A manufacturer’s list price for an item is 40% more than its cost. If he allows a discount of 10% on the list price, what will be his percentage profit?

1. 30% 2. 28% 3. 26% 4. 24%

4. Krishna buys three articles at Rs.800 each. He sells the first at 15% profit and the second at 20% loss. What should be his percentage profit on the third article to ensure that he makes an overall profit of 10%?

1. 30% 2. 35% 3. 40% 4. 25%

5. X buys 6 books for Rs.100 and sells 5 books for Rs.100. What is his percentage profit?

1. 30% 2. 35% 3. 40% 4. 25%

6. Shyam buys X eggs to resell them at a profit of 10% but loses 10% of the eggs due to transit damage. By how much should he mark up the selling price of the eggs in order to retain a 10% profit?

1. 30% 2. 40% 3. 33.33% 4. 22.22%

7. A man sells 2 goats at Rs.120 each. By doing so, he gains 25% on one goat and loses 25% on the other. What is his loss/gain percent if any?

1. 0 2. 6.25 3. 4.3 4. 2

8. A merchant loses 7% on a certain tea, but he uses a weight of 900 gm instead of 1 kg. Find his profit %.

68

1. 3.33 2. 2.05 3. 4.67 4. 4.21

9. A sells a bicycle to B at a profit of 20% and B sells it to C at a profit of 25%. If C pays Rs.225 for the bicycle, how much did A pay for it?

1. 100 2. 125 3. 150 4. 200

10. How much percent above the C.P. should a shop keeper mark his goods so that he gains 26% after allowing a discount of 10%?

1. 40 2. 50 3. 70 4. 30

11. A trader marks his goods 25% above the C.P. and allows his customers a discount of 12%. What is his profit%?

1. 6 2. 8 3. 10 4. 9

12. A saree was priced at Rs.1,200/-. After two successive discounts, the selling price was Rs.816. If the first discount was 15%, find the rate of the second discount.

1. 10% 2. 15% 3. 20% 4. 33 %

13. By selling 25 metres of cloth at Rs.50 per metre, a merchant earns a profit equivalent to the cost price of 5 metres of cloth. The percentage profit is

1. 15% 2. 25% 3. 20% 4. 18%

14. A man buys 50 chairs for Rs.5,000/- but 20 of them are damaged. He decides to sell the

damaged chairs each at th the price of the normal one. What should be the price of a

normal chair in order that the person makes a profit of 35% for the whole lot?

1. Rs.125 2. Rs.170 3. Rs.130 4. Rs.175

15. An article was bought for Rs. 1,300 and sold immediately for Rs. 1,436.50 payable in 10 months time. Find the loss or gain accrued if the rate of interest is 10% per annum?

1. 2% gain 2. 2 % gain 3. 2% loss 4. 5% loss

Answer Keys for Practice Exercise 1

1 2 6 1 11 12 3 7 3 12 43 1 8 4 13 14 4 9 1 14 15 1 10 2 15 2

Answer Keys for Practice Exercise 2

69

1 2 6 4 11 32 3 7 2 12 33 2 8 1 13 34 3 9 3 14 45 2 10 1 15 1

CHAPTER-5

Simple & Compound InterestInterest

The word “interest” has been taken from the Latin word Interesse.

1. Interest is the money paid for the privilege of using another person’s money. When a lender lends the money to the borrower, the money that is lent is called the PRINCIPAL and the sum of money paid for the use of it is called the INTEREST.

2. In other words, the principal is the amount of money that is borrowed and interest is the proportionate sum of money paid for the use of it. The sum of the interest and principal at the end of any time is called the AMOUNT.

Interest: It is money paid for the privilege of using another person’s money.

Principal: When Lender lends the money to the borrowers, the money that is lent is called the principal.

Amount: Amount = Principal + Interest.

The sum of the interest and principal at the end of any time is called the Amount.

Simple Interest1. Simple interest is the interest calculated on the original P rincipal . It is sum of money paid to the

lender by the borrower for particular period.

2. Simple Interest on Principal “P” for “T” number of years, at then Rate of R% per annum is given by,

And the Amount,

A = P + I where I =

70

A = P +

The principal amount does not change for any time period, while calculating Simple Interest.

Compound Interest1. In compound interest calculations, the interest for each period is added to the principal

before interest is calculated for the next period. With this method the principal grows as the interest is added to it. This method is used in investments such as savings account and bonds.

2. When the compound interest is calculated Yearly:

1. Let Principal = P

Rate = R%

Time = n years

2. Let the rates of interest for 2 successive years be R1% & R2% per annum, respectively. Let A be the amount after 2 years, then

If R1%, R2% and R3% be the interest rates for 3 successive years then the amount after 3 years will be

3. If the period is not complete number of year:

If it is n years and m months and interest is R% p.a. compounded yearly, then

3. Simple Interest and Compound Interest for 1 year at a given Rate of Interest per annum are always Equal.

4. When the compound interest is calculated Half-Yearly:

Principal = P, Rate = R% per annum, Time = n years.

71

Since interest is earned half yearly, the rate of interest will be for every half year, and time

will be 2n half years.

Let A be amount after n years, then

5. When the compound interest is calculated Quarterly:

Rate of interest is per quarter.

Time = 4n (quarterly)

Then the Amount,

Worked Examples

1. Mr. Christy borrowed Rs. 7,500 from a bank on 1st June 1994 at 63

4% p.a. S.I. and paid the amount

back on 3rd June 1995. Find the amount paid.

Solution:

Amount = Principal + S.I.

Principal = Rs. 7500

S.I. = , R = 27

4% p.a.

N = 12 months + 3 days 1 year

S.I. = = Rs. 506.25

Amount = Rs. 8006.25.

2. Find S.I on Rs. 8000/- at 161

2% p.a. for the period from 7th June 1997 to 8th September 1997.

Solution:

P = Rs. 8000, N = June 23 days + September 8 days + 2 months

N =( + ) years = 0.252 years

72

R = % p.a.

So, I = = Rs. 332.64

3. On what capital will S.I. for 3 weeks and 4 days at 12% p.a. amount to Rs. 16?

Solution:

S.I. = ; R = 12% ; S.I. = 16

N = 3 weeks + 4 days = years

N = 0.0685 years

16 =

P = = Rs. 1946.5

Capital or Principal amount = Rs. 1946.5

4. A sum of money trebles itself in 15 years 6 months. In how many years would it double itself?

Solution:

Let the sum i.e. P = x

When the sum trebles itself, S.I. = 2x, N = 151

2 years = years

2x = x×R 31

×100 2

R = %

When the sum doubles itself, S.I. = x

x = x N N = = 7.75 years

i.e., 7 years 9 months.

5. Find the compound interest on Rs.6500 for 3 years at 10% p.a. compounded annually.

Solution:

C.I. = A - P

A = P (1 + r)n

P = 6500 A = 6500

73

A = 6500(11)3 = 8651.5

C.I. = 8651.5 - 6500= Rs. 2151.5

6. At what rate percent of compound interest will Rs.970 earn Rs.250 as interest in 10 years?

Solution:

C.I = Rs. 250

P = Rs. 970

A = P + C.I = 970 + 250 = Rs. 1220

A = P (1 + r)n

1.258 = (1+r)10

1 + r =

7. A certain sum put at S.I. amounts in 4 years to Rs.780 and double the same sum amounts to Rs.2000 in 6 years. Find the sum and rate of interest.

Solution:

Amount = P +

780 = P + .......... (1)

and,

2000 = ......... (2)

Equation (2) Equation (1)

2000

R = -22

R = = 32.4% p.a.

74

8. Mr. Christy and Mr. Clint lend Rs.400 for 2 years at 4% p.a. compound interest. Mr. Christy’s interest is paid half yearly and Mr. Clint’s interest is paid yearly. Find the difference between the amounts.

Solution:

Christy’s case

P = Rs. 400

n = 2 years

R = 4%

(Because interest is payable half yearly.)

A = P = 400 = 400 (1.02)4 = Rs. 432.97

Clint’s case

A = P = 400 = 400 (1.04)2 = Rs. 432.64

Difference between the amounts is Rs.0.33/-

9. Find the compound interest on Rs.5600 for 11

2 year at 10% p.a. compounded annually.

Solution:

A = 5600 = Rs.6468

C.I = 6468 - 5600 = Rs.868

10. Find the CI on Rs. 10,000 in 10 years if the rate of interest is 5% for the first 2 years, 8% for the next 3 years, 9% for the 6th and 7th years and 10% for last 3 years.

Solution:

CI = A - P

A = 10,000 = Rs. 21,965

CI = 21,965 - 10,000 = Rs. 11,965

{Note using varying rate of interest of interest formula}

Practice Exercise 1

75

1. A sum of money placed at CI doubles itself in 5 years. It will amount to eight times itself in how

many years?

1. 15 years 2. 20 years 3. 33 years 4. 21 years

2. In how many years will a sum at Rs.800 at 10% per annum compounded semi-annually become

Rs.926.10?

1. 4.5 years 2. 1.5 years 3. 4 years 4. 6 years

3. A sum of money, compounded, amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years.

Find the sum.

1. Rs.1110 2. Rs.5560 3. Rs.5432 4. Rs.4460

4. A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years. Find the sum and rate

percent.

1. Rs.5900, 15% 2. Rs.5200, 16.6% 3. Rs.5400, 16.7%. 4. Rs.5500, 5.6%

5. A man deposited in a bank Rs.5178 for his 2 sons aged 12 years and 15 years, so that the sum

should be divided in such a way that the 2 sons get the same amount when they attain the age of

18 years. The rate of CI is 5% p.a. What should be the share of the elder son at the beginning?

1. Rs.2400 2. Rs.5600 3. Rs.3450 4. Rs.2100

6. If the simple interest on a certain sum of money at 5% per annum for 3 years is Rs.1200, find the

compound interest on the same sum for the same period at the same rate?

1. Rs.1006 2. Rs.3210 3. Rs.990 4. Rs.1261

7. If the compound interest on a certain sum for 2 years at 3% be Rs.101.50, what would be the

simple interest?

1. Rs.98 2. Rs.100 3. Rs.110 4. Rs.220

8. Find the effective rate of compound interest when the money is lent at 62

3% interest payable 6

monthly.

1. 2.3% 2. 8% 3. 2.89% 4. 6.78%

9. Find the effective rate of interest, when money is lent at 10% interest payable quarterly.

1. 10.43% 2. 10.38% 3. 9.98% 4. 11.11%

10. Find the effective rate of compound interest when Rs.8000 is lent at 6% p.a. interest payable

monthly.

1. 3.6% 2. 1.6% 3. 6.17% 4. 5.2%

11. What would be the value of a machine that had an original cost of Rs.3000, at the end of 4 years, if

the depreciation rate is 16% p.a.?

1. Rs.1494 2. Rs.2564 3. Rs.980 4. Rs.1878

76

12. Machinery costing Rs.32000 has an estimated life of 10 years. If the depreciation is 8% p.a. what

will be the value of the machine at the end of its life time.

1. 32000(1 - .06)10 2. 32000(1 - .12)10 3. 32000(1 - .08)10 4. 32000(1 - .08)9

13. A machine costing Rs.16000 is being depreciated at 10% p.a. After how many years will its value

be reduced to Rs.5000?

1. n = 2. n = 3. n = 4. n =

14. If the value of the machine after 5 years is Rs.17910 and at the end of 7 years is Rs.12500, find the

rate of depreciation.

(Note: log(0.6979) = -0.1562), antilog(-0.078) = 0.8354)

1. 33.33% 2. 16.46% 3. 83.98% 4. 32.98%

15. A property is worth Rs.20000 today. It is depreciated by of its book value every year. Find

depreciated amount by the end of the 10th year. (Note: (0.95)10 = 0.5987)

1. Rs.6890 2. Rs.2870 3. Rs.5906 4. Rs.8030

Practice Exercise 2

1. The population of a town is 10,00,000 today. What will be it population after 2 years if the rate of

increase is 10% p.a. of that of the preceding year?

77

1. 1,210,000 2.1,45,890 3.1,12,100 4.2.98.000

2. A tree increases annually by 1

8th of its height. By how much will it increase after 2 years, if it

stands 64 cm high today?

1. 66 cm 2. 98 cm 3. 22 cm 4. 81 cm

3. In a certain village the annual birth rate for 1000 is 44.8 and annual death rate for 1000 is 24.8. In

how many years will the population of the village double itself?

Note: log(2) = 0.3010, log(1.02) = 0.0086

1. 6.4 years 2. 12.5 years 3. 35 years 4. 25 years

4. Find the annual installment if the amount of a loan is Rs.500 at 12% p.a. for 4 equal installments.

1. Rs.164.61 2. Rs.133.10 3. Rs.196.69 4. Rs.120

5. Ramaswami borrowed Rs.33100 at 10% p.a. compound interest and agreed to repay in three equal

installments. Find the approximate amount of the annual installment.

1. 14,685 2. 13,780 3. 9,900 4. 13,310

6. A sum of Rs.1100 was taken as a loan. This is to be repaid in 2 equal installments. If the rate of

interest be 20% compounded annually, then find the value of each installment.

1. Rs.660 2. Rs.720 3. Rs.780 4. Rs.210

7. Mr. Ravi, Mr. Ram and Mr. Sham borrowed Rs.11730 from a money lender at 5% p.a. simple

interest and paid back equal amounts in full settlement of their respective debts after 4, 10 and 12

years. Find the sum borrowed by Ravi?

1. Rs.890 2. Rs.2600 3. Rs.4600 4. Rs.5200

8. A sum of money doubles itself in 121

2 years at a certain rate percent on SI. How long will it take

to double at the same rate on compound interest?

1. 6.5 years 2. 5 years 3. 1.5 year 4. 9 years

9. The difference between CI and SI on a sum of money for 5 years at 13% p.a. was Rs.46.80. Find

the approximate value of the sum.

1. Rs.3500 2.Rs.9890 3. Rs.5050 4. Rs.1200

10. The compound interest on a sum of money for 3 years at 10% is Rs.827.50. What is the SI,

calculated at the same rate, for the same period?

1. Rs.120 2. Rs.750 3. Rs.640 4. Rs.650

11. The SI on a certain sum of money for 2 years is Rs.1550 and CI is Rs.1588.75. Find the sum and

rate of interest.

1. Rs.15500 2. Rs.18500 3. Rs.15321 4. Rs.21500

78

12. A claim of Rs.19720 due after a certain period is worth only Rs.9000 today. Find the period if the

rate of CI is 4% p.a.

1. 13 years 2. 11 years 3. 20 years 4. 15 years

13. A lends to B a sum of Rs.1200, partly at 5% p.a. SI and partly at 6% p.a. SI. If at the end of 21

2

years A received Rs.160 as interest, find the portion lent out at 5% p.a.

1. Rs.300 2. Rs.600 3. Rs.500 4. Rs.800

14. A sum of money doubles itself at compound interest in 15 years. In how many years will it become

eight times?

1. 25 years 2. 45 years 3. 15 years 4. 5 years

15. If the CI on a certain sum for 2 years at 12% p.a. is Rs.159, find SI for the same period, rate and

sum.

1. Rs.56 2. Rs.90 3. Rs.120 4. Rs.150

Answer Keys for Practice Exercise 1

1 1 6 4 11 12 2 7 2 12 13 4 8 4 13 14 3 9 2 14 25 1 10 3 15 4

Answer Keys for Practice Exercise 2

1 1 6 2 11 12 4 7 2 12 33 3 8 4 13 44 1 9 3 14 25 4 10 2 15 4

Chapter 6

Average, Alligation & MixturesAverages

79

Examples:

1. The average age of 30 kids is 9 years. If the teacher’s age is included, the average age becomes

10 years. What is the teacher’s age?

Total age of 30 children = 30 9 = 270 yrs.

Average age of 30 children and 1 teacher = 10 yrs

Total of their ages = 31 10 = 310 yrs

Teacher’s age = 310 – 270 = 40

2. The average of 6 numbers is 8. What is the 7th number, so that the average becomes 10?

Let x be the 7th number

Total of 6 numbers = 6 8 = 48

We are given that =

x = 22

3. Five years ago, the average of Raja and Rani’s ages was 20 yrs. Now the average age of Raja, Rani

and Rama is 30 yrs. What will be Rama’s age 10 yrs hence?

Total age of Raja and Rani 5 years ago = 40

Total age of Raja and Rani now = 40 + 5 + 5 = 50

Total age of Raja, Rani and Rama now = 90

Rama’s age now = 90 – 50 = 40

Rama’s age after 10 years = 50

4. Out of three numbers, the first is twice the second and thrice the third. If their average is 88, find

the numbers.

Third number = x (say)

First number = 3x

Second number =

Total =

80

Average = (given)

i.e.,

x = 48

48, 144, 72 are the numbers.

5. The average of 8 numbers is 21. Find the average of new set of numbers when 8 multiplies every

number.

Total of 8 numbers = 168

Total of new 8 numbers = 168 8 = 1344

Average of new set =

MixturesTypical Problems related to Mixtures

1. The proportion of prices and quantities of the constituents are known and the effective price of the

mixture is to be established.

2. The prices of the constituents are known and it is required to determine the proportion in which

the constituents must be mixed in order to achieve the required profit percentage.

When different quantities of same or different ingredients, of different costs are mixed together to

produce a mixture or mean cost, the ratio of their quantities are inversely proportional to

difference of the mean cost.

Alligation

81

1. Alligation is a rule to find the proportion in which two or more ingredients at the given price must

be mixed to produce a mixture at a given price.

2. Cost per Price of unit Quantity of the mixture is called the Mean price.

Rule of Alligations

If two ingredients are mixed in a ratio, then

Cheaper Quantity: Dearer Quantity = (d – m) : (m – c)

Worked Examples1. How many kg of wheat costing Rs.6.10 /kg must be mixed with 126 kg of wheat costing Rs.2.85/kg,

so that 15% may be gained by selling the mixture at Rs.4.60/kg.

C.P. of mixture =

=

82

Cost price of a unit C.P. of a unit Cheaper quantity (c) quantity of dearer (d)

Mean Price (m)

d – m m – c

C.P. of 1kg of cheaper C.P. of 1 kg of Wheat dearer wheat(285 paise) (610 paise)

Mean Price (400 paise)

210 115

If cheaper wheat is 42 kg then dear one is 23 kg.

If cheaper wheat is 126 kg, then dearer one =

2. A waiter stole wine from a bottle of sherry, which contained 30% of spirit, and he filled the join with

wine, which contains only 15% of spirit. The strength of the jar then was only 22%. How much of

the quantity did he steal?

By alligation rule

They must mixed in the ratio 7:8.

The waiter removed of the jar.

3. ‘x’ covers a distance of 60 km in 6 hrs partly on foot at the rate of 4 kmph and partly on a cycle at

14 kmph. Find the distance traveled on foot.

Average distance traveled in 1 hr = 10 km

Suppose x travels y hrs at 4

He travels (6 – y) hrs at 14

Total distance travelled = 4y + (6 – y) 14 km

4y + (6 – y)14 = 60

y = 2.4

4. A mixture of 40 litres of milk contains 20% of water. How much water must be added to make the

water 25% in the new mixture?

Water quantity initially = litres.

Quantity of milk initially = 40 – 8 = 32 litres

Milk + Water = (32) + (8 + x) = 40 + x

We are given that

83

Wine with 30% Wine with 15% Spirit Spirit

22

7 8

litres of water must be added.

5. Four litres of phenol is drawn from a can. It is then filled water. Four litres of mixture is drawn

again and the bottle is again filled with water. The quantity of phenol now left in the bottle is to

that of water is in the ratio 36:13. How much does the bottle hold?

Hint:

Amount of liquid left after n operation, if x is the capacity of the container from which y units are

taken out each time is

units.

Here y = 4 n = 2

x = 7

The bottle holds 7 litres

84

Practice Exercise 11. The temperatures are recorded on June 7th every year at a certain place during five consecutive

years. Temperatures are 98F, 103F, 95F, 102F and 97F. Find the average temperature on June

7th.

1. 98F 2. 99F 3. 91F 4. 97F

2. The following table gives the monthly income of 6 lecturers in Bangalore City. Find the average

income of the lecturers.

Sl. No. 1 2 3 4 5 6

Income (Rs.) 3500 3800 4200 4800 5200 5100

1. 4433.3 2. 3549.7 3. 4233.3 4. 5366.7

3. The height of students in inches is given below. Find the average height (in inches) of the students.

61, 62, 66, 68, 69, 67, 61, 67, 68, 66, 65, 66

1. 61 2. 65.5 3. 66 4. 62

4. A motor cyclist travels at an average speed at 50 kilometers per hour for 40 minutes and then

travels at an average speed at 80 kmph for the next 50 minutes. Find the average speed for the

whole period.

1. 66 2. 64 3.66 4. 75

5. An automobile went up a hill at a speed of 20 km per hour and down the hill at a speed of 30 km

per hour. Find the average speed for the round trip.

1. 24 2. 23 3. 22 4. 21

6. An electric train passes 12 poles in 30 seconds. If the distance between 2 poles is 50 meters, find

the speed of the electric train (in km per hour).

1. 66 2. 65 3. 32 4. 34

85

7. The average age of students in a class of 40 is 16 years. 12 more students of average age 19

years are admitted to the class. Find the average age, in years, of the class.

1. 16.7 years. 2. 16 3. 14 4. 15

8. A furniture shop sells 40 chairs at an average cost of Rs.14 each per day and 20 tables at an

average cost of Rs.18 each per day. Find the approximate average income of the furniture shop

per day.

1. 23 2. 32 3. 24.8 4. 15.3

9. A cycle shop sells 3 types of cycles. The average per day income of the shop is Rs.200. This shop

sells 10 type I cycles at an average cost of Rs.150 per day. Again it sells 20 type 2 cycles at an

average cost of Rs.180 and type 3 cycles at an average cost of Rs.210 per day. Find the total

number of cycles that the shop sells per day.

1. 100 2. 120 3. 140 4. 160

10. The average age of a class of 35 students is 14.35 years. By admitting one more student the

average age increased by 0.05 years. Find the age of the new student.

1. 16.15 2. 18.25 3. 17.15 4. 15.75

11. If 5 kg of salt costing Rs.5/kg and 3 kg of salt costing Rs.4/kg are mixed, find the average cost of

the mixture per kilogram.

1. Rs.4.50 2. Rs.4.45 3. Rs.4.75 4. None of these

12. A mixture of 125 gallons of wine and water contains 20% water. How much water must be added

to the mixture in order to increase the percentage of water to 25% of the new mixture.

1. 8.33 gallons 2. 10 gallons 3. 8.67 gallons 4. None of these

13. The average salary per head of all employees of a company is Rs.600. The average salary of 120

officers is Rs.4000. If the average salary per head of the rest of the employees is Rs.560, find the

total number of workers in the company.

1. 1000 2. 1050 3. 1032 4. None of these

14. 400 students took the CP Mock CAT in Vijayanagar. 60% of the boys and 80% of the girls clear the

cut-off in the examination. If the total percentage of the students qualifying is 65%, how many

girls appeared in the examination?

1. 120 2. 100 3. 150 4. 180

15. A merchant purchased two qualities of pulses at the rate of Rs.200 per quintal, and Rs.260 per

quintal. In 52 quintals of the second quality how much pulse of the first quality should be mixed,

so that by selling the resulting mixture at Rs.300 per quintal, he gains a profit of 25%?

1. 26 quintals 2. 104 quintals 3. 58 quintals 4. None of these

Answer Keys for Practice Exercise 1

1 2 6 1 11 4

86

2 1 7 1 12 1

3 2 8 4 13 3

4 4 9 2 14 2

5 1 10 1 15 1

Practice Exercise 21. The average weight of patients on a particular day in a hospital is recorded as 50 kg. Of these

there were 20 children of average weight 30 kg, 40 males of average weight 60 kgs and 50 female

patients. Find the average weight of the female patients.

1. 40 2. 50 3. 30 4. 25

2. The average income in a cake shop is Rs.440 per day. That shop sells 3 types of cakes. The shop

sells 200 cakes of the first type at an average cost of Rs.250, 300 cakes of the second type at an

average cost of Rs.300 and the third type of cakes at an average cost of Rs.600. Find the number

of the third type of cakes that the shop sells per day.

1. 500 2. 100 3. 5 4. 1000

3. In a factory the average hourly wage of 15 workers was $4, the average hourly wage of 5 workers

was $10 and that of remaining workers was $15 each. What was the average wage paid to 30

workers?

1. 6.7 2. 7.7 3. 8.7 4. 5.7

4. A juice shop sells 20 glasses of orange juice for Rs.6 each, 15 glasses of apple juice for Rs.10 each

and 25 glasses of grape juice for Rs.8 each per day. Find the average income per glass.

1. 1.83 2. 9.83 3. 8.83 4. 7.83

5. In a car race there are 25 signal points. The average distance between 2 signal points is 2 kms and

a particular car takes min. to complete the distance between 2 signal points. Find the average

speed of the car.

1. 240 kmph 2. 210 kmph 3. 220 kmph 4. 230 kmph

6. In a cycle rally, a particular cyclist’s speed was 20 kmph to cross the first part of a distance of 3

kms. His speed was 30 kmph for the second part of the distance 4 kms and his speed was 40

kmph for the remaining part. Find the average speed of the cyclist if the total distance is 10 kms.

1. 23.7 kmph 2. 27.9 kmph 3. 24.7 kmph 4. 29.9 kmph

7. Jaya bought 4 sarees in all, the average price being Rs.150. The average price of the best 2 sarees

is Rs.170. If the price of the other 2 sarees are in the ratio 7: 6. Find the price of the cheapest

saree.

1. 120 2. 140 3. 160 4. 18

8. Mr. Thomas had taken 8 subjects at B.E Examination in March 1971. In the first 5 subjects he

scored 52 marks on an average while he scored 16 marks on an average in the last five subjects. If

87

he scored 480 marks in all and ten marks more in the fifth subject than he scored in the fourth,

find the number of marks scored by him in the fifth subject.

1. 40 2. 45 3. 50 4. 60

9. A polish boy earned Rs.36 in 6 days. His average earning for the first 5 days was Rs.5.66 per day

while his earning for the last 4 days was Rs.6.25 per day. On the 4th day he earned Re.1 more

than what he earned on the 3rd day but Rs.3 less than what he earned on the fifth day. What was

his per day earning for the last two days?

1. 8 2. 7 3. 6 4. 5

10. The daily average wages of a group of 25 males, 15 females and 10 children working together

amount to Rs.2.80. If the female worker gets twice what a child is paid and a male worker gets

twice what a female worker is paid as daily wages, find the daily wages paid to a male worker, a

female worker and a child.

1. 1, 2, 3 2. 4, 2, 1 3. 5, 6, 8 4. 1, 3, 5

11. Two vessels contain spirit and water, mixed respectively, in the ratio of 1:3 and 3:5. Find the ratio

in which the contents of the vessels are to be mixed so as to get a new mixture in which the ratio

of spirit to water is 1:2.

1. 1:1 2. 1:2 3. 2:1 4. None of these

12. In what proportion must water be mixed with milk so as to gain 20%, by selling the mixture at the

cost price of the milk? (assume that water is freely available).

1. 2:7 3. 1:5 3. 3:4 4. None of these

13. If a mixture of 20 litres of brandy and water contains 10% water, how much water should be added

to it to increase the percentage of water to 25%.

1. 5 litres 2. 6 litres 3. 2 litres 4. None of these

14. If a man decides to travel 80 kilometers in 8 hours, partly by bicycle and partly by foot. His speed

on foot is 8 kmph and that on bicycle is 16 kmph. What distance would he cover on foot?

1. 48 km 2. 30 km 3. 64 km 4. None of these

15. Two solutions of 90% and 97% purity are mixed, resulting in 21 litres of mixture of 93% purity.

How many litres of the first solution were used?

1. 9 2. 18 3. 12 4. None of these

Answer Keys for Practice Exercise 2

1 2 6 2 11 1

2 1 7 1 12 2

3 3 8 4 13 4

4 4 9 1 14 1

5 1 10 2 15 3

88

Chapter-7

Time, Speed & DistanceImportant points to remember

1. The rate at which the distance is traveled is called SPEED.

2. Average Speed is the ratio of the total distance traveled to the total time taken.

i.e., Average Speed =

Speed = where s = Total distance, t = Total time

Hence, (a) s = vt

(b)

3. Relative Velocity

The relative velocity of two bodies moving at velocities u and v (u>v) in the same

direction is u – v.

The relative velocity of two bodies moving in opposite directions is u + v.

4. Motion downstream or upstream:

Velocity of boat downstream = u + v

Velocity of boat upstream = u – v

where ‘u’ is the velocity of the boat in still waters and ‘v’ is the velocity of the

stream.

5. Speed Change

If a man changes his speed in the ratio u : v, the corresponding ratio of traveled times

will be v : u.

6. Races

Race: A contest of speed in running, riding, driving, sailing or rowing is called a race.

Dead-heat race: If all the persons contesting a race reach the goal at the same time,

then it is called a dead-heat race.

7. Useful Points:

89

1. A train or a moving body of known length has to travel its own length in passing a

lamppost or a fixed body of insignificant size.

2. A train or a moving body must travel its own length plus the length of the

stationary body in question, if the train or the moving body has to pass a

stationary body i.e. a bridge, a railway platform etc.

3. Time taken by the moving train to pass a stationary object is the time taken to

cover its own length in the speed it is moving.

If a train is moving at speed at R kmph of length L mts, then the time

taken to pass the stationary object is the time taken to cover distance L

mts in R kmph.

4. Time taken by the moving train of a meters to pass a stationary object of b

meters is the time taken by the train to cover (a+b) metrers in its own speed.

5. If two trains of length x meters and y meters move along the same direction at ‘a’

m/s and ‘b’ m/s, then the time taken by them to move across each other form the

time they meet is

t =

Worked Examples

90

1. A train travels at 18 km/hr. How many metres will it travel in 12 minutes?

Solution:

Distance traveled in 1 hour, i.e., 60 minutes = 18 km = 181000 metres.

Distance traveled in 12 minutes = = 3600 metres.

2. A passenger train running at 60 km/hr leaves the railway station 5 hours after a goods

train had left. The train overtakes the latter in 4 hours. What is the speed of the goods

train?

Solution:

Let the speed of the goods train be x km/hr.

Distance traveled by goods train before the passenger train overtakes it

= Speedtime

= x (5 + 4) = 9x km -------(1)

Distance traveled by passenger train before it overtakes the goods train

= 60 4 = 240 km -------(2)

Equate (1) and (2), We get 9x = 240

x =

Alternatively:

The slower train covers in 9 hours the distance of 240 km covered by the faster train in 4

hours.

Hence speed of slower train = or 26 km/hr.

3. A certain distance is covered with a certain speed. If th of the distance is covered in

twice the time, find the ratio of this speed to that of the original speed.

Solution:

Let the distance be x and the speed be y,

Time taken =

91

In the second instance, distance = and time=

Speed =

Ratio of speed = = 1 : 12

4. A train 300 m. long passes a pole in 15 sec. Find the speed.

Solution:

Distance = 300 m, Time = 15 sec

Speed =

5. How many seconds will it take for a train 120 metres long moving at the rate of 15 m/sec

to overtake another train 150 m long running from the opposite side at the rate of 90

km/hr?

Solution:

Speed of the first train = 15 m/s.

Speed of the second train =

Since both the trains are moving in opposite directions,

Relative speed = sum of the speed of the two trains.

= (15 + 25) = 40 m/ sec.

Total distance = sum of the length of the two trains

= 120 + 150 = 270 metres.

Time taken =

Note: From km/hr into m/sec, the conversion ratio is . From m/sec to km/hr conversion

ratio is .

6. A train running between two stations A and B arrives at its destination 15 minutes late,

when its speed is 45 kmph and 36 minutes late when its speed is 36 kmph. Find the

distance between the stations A and B.

92

Solution:

Let ‘x’ km be the distance between the station A and B.

Speed of the train = 45 km/hr

Time taken =

Since the train is late by 15 minutes (= ),

Actual time =

Time taken when the speed is 36 km/hr =

Now, since the train is late by 36 min i.e., hrs

Actual time =

Equating (1) and (2), we get,

x = 63 km

7. If a man travels at a speed of 20 km/hr, then reaches his destination late by 15 minutes

and if he travels at a speed of 50 km/hr. Then he reaches 15 minutes earlier. How far is

his destination?

Solution:

Let ‘x’ be the distance of his destination from his starting point.

When his speed is 20 km/hr,

Time taken = hrs

93

Since, he is late by 15 minutes ie., hrs

Actual time = hrs …………………………(1)

When his speed is 50 km/hr,

Time taken = hrs

Now, since he reaches early by 15 minutes ie., hrs.

Actual time = + hrs ………………………(2)

Equating (1) and (2), we get, = +

- = + =

8. A person is standing on a railway bridge 70m long. He finds that a train crosses the

bridge in 6 secs., but himself in 4 secs. Find the length of the train and its speed in

km/hr.

Solution:

Let the speed of the train be x m/sec and the length of the bridge = y metres.

When the train crosses the bridge, speed = x m/sec

Distance = (90 + y) m

Time =

= 6

90 + y = 6x 6x – y = 90 ----------1

94

When the man is crossed by the train, speed = x m/sec and distance = y m

Time =

= 4 y = 4x ---------2

Solving 1 and 2, 6x – 4x = 90 2x = 90

x = 45 m/sec, y = 4x = 4 45 = 180 metres

Speed = 45 m/sec = 45 = 9 18 = 162 km/hr

Length of the train = 180 metres.

9. A wheel rotates 12 times in a minute and moves 5 metres during each rotation. What is

the time taken for the wheel to move through 930 metres?

Solution:

For one rotation, distance moved = 5 metres

For 12 rotations, distance moved = 12 5 = 60 metres

Distance covered in 1 minute = 60 metres

Total distance covered by the wheel = 930 minutes

Total time taken = minutes.

10. A boat sails 6 km upstream at the rate of 4 kmph. If the stream flows at the rate of 3

km/hr., how long will it take for the boat to make the return journey?

Solution:

Speed of stream = 3 km/hr

Speed of boat in still water = speed of stream + speed of boat upstream

= 3 + 4 = 7 km/hr.

Distance to be covered during the return journey downstream = 6 km

Speed of the boat downstream = Speed of the boat in still water + speed of the

stream

= 7 + 3 = 10 km/hr.

Time taken for return journey = Distance downstream/Speed of boat Downstream

95

= min.

11. A car covers a distance PQ in 32 minutes. If the distance between P and Q is 54 km., find

the average speed of the car.

Solution:

Average speed of the car =

km/hr

12. A train traveling at y km/hr arrives at its destination 1 hour late after describing a

distance of120 km. What should have been its speed in order that it arrives on time?

Solution:

Normal time + 1 hour =

Normal time =

Normal speed = = km/hr.

13. A boy rides his motorcycle 45 km at an average rate of 25 km/hr and 30 km at an

average speed of 20 km/hr. What is the average speed during the entire trip of 75 km?

Solution:

Total distance traveled = 45 + 30 = 75 km

Total time taken = Time for the first 45 km + Time for the next 30 km

= hours.

Average speed for the entire journey,

= km/hr.

96

14. A motorist travels from P to Q in the rate of 30 km/hr and returns from Q to P at the rate

of 45 km/hr. If the distance PQ = 120 km, find the average speed for the entire trip.

Solution:

Time from P to Q =

Time from Q to P =

Average speed for the entire journey,

= km/hr

15. A wheel rotates 15 times each minute. How many degrees will it rotate in 12 seconds of

time?

Solution:

In 12 seconds or of a minute, the wheel rotates or 3 times. Angle through which it

turns = 3 360 = 1080

16. The distance between the cities M and N is 275 km along the rail-route and 185 km,

along the aerial route. How many hours shorter is the trip by plane traveling at 250

kmph than by train traveling at 80 kmph?

Time taken along the rail-route = hours ________ (1)

Time taken to travel by air = hours _________ (2)

(1) – (2) gives:

= 2 hours 41 min. 51 sec.

17. An elevator in a 12-storey building travels at the rate of 1 floor every 15 seconds. At the

ground floor and the top floor, the lift stops for 50 sec. How many round trips will the

elevator make during a 5 ½ hour period?

97

Solution:

Time taken for the elevator to make 1 round trip starting from the ground,

= 50 sec + 50 sec + 12 15 sec + 12 15 sec

= 100 + 180 + 180 = 460 sec

Total number of round trips,

=

= approx.

18. A can run a km in 3 min and 54 sec and B can run the same distance in 4 min and 20

sec. By what distance can A beat B?

A beats B by 26 sec.

Distance covered by B is 260 sec = 1000 m

Distance covered by B is 26 sec = m A beats B by 100 m

19. A and B run a km and A wins by 1 min. A and C run a km and A wins by 375 m. B and C

run a km and B wins by 30 sec. Find the time taken by B to run a km.

Solution:

A beats B by 60 sec.

B beats C by 30 sec.

Hence A beats C by 90 sec. But we are given that A beats C by 375 m, i.e., C covers 375

m in 90 sec.

Hence, time taken by C to cover 1 km = sec

Time taken by B to cover 1 km = 240 – 30 = 210 secs.

20. In a race A has a start of 100mtrs and sets off 6 min before B, at the rate of 10 kmph.

How soon will B overtake A, if his speed of running is 12 kmph.

Speed of A =

Distance run by A is 6 min = m.

98

A has a start of 1000 + 100 = 1100 m. In order to overtake A, B should gain 1100 m. But

B gains 2000 m in 60 min.

The time taken by B to gain 1100 m = min.

Practice Exercise 11. A & B are two stations, 660 km apart. A train starts from A at 8 am and travels towards B

at 120 kmph. Another train starts from B at 9 am and travels towards A at 150 kmph. At

what time do they meet?

1. 11 am 2. 12 pm 3. 1 pm 4. 2 pm

2. Running at of its usual speed, a train is 20 minutes late. Find its usual time to cover

the journey.

1. 1000 min 2. 100 min 3. 10 min 4. 1 min

3. A car can finish a certain journey in 20 hours at a speed of 96 kmph. In order to cover

the same distance in 16 hours, the speed of the car must be increased by what speed?

1. 6 kmph 2. 12 kmph 3. 24 kmph 4. 18 kmph

4. A train covers a certain distance in 25 min, if it runs at a speed of 24 kmph on an

average. What will be the speed at which the train must run to reduce the time journey

to 20 min?

1. 1 kmph 2. 10 kmph 3. 21 kmph 4. 30 kmph

5. Walking at of his usual speed, a man is late by 5 hours. What would have been his

usual time?

1. 10 hrs 2. 5 hrs 3. 15 hrs 4. 25 hrs

99

6. If a boy walks from his house to school at the rate of 16 kmph, he reaches the school 40

minutes earlier than the scheduled time. However, if he walks at the rate of 12 kmph, he

reaches 40 minutes late. What is the distance between his school and house?

1. 74 kms 2. 64 kms 3. 54 kms 4. 44 kms

7. A car travels a distance 1430 km at a uniform speed. If the speed of the car is 10 kmph

more, it takes 4 hours less to cover the same distance. What is the original speed of the

car?

1. 55 kmph 2. 44 kmph 3. 33 kmph 4. 22 kmph

8. Two trains start from stations A & B at the same time and travel towards each other at

speeds of 60 kmph and 50 kmph respectively. At the time of their meeting the second

train has traveled 240 km less than the first. Find the distance between A & B.

1. 2460 km 2. 2640 km 3. 2600 km 4. 2400 km

9. A walks at 8 kmph and 8 hours after his start, B cycles after him at 20 kmph. How far

from the start does B catch up with A?

1. 320 km 2. 3 km 3. km 4. km

10. In covering a certain distance, the speeds of A & B is in the ratio of 3 : 4. A takes 60 min

more than B to reach the destination. What is the time taken by A to reach the

destination?

1. 1 hr 2. 2 hrs 3. 3 hrs 4. 4 hrs

11. A train starts from station X at the rate of 240 kmph and reaches station Y in 180

minutes. If the speed is reduced by 24 kmph, how much more time will the train take to

return from station Y to station X?

1. 5 minutes 2. 10 minutes 3. 20 minutes 4. 30 minutes

12. From point A, Ashok drives 300 km to point B, in 4 hours and returns to point A in 6

hours. Then, what is the difference in the average speed and the speed with which he

traveled the distance from A to B?

1. 10 kmph 2. 15 kmph 3. 20 kmph 4. 25 kmph

13. A bullock cart has to cover a distance of 40 km in 5 hrs. If it covers half of the journey in

th of the time, then what should be its speed to cover the remaining distance in the

time left?

1. 10 kmph 2. 15 kmph 3. 20 kmph 4. 25 kmph

100

14. In covering a certain distance, the speeds of A & B are in the ratio of 3 : 4. If A takes 30

minutes more than B to reach the destination, what is the time taken by A to reach the

destination?

1. 1 hours 2. 2 hours 3. 3 hours 4. 4 hours

15. A is twice as fast as B & B is thrice as fast as C. In how much time would a particular

distance, that is covered by C in 48 min, be covered by B?

1. 4 min 2. 8 min 3. 16 min 4. 12 min

Practice Exercise 21. Anita covers two-third of a certain distance at 8 kmph and the remaining at 10 kmph.

What distance will be covered by him, if he takes 30 minutes in all?

1. 4.28 2. 5.28 3. 6.28 4. 7.28

2. Sudarshan traveled 2400 km by air which formed of his trip. One-third of the whole

trip, he traveled by car and the rest of the journey he performed by train. Find the

distance, in km, traveled by him by train.

1. 1500 2. 1600 3. 1700 4. 1800

3. Soumya covers 200 km in 3 hours and 15 min, Radha covers 300 km in 4 hours, and

Sindu covers 150 km in 2 hours and 25 min. Which two are faster than the third one?

1. Soumya & Radha 2. Sindu & Soumya 3. Radha & Sindu 4. Can’t determine

4. A man takes 6 hours in walking to a certain place and riding back. He would have

gained 2 hours by riding both the ways. What time would he take to walk both ways?

1. 5 2. 6 3. 7 4. 8

5. The ratio between the rates of walking of A & B is 4 : 5. If the time taken by B to

cover a distance is 40 min, what time will A take to cover the same distance?

1. 10 2. 20 3. 50 4. 60

6. A man is standing on a railway bridge which is 200 m long. He finds that a train

crosses the bridge in 20 seconds but himself in 10 seconds. Find the length of the

train and its speed.

1. 52 2. 72 3. 62 4. 32

101

7. A train 300 m long is running with a speed of 60 kmph. In what time will it pass a

man who is running at 10 kmph in the same direction in which the train is going?

1. 18 sec 2. 12 sec 3. 16 sec 4. 20 sec

8. A train running at 60 kmph takes 30 seconds to pass a platform. Next it takes 18

seconds to pass a man walking at 5 kmph in the opposite direction in which the train

is going. Find the length of the train and the length of the platform.

1. 150 m 2. 175 m 3. 200 m 4. 225 m

9. A train crosses a platform 200 m long in 120 seconds at a speed of 90 kmph. How

much time does the train take to cover an electric pole, of negligible width?

1. 28 sec 2. 26 sec 3. 82 sec 4. 62 sec

10. The length of a running train A is 20% less than the length of another train B running

in the opposite direction. If the length of train B is 240 meters, find the speed of the

train B, if the speed of train A is 100 kmph & they take 11 seconds to cross each

other.

1. 11 kmph 2. 22 kmph 3. 33 kmph 4. 44 kmph

11. A man sees a train passing over a bridge 2 km long. The length of the train is half

that of the bridge. If the train clears the bridge in 4 min, what would have been the

speed of the train?

1. 15 kmph 2. 30 kmph 3. 45 kmph 4. 60 kmph

12. A man can row 24 kmph in still water. If it takes him twice as long to row up as to row

down the river, find the rate of the stream?

1. 6 kmph 2. 8 kmph 3. 4 kmph 4. 10 kmph

13. In a stream running at 5 kmph, a motor boat goes 12 km upstream and back again to

the starting point in 60 minutes. Find the speed of the motor boat in still water.

1. 25 kmph 2. 50 kmph 3. 75 kmph 4. 100 kmph

14. A man can row 40 km upstream and 20 km downstream in 30 hours. Also he can row

30 km upstream & 45 km downstream in 30 hours. Find the ratio speed of the man in

still water and the speed of the current?

1. 2 : 9 2. 7 : 5 3. 5 : 7 4. 9 : 2

15. A boat covers 24 km upstream & 36 km downstream in 6 hours while it covers 36 km

upstream & 24 km downstream in 6 hours. Find the velocity of the current.

1. 1 kmph 2. 3 kmph 3. 2 kmph 4. 8 kmph

Answer Keys for Practice Exercise 1

102

1 1 6 2 11 32 2 7 1 12 23 3 8 2 13 14 4 9 3 14 25 3 10 4 15 3

Answer Keys for Practice Exercise 2

1 1 6 2 11 32 2 7 1 12 23 3 8 2 13 14 4 9 3 14 25 3 10 4 15 3

Chapter-8

Time and WorkWork

Work is of many kinds. It may be traveling from one place to another, constructing a bridge or wall,

eating the food, sitting at a place for some time, etc….

But what ever may be the work it is always considered as one unit.

Basic assumptions made in the study of time and work

Following are the assumptions, which are made in the study of time and work. These assumptions may

not be specifically mentioned in every problem but they are taken for granted and we proceed unless

and until it’s specifically mentioned.

Assumption 1.

If a one person completes the work in certain number of days then we assume that he does

the same amount of work everyday, i.e. work get distributed uniformly.

For example, let us say a person completes the work in 10 days. Then per day he does th of the

work.

In general a person, who completes the work in n number of days, will be doing th of the work per

day.

Assumption 2.

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If two persons or more than two people complete the work in certain number of days then

we assume that both of them do the work uniformly.

For example, if two people complete the work in ten days then they do the same amount of work

everyday, i.e. they do th of the work per day.

This in turn means that if they are doing th of the work per day they will finish of the work in 10

days.

Concept of Unitary Method

The method of converting the total time taken by a person to do the work, to per-day-work is known as

the unitary method.

Variables in the study of Time and Work

In the study of time and work, we consider Time, Work and Man {or Woman} as the variables, as the

concept of time and work is dependent entirely on these three entities.

Relations between the VariablesRelation 1:

Men are directly proportional to work when the time is kept constant.

Explanation:

Let the work be construction of road of length 2 kilometers.

There are 5 worker involved in completing the work taking 10 days.

But if the work increases i.e. instead of 2 kilometers it becomes 4 kilometers and the time limit is

given as same 10 days then we have to increase the number of workers so as to complete the work

in given time limit or vice versa.

In general as work increases the number of men required to complete the given work in the same time

limit, also increases and vice versa.

Relation 2:

Time is directly proportional to work when the number of men is kept constant.

Explanation:

Let the work be construction of a house.

There are 5 worker involved in completing the work taking 10 days.

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But if the work increases i.e. instead of 1 house construction it becomes construction of 2 houses with

the limitation that it has to be done with the same 5 workers then we have to increase the number of

days so as to complete the work with given number of people {which is 5 in this case} or vice versa.

In general as work increases the number of men required to complete the given work in the same time

limit, also increases and vice versa.

Relation 3:

Time and men are inversely proportional.

Explanation:

Consider the same example of constructing a road by certain number of people in certain number of

days. In this example we are keeping the work as constant. Let us say construction of a road of length

2km. If there are more number of men are involved in the work the gets finished in less number of

days on the other hand if there are less number of people who are doing the work then it takes a long

time for them to complete the same work.

t

In general more the number of men less will be the time taken to complete the work and less the

number of men more will be the people required to complete the same work

Concept of Man days

It is the total time taken multiplied by the total number of people to complete the work.

If 10 people take 30 days to complete the given task then the time taken by them is given as 10 30.

In general if ‘n’ number of persons do the work in ‘m’ number of days then the total time taken is

given as m*n.

In the study of time and work we encounter the problems on regular types which are based on the

basic concepts, people working on alternate days, problems on efficiency, two people out of three

people working, etc.

105

Worked Examples1. A can do a piece of work in 20 days which B alone can do in 24 days. In how many days will they

finish the work, both working together?

Solution:

A’s 1 day’s work = B’s 1 day’s work =

(A + B)’s 1 day work =

Both will finish the work in days.

2. Two persons A & B working together can dig a trench in 24 hours while A alone can dig it in 36

hours. In how many hours can B alone dig such a trench?

Solution:

(A + B)’s 1 hour’s work =

A’s 1 hour work =

B’s 1 hour’s work =

Hence B alone can dig the trench in 72 hours.

3. A & B can do a piece of work in 48 days; B & C can do it in 60 days; A & C can do it in 80 days. In

how many days will A, B & C finish it, all working together?

Solution:

(A + B)’s 1 day’s work =

(B + C)’s 1 day’s work = and

(A + C)’s 1 day’s work =

Adding, we get 2(A + B + C)’s 1 days work =

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(A + B + C)’s 1 day work =

Thus A, B & C together can finish the work in 40 days.

4. A can do a piece of work in 30 days and B alone can do it in 20 days. B works at it for 10 days and

then leaves. In how many days will A alone finish the remaining work?

Solution:

B’s 10 days work =

Remaining work =

A can do 1

2 the work in 15 days.

5. A is four times as good a workman as B and together they finish a piece of work in 28 days. Find

the number of days A alone takes to finish the work?

Solution:

(A’s 1 day’s work) : (B’s 1 day’s work) = 4 : 1

(A + B)’s 1 day work =

Divide in the ratio of 4 : 1

A’s 1 day’s work =

Hence, A alone can finish the work in 35 days.

6. A & B can do a piece of work in 20 days. B & C can do it in 28 days. A & C can do it in 16 days.

Who among these will take the least time if put to do it alone?

Solution:

(A + B)’s 1 day’s work =

(B + C)’s 1 day’s work =

& (A + C)’s 1 day’s work =

2(A + B + C)’s 1 day’s work =

(A + B + C)’s 1 day’s work =

107

A’s 1 day’s work =

B’s 1 day’s work =

C’s 1 day’s work =

Thus time taken by A, B, C is respectively.

Clearly time taken by A is the least.

7. If 20 men or 36 boys can do a piece of work in 15 days, then 50 men and 30 boys together will do

twice the work in how many days?

Solution:

20 men 36 boys

1 man = boys

50 men + 30 boys = boys = 120 boys

Now, more work more days

More boys, less days

36 boys can do one work in 15 days

then 120 boys will do twice the work in days = 9 days.

8. Two pipes can fill a cistern in 7 hours & 8 hours respectively. The pipes are opened simultaneously

and it is found that due to leakage in the bottom 16 min extra are taken for the cistern to be filled

up. In what time will the leak empty it?

Solution:

Work done by the two pipes in 1 hour =

Time taken by these pipes to fill the tank = hours

Due to leakage, time taken = hours = = 4 hours.

Work done by (two pipes + leak) in hour =

Work done by the leak in 1 hour =

Work done by the leak in 1 hour = hours

108

The leak will empty the full cistern in 56 hours.

9. Two pipes A & B can fill a tank in 12 min & 16 min respectively. If both the pipes are open,

simultaneously, after how much time B should be closed so that the tank is full in 9 minutes?

Solution:

Let B be closed after x minutes. Then part filled by (A + B) in x minute + part filled by A in (9-x)

min =1

x =1

or x

  or

= 1

3x + 36 = 48

3x = 12 x = 4

Hence, B must be closed after 4 minutes.

10. Two pipes A & B can fill a tank in 72 minutes & 90 minutes respectively. A water pipe C can empty

the tank in 60 min. First A & B are opened. After 14 minutes, C is opened. In how much time, the

tank is full?

Solution:

Let it take x minutes after the first 14 minutes for the tank to fill, then

x = 78 minutes

Total time taken to fill the tank = (78 + 14) minutes = 92 minutes

= 1 hour 32 minutes.

Alternative method of Solving Problems on TIME AND WORK.

Example:

Sangam and Krishna are two friends. They work together and finish off the work in twelve days. If

Sangam has to work alone, he takes 16 days to finish off the work. They work together for 4 days and

Sangam quits. In how many more days Krishna completes the work?

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Solution:

Step I: Write down the data given.

Find out the size of the work.

This is nothing but the LCM of given two numbers.

In this case, it is 48 (LCM of 12 and 16) 48 units of work.

This is the amount of work which must be completed irrespective of people working in a group or

together or alone.

So let us re-write the equation.

If together they are completing 48 units of work in 12 days, then per day they together do 4 units of

work.

Similarly, Sangam is completing the work in 16 days. So, per day, he will be doing 3 units of work.

That means units of work done by Krishna will be

= [(Sangam+Krishna)work units/day] – [Sangam work units/day]

= 4 – 3 = 1 unit.

So Krishna will be doing 1 unit of work per day, which means, in order to complete 48 units of work, he

will take 48 days.

Step III:

According to problem, we know that together they have worked for 4 days, together they can do 4

units of work. So in 4 days, they will finish off 4 x 4 = 16 units of work.

Step IV:

But we know that total amount of work which needs to be completed is 48 units. But already 16 units

of work got over. So amount of work remaining is 48 – 16 = 32 units.

Step V:

Krishna is doing 1 unit of work per day. So in order to complete remaining 32 units of work, he takes 1

x 32 = 32 days

(Sangam+Krishna) Sangam Krishna

12 16 ?

(Sangam+Krishna) Sangam Krishna

12 16 32

110

Practice Exercise 1

1. A can do of the work in 10 days & B can do of the work in 20 days. In how many days can

both A & B together do the work?

1. 2. 3. 4.

2. A can do a piece of work in 40 days. He works at it for 5 days and B alone finishes the remaining

work in 21 days. How many days will the two together take to complete the work?

1. 12 2. 13 3. 14 4. 15

3. A is four times as good a workman as B and together they finish a piece of work in 28 days. Find

the number of days A alone takes to finish the work?

1. 20 2. 15 3. 35 4. 25

4. A can do a piece of work in 28 days which B can do in 42 days. They begin together but 6 days

before the completion of the work, A leaves off. Find the total number of days to complete the

work.

1. 2. 3. 4.

5. 24 men can complete a work in 36 days. Twelve days after they started working, 8 men joined

them. How many days will all of them take to finish the remaining work?

1. 18 2. 20 3. 22 4. 24

6. 8 men & 12 women finish a job in 16 days while 6 men and 14 women finish in 20 days. In how

many days will 20 women working together finish it?

1. 60 2. 70 3. 80 4. 90

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7. 12 children take 16 days to complete a work which can be completed by 8 adults in 12 days. 16

adults started working and after 3 days 10 adults left and 4 children joined them. How many days

will it take them to complete the remaining work?

1. 6 2. 7 3. 8 4. 9

8. Ram can do a piece of work in 16 days which Shyam can finish in 24 days. If they work at it on

alternate days with Ram beginning, in how many days, the work will be finished?

1. 17 2. 18 3. 19 4. 20

9. A can do a piece of work in 24 days & B alone can do it in 32 days. A & B undertook to do it for Rs.

2560. With the help of C, they finished in 12 days. How much is paid to C?

1. 200 2. 320 3. 480 4. 209

10. A does half as much work as B & C does half as much work as A & B together. If C alone can finish

the work in 80 days, in how many days will they finish the work together?

1. 2. 3. 4.

11. A can do a piece of work in 20 days while B can do it in 30 days. A starts the work and works for 10

days when B joins him and both finish the work. If they together receive Rs. 480 for the job, find

the share of A?

1. Rs.400 2. Rs.384 3. Rs.340 4. Rs.320

12. A, B and C together can complete a piece of work in 20 days. A and B together complete half the

work in 12 days and the rest is finished by B and C in 15 days. How long will it take for B alone to

complete the work?

1. 40 days 2. 25 days 3. 36 days 4. 48 days

13. Two pipes can fill a cistern in 14 and 16 hours respectively. The pipes are opened simultaneously

and it is found that due to leakage in the bottom 32 minutes extra are taken for the cistern to be

filled up. When the cistern is full, in what time will the leak empty it?

1. 114 hrs 2. 112 hrs 3. 100 hrs 4. 80 hrs

14. A tank has a leak which would empty it when full in 8 hrs. A tap is turned on which admits 6 litres a

minute into the tank and it now requires 12 hrs. to be emptied. How many litres does the tank

hold?

1. 8640 litres 2. 5760 litres 3. 7200 litres 4. 6480 litres

15. If 15 women or 10 men can complete a project in 55 days, in how many days will 5 women and 4

men working together complete it?

1. 75 2. 95 3. 55 4. 85

112

Practice Exercise 21. Ram and Shyam are constructing roads, working individually. Ram alone can do it in 50 days

where as Shyam is two times efficient than Ram. If they work together, how long they take to

complete the work?

1. 10 hrs 2. 12 hrs 3. 15 hrs 4. 16 hrs

2. There are two pipes X and Y involved in filling a tank. There is also a drain Z that is emptying the

tank. X and Y alone can fill the tank in 20 and 30 hours respectively, while Z empties the tank in 40

hours. After how long from starting, the tank will be filled?

1. 17 hrs 2. 15 hrs 3. 15 hrs 4. 18 hrs

3. There is a pipe filling the tank in 8 hours. Then there is a drain created because of which, now the

tank is filled in 12 hours. The flow rate is 4 ltr/min. Find the size of the tank.

1. 3457 litres 2. 5760 litres 3. 3567 litres 4. 6546 litres

4. A man can paint a room in 6 hours. His son can paint the same room I in 8 hours alone. How

long will it take for the man and his son to paint the room if they work together?

1. hrs 2. hrs 3. hrs 4. hrs

5. Sangam and Krishna are two friends. They work together and finish off the work in twelve days. If

Sangam has to work alone, he takes 16 days to finish off the work. They work together for 4 days

and Sangam quits. In how many more days Krishna completes the work?

1. 40 days 2. 45 days 3. 48 days 4. 52 days

6. If 10 men can do a piece of work in 20 days, how long will 8 men take to do the job if they work at

the same rate?

1. 30 days 2. 35 days 3. 40 days 4. 45 days

7. If 20 cows eat 380 bales of hay in 15 days, how man bales of hay will 15 cows eat in 10 days?

113

1. 170 2. 180 3. 190 4. 200

8. A garrison of 60 men has food for 28 days. Eight days later, reinforcement arrives reducing the

number of days the food would last to 15. What was the strength of the reinforcement?

1. 20 2. 30 3. 40 4. 50

9. A is thrice as good a workman as B and is therefore able to finish a piece of work 60 days earlier

than B. Find the time in which they can finish the job by working together?

1. 33 days 2. 35.5 days 3. 40 days 4. 22.5 days

10. Two pipes can fill a cistern in 7 hours & 8 hours respectively. The pipes are opened simultaneously

and it is found that due to leakage in the bottom 16 min extra are taken for the cistern to be filled

up. In what time will the leak empty it?

1. 56 hrs 2. 58 hrs 3. 62 hrs 4. 70 hrs

11. Two pipes A & B can fill a tank in 12 min & 16 min respectively. If both the pipes are open,

simultaneously, after how much time B should be closed so that the tank is full in 9 minutes?

1. 4 mins 2. 6 mins 3. 8 mins 4. 10 mins

12. Two pipes A & B can fill a tank in 72 minutes & 90 minutes respectively. A water pipe C can empty

the tank in 60 min. First A & B are opened. After 14 minutes, C is opened. In how much time, the

tank is full?

1. 45 mins 2. 56 mins 3. 92 mins 4. 108 mins

13. Pipe A can fill a tank in 6 hours and pipe B can fill it in 4 hours. If they are opened at alternate

hours and if pipe A is opened first, in how many hours will the tank be full?

1. 5 hrs 2. 8 hrs 3. 6 hrs 4. 7 hrs

14. A thin candle 8 inches long, which burns at the rate of 3 inches in 2 hours and a thick candle 5.5

inches long which burns at the rate of 2 inches in 3 hours are lighted at the same time. After how

long will (i) they be the same length? (ii) the thinner candle 1 inch longer than the thicker?

1. 1.5 hrs 2. 1.8 hrs 3. 2.6 hrs 4. 4hrs

15. There are two people Sita and Gita capable of finishing the work on 20 days and 30 days,

respectively. If they work alternatively, how long do they take to complete the entire work?

1. 24 days 2. 35 days 3. 40 days 4. 28 days

Answer Keys for Practice Exercise 1

1 1 6 3 11 22 4 7 1 12 13 3 8 3 13 24 2 9 2 14 15 1 10 1 15 1

Answer Keys for Practice Exercise 2

114

1 1 6 4 11 12 1 7 3 12 33 2 8 1 13 14 1 9 4 14 25 3 10 1 15 1

CHAPTER-9

Theory of IndicesIndices

If m is a positive integer then a a a ......... m times is symbolically written as am here ‘a’ is called

the base and ‘m’ is called the power, index or exponent.

The above statement leads us to questions like, what is a0 ? What does a4 mean, how do we

understand a

34 etc. The basic laws of indices help us to understand these numbers better.

Laws of Indices:1. am. an = am+n

2. a

a

m

n = amn if m > n and

a

a a

m

n n m

1

if m < n

3. am n = amn

4. a0 = 1 whenever a 0

5. am = 1

am

6.

7. ab a bm m m ; a

ba

b

m m

m

Rules used in solving problems on Indices:

Rule 1:

115

When the bases of two equal numbers are equal then their powers also will be equal.

(If the bases are non- zero and non-unity).

For ex: If 3n = 34, then it means, n = 4.

Rule 2:

When the powers of two equal numbers are equal (and not equal to zero), two cases arise:

a) If the powers are odd numbers, then the bases are equal

For ex: If x5 = 25, then x = 2.

b) If the powers are even numbers, then the bases are numerically equal but can have different

signs.

For ex: If x4 = 24, then x =+2 or –2

Types of Questions on IndicesThe questions associated with indices are normally of three types:

1. Simplification:

In this type the problems will contain different bases and powers which have to be simplified using

the Rules / Laws discussed above.

Example: Simplify

Solution: =

=

=

=

2. Solving for the value of an unknown:

In this type, the problem will have an equation where an unknown (like x or y) will appear in the

power and using Rule 1 and Rule 2 discussed above, values of unknown are to be determined.

116

Example: Simplify

Solution: =

=

=

=

3. Comparison of Numbers:

In this type two or more quantities will be given each being a number raised to a certain power.

These numbers have to be compared in magnitude either to find the largest or smallest of the

quantities or to arrange the given quantities in ascending or descending order.

Example: Solve for y,

Solution:

Worked Examples1. A certain number of persons agree to subscribe as many rupees each as there are subscribers.

The whole subscription is Rs.2582449. Find the number of subscribers?

Solution:

Let the number of subscribers be x, since each subscriber agrees to subscribe x rupees.

The total subscription = no. of persons subscription per person

= x x = x2

117

given x2 = 2582449

x = 1607

2. Simplify:

Solution:

Use the relations

3. Simplify

Solution:

4. Find the number whose square is equal to the difference between the squares of 75.12 and 60.12.

Solution:

Let ‘x’ be the number required

x2 = (75.12)2 – (60.12)2

= (75.12 + 60.12) (75.12 – 60.12)

= 135.24 15 = 2028.60

45.0399

5. Reduce to an equivalent fraction with a rational denominator.

Solution:

118

6. Find the value of

Solution:

approx.

7. An army general trying to draw his 16,160 men in rows such that there are as many men as there

are rows, found that he had 31 men over. Find the number of men in the front row.

Solution:

Let ‘a’ be the number of men in the front row.

a2 + 31 = 161610

No. of men in the front row = 127

a2 = 161610 – 31 = 16129

a = 127

8. A man plants his orchid with 5625 trees and arranges them so that there are as many rows as

there are trees in a row. How many rows are there?

Solution:

Let ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say

x2 = 5625

x = 75

There are 75 trees in a row and 75 rows are arranged.

9. Evaluate

Solution:

The given expression is

= =

119

= = x0 = 1.

10. If P = 2x, q = 2y and = 4, find xyz.

Solution:

= = = (given).

2xyz = 2 or xyz = 1.

Practice Exercise 1

1. Find

120

1. 2. 3. 4.

2. is equal to:

1. 2. 3. 4. 1

3. Solve for x and y: ,

1. 3, 1 2. 2,3 3. 2,1 4. 5,3

4. , the value of a is:

1. 4 2. 5 3. 6 4. 8

5. Solve the equations: and .

1. , 2. , 3. , 4. ,

6. If =64, then the value of n is:

1. 2 2. 4 3. 6 4. 12

7. is equal to:

1. 2. 3. 4.

8. If then n equals:

1. 0 2. 2 3. 3 4. 4

9. If then is equal to:

1.0 2.1 3. 4.

121

10. The simplified form of

1. 2. 3. 4. xy

Practice Exercise 2

1. Solve if .

1.0 2.1 3.2 4.x

2. is equal to:

122

1. 0 2. 1 3. 4.

3. If x, y, z are real numbers, then the value of is:

1. xyz 2. 3. 4. 1

4. Find the value of

1. 2. 1 3. 0 4. 3

5. Find . .

1. 2. 3. 4. 1

6. If and , the values of a and b respectively are:

1. 2,3 2. -2,3 3. 2,-3 4. 3,2

7. Find

1. 0 2. 3. 4. 1

8. If , then the value of x is:

1.1 2. 3.2 4.

9. If , then the value of x is:

1. 2. 3. 4.

10. Find the value of x .

1. 0, 2 2. 1, 3 3. 0, 1 4. 2, 1

123

Answer Keys for Practice Exercise 1

1 3 6 42 4 7 13 1 8 34 1 9 45 1 10 4

Answer Keys for Practice Exercise 2

1 2 6 42 2 7 43 4 8 34 2 9 25 4 10 1

CHAPTER-10

SurdsRational numbers are the numbers which can be expressed in the form of where p and q are

integers are called Rational Numbers .

When the root of number is not a rational number, the Root is called a SURD.

124

For example are surds.

Important points to remember1. Order and Radicand of a Surd:

Let ‘a’ be a rational number and be a positive integer such that is irrational, then is called

a surd or a radical of the order n, and a is called the radicand and n is the called the order.

Example: Find the order and radicand of the following: , , ,

1. in this order is 3 and radicand is 4

2. in this order is 2 and radicand is 3

3. in this order is 4 and radicand is 7

4. in this order is 6 and radicand is 8

2. Pure Surds: A pure surd is one which is made up of only the irrational number

, etc.

3. Mixed Surds: A mixed surd comprises of one rational and other irrational

, etc.

4. Similar Surds: When two or more surds are similar they can be reduced so as to have the same

irrational parts.

and are similar surds.

The irrational part is the same.

5. Order of Surds:

If a1/m is a surd then a1/m is said to be a surd of order ‘m’.

6. Similar and Dissimilar Surds:

Surds having the same base and order are called similar surds. , , are similar surds.

Surds which are not similar are dissimilar surds-clearly, , 2 , are dissimilar surds.

7. To convert a Surd of order n into a Surd of the order m:

125

Let be a given surd of order n.

Then,

, i.e.,

8. Comparison of Two Surds:

If two surds are of same order then the one whose radicand is larger is the larger of two.

Examples: (i) , (ii) ,(iii)

Note:

If any two surds are of different orders are to be compared then we reduce them to the same but

smallest order and then compare them as above.

Operations on Surds:1. Addition and Subtraction of Surds

Similar surds can be added or subtracted by adding on subtracting the rational part and prefixing

the sum or difference obtained to their common irrational parts. In order to add or subtract surds

the first step is to reduce them to their lower terms before adding or subtracting their rational

coefficients.

For example, , , , can be added but , and can be neither be added nor

subtracted.

Example: Solve:

Solution:

126

2. Multiplication and Division of Surds

When multiplying or dividing surds, a convenient method used, to convert the surds into quantities

with fractional indices and to then use the rules of indices to multiply or divide

Multiplication and division of surds are governed by the laws of indices.

i.e., am. an = a m+n

= am-n

Examples

1. Multiply: =

2. Multiply:

3. Divide: By

3. Rationalisation of Surds

If the product of two irrational quantities is rational each is called the rationalizing factor of the

other.

1. If on multiplying two surds we get a rational product, then each of the surds is said to be

rationalised.

2. A surd consisting of two terms is known as Binomial Surd. Two Binomial Surds are said to be

conjugate if they differ from each other only in the sign which connects the two pairs.

Example:

is conjugate to .

3. The product of 2 conjugate surds is rational. , which is a rational

number.

4. When the denominator of a fraction is a binomial quadratic surd, it can be rationalised by

multiplying the denominator and numerator by its conjugate surd.

127

4. To find the square roots of the surd +

Example: Express the given surd in the form + i.e. + = +

Solution:

Let = + squaring both sides,

+ = x + y + .

Therefore x + y = a, and xy = c which can be solved for the values of x and y.

Worked Examples

1. Convert into a Surd of the order 6

Solution:

We have,

2. Convert into a Surd of the order 6

Solution:

We have,

Now,

3. Convert into a Surd of the order 4

Solution:

We have,

128

Now,

4. Convert and into Surds of same but smallest order.

Solution:

Given surds are of order 4 and 6 respectively.

LCM of 4 and 6 is 12.

So, we shall convert each of the given surds into a surd of the order 12.

Now,

And,

5. Which surd is larger: or

Solution:

The given surd are of order 2 and 3.

The LCM of 2 and 3 is 6.

So, we reduce each to a surd of the order 6.

Now, =

And, =

Clearly,

Hence,

6. Write the following surds in ascending order: and .

129

Solution:

The given surds are of the order 4, 2 and 3

The LCM of 4, 2,3 is 12

So, we reduce each to a surd of the order 12.

Now,

And,

Clearly,

Hence,

7. Arrange the following surds in descending order: and

Solution:

The given surds are of the order 3, 6 and 9

The LCM of 3, 6, 9 is 18

So, we reduce each to a surd of the order 18.

Clearly,

Hence,

8. Which is greater: or

Solution:

130

The given surds are of order 2 and 3

The LCM of 2 and 3 is 6

So, we reduce each to a surd of the order 6.

Now,

And,

Clearly, ,since

Hence,

9. Rationalise the denominator of .

Solution:

10. Find the square root of, 12 + .

Solution:

12 + =

Let = +

Squaring both sides,

= x + y +

Therefore, x + y = 12, xy = 32. i.e., x = 4, y=8

131

Hence, = + = 2 + .

Practice Exercise 1

1. Convert into a surd of order 10;

1. 2. 3. 4.

2. Convert and into surds of the same but smallest order and find which one is smaller?

1. and the smaller one is .

2. and the smaller one is .

3. and the smaller one is .

4. and the smaller one is .

3. Which one of the two or is greater?

1. 2. 3. equals 4. Can’t determine

4. Arrange in ascending order

132

1. 2. 3. 4.

5. Arrange in descending order

1. 2. 3. 4.

6. Rationalize the denominators of

1. 2. 3. 4.

7. Rationalize the denominator of

1. 2. 3. 4.

8. Simplify:

1. 2. 3. 4.

9. Simplify:

1. 2. 3. 4.

10. If and , find the value of

1. 2. 3. 4.

133

Practice Exercise 2

1. Convert into a surd of order 9;

1. 2. 3. 4.

2. Convert and into surds of the same but smallest order and find which one is larger?

1. and the larger one is .

2. and the larger one is .

3. and the larger one is .

134

4. and the larger one is .

3. Which one of the two or is greater?

1. 2. 3. equals 4. Can’t determine

4. Arrange in ascending order

1. 2. 3. 4.

5. Arrange in descending order

1. 2. 3. 4.

6. Rationalise the denominators of

1. 2. 3. 4.

7. Find the square roots of 6 -

1. 2. 3. 4.

8. Rationalize the denominator of

1. 2. 3. 4.

9. Simplify:

1. 2. 3. 4.

10. Simplify:

1. 2. 3. 4.

135

Answer Keys for Practice Exercise 1

1 3 6 42 2 7 13 1 8 14 4 9 45 1 10 2

Answer Keys for Practice Exercise 2

1 4 6 42 3 7 13 1 8 44 1 9 15 2 10 4

CHAPTER-11

Linear EquationsSimple Equations

A simple equation is an equation in a single variable, whose value must be determined.

a. ax + b = 0 is called a Simple Equation in one Unknown.

b. ax + by + c = 0 is the general form of a Linear Equation in Two Variables.

ax + by + c = 0 is a Single Linear Equation in two variables which admits infinite number of

solutions.

a1x + b1y + c1 = 0 ……. (1)

136

a2x + b2y + c2 = 0 ……. (2)

For the above equations (1) and (2) in two variables x and y, the solution is and

(i). The above system of equations has a unique solution if . Such a system is a

consistent system. The graph consists of two intersecting lines.

(ii). If , then there is no solution to the system. It is an inconsistent system. The

graph would consist of two parallel lines.

(iii). If , the system has infinite solutions. The graph would consist of two

coincident lines.

Linear EquationsGeneral linear equation in two variables, ax + by + c = 0 where a, b and c are constants always

represents a straight line in a plane.

If b = 0 then the equation takes the form, ax + c = 0. Clearly, x = ca

satisfies this equation and

hence is a root.

Simultaneous Linear EquationsIf two or more equations are satisfied by the same values of variables (two or more

variables) the equations are said to be simultaneous equations.

For example, x + y = 2 and 3x + y = 4 are both satisfied for x = y = 1 and hence are simultaneous

equations.

To solve simultaneous equations in two variables, we eliminate one of the unknown quantities, find

the value of the other unknown and substitute this value in any of the given equations which

results in the solution.

Example

Solve 2x + y = 16 .... (1) x y = 2 ...... (2)

From (2), x = y + 2, substitute in (1), 2(y + 2) + y = 16

2y + 4 + y = 16 or 3y = 12 y = 4

Since x = y + 2 and y = 4 we have, x = 6

137

Hence x = 6, y = 4.

Note :

Geometrically the solution of simultaneous equations represents the point of intersection of the

two lines represented by the two given equations.

Solving Linear Simultaneous Equations

Consider the equations x + y = 8 and x y = 6.

Both are simple simultaneous equations in two unknowns x and y and both the equations are

satisfied by the same values, x = 7 and y = 1 which are known as the solutions (or roots) of the

equation.

When a pair of simple simultaneous equations in two unknowns, say x and y, is given, we first

eliminate any one of the two unknowns, say x, and find the value of other unknown y. Then by

substituting the value of y in any of the equations the value of x is found.

Using the Rule of Cross Multiplication

The solutions of linear simultaneous equations

and are given by:

xb c b c

yc a c a a b a b1 2 2 1 1 2 2 1 1 2 2 2

1

where a1b2 - a2b1 0

x = b c b c

a b a by

c a c a

a b a b1 2 2 1

1 2 2 1

1 2 2 1

1 2 2 2

,

Example: Solve 3x 7y + 10 = 0 and 2x y + 3 = 0

We have, 3x 7y + 10 = 0 ......... (i)

and 2x y + 3 = 0 ......... (ii)

from (i) and (ii) by method of cross multiplication, we get

xb c b c

yc a c a a b a b1 2 2 1 1 2 2 1 1 2 2 1

1

orx y

21 10 20 91

3 14

Hence the required solutions are x = 1 and y = 1.

138

Linear Systems with Two Variables

A linear system of two equations with two variables is any system that can be written

in the form.

ax + by = p

cx + dy = q

where any of the constants can be zero with the exception that each equation must

have at least one variable in it.

Also, the system is called linear if the variables are only to the first power, are only in the

numerator and there are no products of variables in any of the equations.

Here is an example of a system with numbers.

3x – y = 7

2x + 3y = 1

Before we discuss how to solve systems we should first talk about just what a solution to a system

of equation is. A solution to a system of equations is a value of x and a value of y that, when

substituted into the equations, satisfies both equations at the same time.

For example, above, x=2 and y=-1 is a solution to the system. This is easy enough to check.

3(2)-(-1)=7 and, 2(2)+3(-1)=1

So, sure enough that pair of numbers is a solution to the system. Do not worry about how we got

these values. This will be very first system that we solve when we get into examples.

Note that it is important that the pair of numbers satisfy both equations. For instance x=1 and y=-

4 will satisfy the first equation, but not the second and so isn’t a solution to the system. Likewise,

x=1 and y=1 will satisfy the second equation but not the first and so can’t be a solution to the

system.

Now, just what does a solution to a system of two equations represent? Well if you think about it

both of the equations in the system are lines. So, let’s graph them and see what we get.

As you can see the solution to the system is the coordinates of the point where the two lines

intersect. So, when solving linear systems with two variables we are really asking where the two

lines will intersect.

We will be looking at two methods for solving systems in this section.

The first method is called the method of substitution. In this method we will solve one of the

equations for one of the variables and substitute this into the other equation. This will yield one

equation with one variable that we can solve. Once this is solved we substitute this value back into

one of the equations to find the value of the remaining variable.

Example:

Solve the equations: 7x – 5y = 30, 12y + 4x = 136.

Solution:

139

7x – 5y = 30 ………… (1)

12y + 4x = 136 …… (2)

From equation (2): y =

Substituting y in equation (1), we get,

7x – 5 = 30

21x – 170 + 5x = 90

26x = 260

x = 10

Substituting x in equation (1), we get,

7 10 – 5y = 30 5y = 40 y = 8

Hence, x = 10 and y = 8.

As with single equations we could always go back and check this solution by plugging it into both

equations and making sure that it does satisfy both equations. Note as well that we really would

need to plug into both equations. It is quite possible that a mistake could result in a pair of

numbers that would satisfy one of the equations but not the other one.

Let’s now move into the next method for solving systems of equations. The method of substitution

will often force us to deal with fractions, which adds to the likelihood of mistakes. The next

method will not have this problem. Well, that’s not completely true. If fractions are going to show

up they will only show up in the final step and they will only show up if the solution contains

fractions.

This second method is called the method of elimination. In this method we multiply one or both

of the equations by appropriate numbers (i.e., multiply every term in the equation by the number)

so that one of the variables will have the same coefficient with opposite signs. Then next step is to

add the two equations together. As one of the variables had the same coefficient with opposite

signs will eliminated when we add the two equations. The result will be a single equation that we

can solve for one of the variables. Once this is done substitute this answer back into one of the

original equations.

Example:

Solve the equations: 7x – 5y = 30, 12y + 4x = 136.

Solution:

7x – 5y = 30 ………… (1)

12y + 4x = 136 …… (2)

(1) 12 + (2) 5 104x = 1040 x = 10

Substituting x in equation (1), we get

140

7 10 – 5y = 30 5y = 40 y = 8

Hence, x = 10 and y = 8.

.

Worked Examples

1. Solve

Solution:

2. Find a number such that the difference between nine times the number and four times the number

is 55.

Solution:

Let the number be ‘x’ say.

9x – 4x = 55

5x = 55

x = 11

3. A father is now three times as old as his son. Five years ago. he was four times as old as his son.

Find their present ages.

Solution:

Let the present age of the son be ‘x’ years, say.

Father’s present age = 3x years

141

Five years ago, age of the son = (x – 5) years

Fathers age = (3x – 5) years

Given that, 5 years ago. Father’s age = 4 son’s age

3x – 5 = 4(x – 5) = 4x – 20

x = 15

Son’s present age = 15 years

Father’s present age = 45 years

4. If ‘A’ gives B Rs.4. B will have twice as much as A. If B gives A Rs.15, A will have 10 times as much

as B. How much each has originally?

Solution:

Let the amounts with ‘A’ and ‘B’ be ‘x’ rupees ‘y’ rupees respectively

A gives 4 rupees to B.

A will have (x – 4)

B will have (y + 4)

Given, y + 4 = 2(x – 4) = 2x – 8

2x – y = 12 …………………(1)

Secondly, if B gives 15 rupees to A.

B will have y – 15

A will have x + 15

Given, x + 15 = 10(y – 15) = 10y – 150

-x + 10y =165 …………..(2)

(2) 2 -2x + 20y = 330 ……….. (3)

2x – y = 12 ………….. (1)

(3) + (1) 19y = 342

Put y = 18 in (1), we get 2x – 18 = 12

2x = 30

x = 15

A’s original possession = Rs.15

B’s original possession = Rs.18

5. Find two numbers which are such that one-fifth of the greater exceeds one-sixth of the smaller by

‘4’; and such that one-half of the greater plus one-quarter of the smaller equals ‘38’.

142

Solution:

Let the two numbers be ‘x’ and ‘y’ say

Let x > y, say

Given

6x – 5y = 120 …………… (1)

Given secondly

2x + y = 152 ………………… (2)

Solve for ‘x’ and ‘y’ from (1) & (2)

(2) 5 10x + 5y = 760 ………… (3)

(1)+ (3) 16x = 880

From (2), 110 + y = 152

y = 152 – 110 = 42

The numbers are 55 and 42.

6. I have a certain number of apples to be divided equally among 18 children. If the number of apples

and the number of children were increased by ‘2’, each child would get 5 apples less. How many

apples have to distribute?

Solution:

Let the number of apples be ‘x’, say

Number of children = 18

Since each should get equal number of apples,

No of apples for each child =

Now, when the number of apples and the number of children were

Increased by 2, each child gets 5 apples less.

20x – 18(x + 2) = 360 5

143

20x – 18x – 36 = 1800

2x = 1800 + 36 = 1836

The number of apples is ‘918’, so that equal distribution among ‘18’ children is quite likely.

7. Two friends A and B start on a holiday together, A with Rs.380 and B with Rs.260. During the

holiday, B spends 4 rupees more than A and when holidays end, A has 5 times as much as B. How

much has each spent?

Solution:

Let ‘x’ rupees be the amount spent by A.

(x + 4) rupees would have been spent by B.

Initially A had Rs.380 and B had Rs.260.

At the end of the holidays ‘A’ had an amount of 380 – x

‘B’ had an amount of 260 – (x + 4)

i.e., 256 – x

Given, 380 – x = 5(256 – x) = 1280 – 5x

4x = 1280 – 380 = 900

Rupees

‘A’ spent Rs.225 and ‘B’ spent Rs.229.

8. The monthly incomes of two persons are in the ratio 7:9 and their monthly expenditures are in the

ratio 12:16. If each saves Rs.100 per month, find their monthly incomes.

Solution:

Let the monthly incomes of the two persons be Rs.7x and Rs.9x. Then their monthly expenditures

are Rs.(7x 100) and Rs.(9x 100).

Given that 7 1009 100

1216

xx

112x 1600 = 108x 1200

112x 108x = 1600 1200

Hence, x = 100. Therefore the monthly incomes are Rs.7x = Rs.700 and 9x = Rs.900.

9. Solve: 3x – y = 7 and 2x + 3y = 1

Solution:

So, this was the first system that we looked at. We already know the solutions, but this will gives

us a chance to verify the values that we wrote down for the solution.

144

Now, the method says that we need to solve one of the equations for one of the variables. Which

equation we choose and which variable that we choose is up to you, but it’s usually best to pick an

equation and variable that will be easy to deal with. This means we should try to avoid fractions if

at all possible.

In this case it looks like it will be really easy to solve the first equation for y so let’s do that.

3x – 7=y

Now, substitute this into the second equation

2x + 3(3x – 7) = 1

This is an equation in x that we can solve so let’s do that

2x + 9x - 21 = 1

11x = 22

x = 2

So, there is the x portion of the solution.

Finally, do NOT forget to go back and find the y portion of the solution. This is one of the more

common mistakes students make in solving systems. To so this we can either plug the x value

into one of the original equations and solve for y or we can just plug it into our substitution that we

found in the first step. That will easier so let’s do that.

y = 3x – 7 = 3(2) – 7 = -1

So, the solution is x = 2 and y = -1, as we noted above.

10. Solve 5x + 4y = 1, 3x – 6y = 2

Solution:

With this system we aren’t going to be able to completely avoid fractions. However, it looks like if

we solve the second equation for x we can minimize them. Here is that work.

3x = 6y + 2

x = 2y + 2/3

Now, substitute this into the first equation and solve the resulting equation for y.

145

Finally, substitute this into the original substitution to find x.

So, the solution to this system is and

Practice Exercise 11. Solve: x + y = 10 and x 2y = 4

1. x=2, y=4 2. x=8, y=2 3. x=6, y=7 4. x=4, y=8

2. Solve: x + y = 6 and x y = 8

1. x=7, y=-1 2. x=5, y=-1 3. x=7, y=5 4. x=6, y=-5

3. Solve:5 9

108

15x y and

3 56

x y

1. , 2. , 3. , 4. ,

4. Solve: 1

161

159

20x y and

120

127

445

x y

1. 2.

3. 4.

5. Solve: 8 9

7x y

and 4 3

1x y

1. x=6, y=8 2. x=2, y=3 3. x=3, y=5 4. x=7, y=5

6. A man travels from A to B at the rate of 4.2 kms an hour. If he had moved at the rate of 4 km an

hour he would have taken 2 hours more to travel that distance. What is the distance between A

and B.

1. 168 km 2. 180 km 3. 200 km 4. 268 km

7. Monthly income of two persons are in the ratio 7 : 8 and their monthly expenditures are in the ratio

11 : 13. If each saves Rs.120 per month, find their monthly incomes.

1. Rs.560 and Rs.600 2. Rs.580 and Rs.640

3. Rs.560 and Rs.640 4. Rs.520 and Rs.650

146

8. A number has two digits. The sum of the two digits is 9. If the digits are reversed and 9 is added to

the new number, the result will be thrice the original number. Find the number.

1. 22 2. 27 3. 25 4. 30

9. Find the fraction which is equal to 35

when both its numerator and denominator are increased by 3

and which is equal to 34

when both are increased by 9.

1. 2. 3. 4.

10. The age of a man is twice the sum of the ages of his two sons, and five years ago his age was

thrice the sum of their ages. Find the present age.

1. 30 years 2. 40 years 3. 50 years 4. 45 years

11. Solve: 2p 3q = 4 and 3p 2q = 11

1. p=5, q=2 2. p=6, q=-2 3. p=4, q=7 4. p=3, q=5

12. Solve: a + 5b = 36 and a ba b

53

1. a=20, b=7 2. a=10, b=5 3. a=16, b=4 4. a=14, b=8

13. Solve: x + y = 3, y + 2 = 5, z + x = 4

1. x=0, y=3, z=6 2. x=0, y=3, z=4 3. x=2, y=4, z=8 4. x=5, y=3, z=2

14. Solve: 8 9

74 3

1x y

andx y

1. x=5, y=7 2. x=3, y=6 3. x=-4, y=-6 4. x=2, y=3

15. Solve: 4x 3y = xy and 6 4

4x y

1. x=7, y=6 2. x=3, y=2 3. x=-4, y=-6 4. x=-5, y=3

147

Practice Exercise 21. Solve 3x + 5y = 19, x + 3y = 9

1. x=3, y=2 2. x=4, y=5 3. x=3, y=5 4. x=6, y=4

2. Solve 5x 2y + 25 = 0, 4y 3x = 29

1. x=-2, y=-4 2. x=-3, y=5 3. x=-4, y=-3 4. x=-3, y=6

3. Solve a b

anda b

2 46

5 20 .

1. a=5, b=8 2. a=4, b=6 3. a=10, b=4 4. a=6, b=2

4. Solve x = 2y 1, y = 5 3x

1. 2. 3. 4.

5. Solve x + y 5 = 0 and y 2 = 2x.

1. x=3, y=5 2. x=1, y=4 3. x=3, y=6 4. x=6, y=2

6. Find four consecutive odd numbers whose sum is 56.

1. 11,13,15,17 2. 9,11,13,15 3. 13,15,17,19 4. 7,9,11,13

7. ‘A’ is 7 years older than ‘B’. 15 years ago, ‘B’ is age was of A’s age. Find their present ages.

1. 45, 34 2. 43, 36 3. 47, 33 4. 46, 32

8. A man left Rs.1750 to be divided among his two daughters and four sons. Each daughter was to

receive three times as much as a son. How much did each son and daughter receive?

1. Rs.175, Rs.525 2. Rs.155, Rs.500 3. Rs.150, Rs.400 4. Rs.145, Rs.450

9. A man travels from A to B at the rate of 4 km an hour. Had he travelled at the rate of 325

km an

hour, he would have taken 3 hours more to cover that distance. Find the distance between A and

B.

1. 69 Km 2. 67 Km 3. 68 Km 4. 63 Km

10. Puja is four times as old as Madhu. 2 years hence she will be twice as old as Madhu. What are their

present ages?

1. 3, 2 2. 4, 2 3. 3, 2 4. 4, 1

148

11. 2 chairs and 3 tables are worth Rs.1900 while 5 chairs and 2 tables are worth Rs.2000. What is the

value of each chair and table?

1. Chair= Rs 200, Table= Rs 500 2. Chair= Rs 200, Table= Rs 500

3. Chair= Rs 200, Table= Rs 500 4. Chair= Rs 200, Table= Rs 500

12. A certain two digit number is four times the sum of its digits and if 27 is added to it, its digits are

reversed. Find the number.

1. 34 2. 38 3. 36 4. 32

13. Sreenivas and Suresh start at the same time on scooters from Bangalore and Mysore towards each

other at the speed of 25 kmph and 35 kmph respectively. At the point where they meet Suresh

realizes that he has traveled 25 kms more than Sreenivas. Find the distance between Bangalore

and Mysore.

1. 150 Km 2. 120 Km 3. 140 Km 4. 130 Km

14. Solve: 5x+4y=1 and 3x-6y=2

1. 2. 3. 4.

15. Solve: 2x+4y=-10 and 6x+3y=6

1. x=2, y=-2 2. x=3, y=-2 3. x=3, y=-4 4. x=2, y=-4

Answer Keys for Practice Exercise 1

1 2 6 1 11 12 1 7 3 12 33 4 8 2 13 24 3 9 1 14 45 2 10 3 15 2

Answer Keys for Practice Exercise 2

1 1 6 1 11 12 2 7 2 12 33 3 8 1 13 14 4 9 3 14 45 2 10 4 15 3

149

CHAPTER-12

Quadratic EquationsDefinition

General quadratic equation in one variable is ax2 + bx + c = 0 where a, b and c are

constants.

The roots of this equation are given by, .

The term b2 4ac determines the nature of the roots and is called the discriminant of the quadratic

equation. The Greek letter (delta used to denote the discriminant i.e., = b2 4ac.)

We know that for any given equation, exactly one of the following is true.

1. < 0 b2 4ac is negative. Clearly b ac2 4 is imaginary. Hence when < 0, the roots are

imaginary.

2. = 0 b2 4ac = 0. Hence b ac2 4 vanishes. Therefore the roots are

b

aba

02 2

. Hence the

roots are real and coincident.

3. When > 0, b2 4ac is positive and hence b ac2 4 is real. In this case the roots are real and

distinct. If in addition to being positive, is a perfect square then, the roots are rational; if is not

a perfect square the roots are irrational.

Note:

1. When b2-4ac is a perfect square, then the roots are rational and incase it is not a perfect square

then the roots are irrational.

2. Imaginary or irrational roots of the equation ax2+bx+c=0, where a, b, c are all real numbers, will

occur in conjugate pairs i.e., if 4+5i is a root then 4-5i will also be a root. But if the coefficients a,

b, c are not all real or any one of them is non real or irrational, then it is not necessary that the

roots to be occur in conjugate pairs.

3. If sum the coefficients is zero, i.e., a+b+c=0 then x=1, is a root of the equation:

ax2+bx+c=0 and if a-b+c=0, then x=-1, is a root of the equation ax2+bx+c=0.

Solving Quadratic Equations by Factorization

If the roots of ax2 + bx + c = 0 are real and rational, the equation can be resolved into linear factors

and hence be easily solved.

150

Example:

Solve x2 11x + 28 = 0

Solution:

We need to express 11 as the sum of two numbers whose product is 28. Clearly 7 and 4 satisfy

these conditions.

Hence, x2 11x + 28 = 0

x2 7x 4x + 28 = 0

x(x 7) 4 (x 7) = 0

(x 4) (x 7) = 0

x 4 = 0 or x 7 = 0

x = 4 or 7

Note:

Quadratic equations whose roots are imaginary or irrational cannot be solved by factorization. We

use the formula in such cases.

Note:

Sum, difference and product of the roots of a quadratic equation. If and are the roots of ax2 +

bx +c = 0, then + = , = and - = .

Forming a Quadratic Equation

So far, we have learnt how to find the roots of a quadratic equation (given the equation). Now we shall

work on the converse problem i.e., to form the quadratic equation given its roots. If and are the

roots of a quadratic equation, we know that (x ) and (x ) are its factors (by factor theorem). Hence

the required equation can be written as

(x ) (x ) = 0

or x2 x x + = 0

or x2 ( + )x + = 0

i.e., x2 (sum of the roots) x + (product of the roots) = 0, gives the required quadratic equation.

Method of Solving Quadratic Equations

1. Solve 3x2 12 = 0

Solution:

3x2 12 = 0 3x2 = 12 or x2 = 4

x = 4 = 2 i.e. x = 2, 2.

151

2. Solve (x + 3)2 = 6(x + 3)

Solution:

x2 + 6x + 9 = 6x + 18 x2 = 9 or x = 3.

Equation Reducible to Quadratic Type

Some equations, which are not quadratic, can be reduced to quadratic equations by proper

substitution and can be solved using any one of the previous methods.

Example: Solve + 12 = 0.

Solution:

Let x + 1x

= y then y2 8y + 12 = 0

y2 2y - 6y + 12 = 0 y(y 2) 6(y 2) = 0

(y 2)(y 6) = 0

y = 2 or 6.

Hence, x + 1x

= 2

xx

2 1 = 2

x2 2x + 1 = 0

x(x 1) 1(x 1) = 0 (x 1)2 = 0 or x = 1.

Also, x + 1x

= 6

xx

2 1 = 6

x2 6x + 1 = 0

Here a = 1, b = 6 and c = 1,

x = b b aca

2 42

By substituting the values of a, b and c, we get

x =

152

x = 6 322

6 4 22

3 2 2

.

Hence the roots are 1, 1, 3 2 2 .

Relationship between Roots and Coefficients: ax 2 +bx+c=0

If and are the roots of the equation, ax2+bx+c=0, then,

Sum of the roots =

Product of the roots =

Example: If and are the roots of , find the equation whose roots are and

Solution: and are the roots of

If the new roots are and

Sum of the roots

Product of the roots

Equation whose roots are and is

Worked Examples

1. Three positive numbers are in the ratio . Find the numbers; if the sum of their squares is

196.

Solution:

Let the numbers be .

153

Then = 196

x2 = 196

x2 = 196

x2 = 196

x2 =

x2 = 144 x 4 x = 144 4x

2. A is older than B by 5 years. Seven years hence, thrice A’s age shall be equal to four times that of

B. Find their present ages.

Solution:

Let x = A’s present age ; y = B’s present age

x = y + 5 x - y = 5 ......... (1)

After 7 years, A’s age = x + 7; B’s age = y + 7

3 ( x + 7 ) = 4 ( y + 7 )

3x + 21 = 4y + 28

or 3x - 4y = 7........ (2)

Multiplying equation (1) by 3 we get

3x - 3y = 15 ............ ( 3)

Subtracting (3) from (2) we get

- 4y + 3y = 7- 15 -y = -8 y = 8

Substituting this value in ( 1), we get

x - 8 =5 x =13

Thus A is 13 years old and B is 8 years old.

3. Show that (x-1)(x-3)(x-4)(x-6) + 10 is positive for real values of x

Solution :

Consider : (x-1)(x-6)(x-3)(x-4) + 10 =(x2-7x+6) (x2-7x+12) + 10

Put x2-7x+6 = t

t(t+6) + 10 = t2+6t+10 = (t2+6t+9) + 1

= (t+3)2+1 = (x2-7x+9)2+1 which is clearly positive for real values of ‘x’.

154

4. A two digit number is less than 3 times the product of its digits by 8 and the digit in the ten’s place

exceeds the digit in the unit’s place by ‘2’.Find the number.

Solution:

Let the numbers have x in ten’s place and y in the unit place

10x+y = 3xy – 8……….(1)

x-y = 2 …………...(2)

From (2) x= y+2

In (1), 10(y+2)+y = 3y(y+2)-8

10y + 20 + y = 3y2+6y-8

3y2-5y-28 = 0

3y2-12y+7y-28 = 0

3y(y-4) + 7(y-4) = 0

(y-4) (3y+7) = 0

y=4 or - ( y = -7/3 is delete4.

x = y+2 => x = 6

The number is ‘64’.

5. A carpet whose length is times its width is laid on the floor of a rectangular room, with a margin

of 1 foot all around. The area of the floor is 4 times that of the margin. Find the width of the

room.

Solution:

Let x and y be the length and breadth of the carpet.

Given, x=

6x = 7y…………(1)

Length of the floor = x+2

Breadth of the floor = y+2

Area of the margin=Area of the floor–Area of the carpet

=(x+2)(y+2) – xy

= xy +2x + 2y + 4 – xy

= 2x+2y+4

Given, Area of the floor = 4 Area of the margin.

(x+2)(y+2) = 4 (2x+2y+4)

155

D C

S R 1 y

P Q x A B

xy + 2x + 2y + 4 = 8x + 8y + 16

xy – 6x – 6y – 12 = 0………..(2)

put, x = in (2)

7y2 - 78y – 72 = 0

7y2 - 84y + 6y – 72 = 0

7y(y - 12) + 6(7y - 12) = 0

(y - 120)(7y + 6) = 0

y = 12; y = - is ignored.

x =

Width of the room = 12 feet

Length of the room = 14 feet

6. The sum of the first ‘x’ natural numbers is . How many numbers must be taken to get 351

as the sum?

Solution:

x2+x = 702 x2+x-702 = 0

x2+27x-26x-702 = 0 x (x+27) – 26(x+27) = 0

(x - 26)(x + 27) = 0

x = 26 (x= -27 is to be deleted)

The first 26 natural numbers must be taken to get a sum of ‘351’.

7. Find the sum and product of the roots of the equation 3x2 4x + 5 = 0.

Solution:

Here a = 3, b = 4, c = 5

Let and be the roots of the equation. Then the sum of the roots, + =

ba

43

43

Product of the roots = ca

53

.

156

8. Discuss the nature of the roots of the equation x2 + 7x + 12 = 0.

Solution:

Here a = 1, b = 7, c = 12

b2 4ac = 72 4.1.12 = 49 48 = 1.

Since b2 4ac > 0, the roots are real and distinct.

Therefore the roots are real and distinct.

9. Find the quadratic equation whose roots are 2 and 3.

Solution:

Here the roots are 2,3. Therefore the required quadratic equation is

(x 2)(x 3) = 0

x2 3x 2x + 6 = 0 or x2 5x + 6 = 0.

10. Find the equation whose roots are {3, 4}.

Solution:

Here the roots are 3, 4. Therefore the required quadratic equation is

(x 3)(x + 4) = 0

x2 + 4x 3x 12 = 0 or x2 + x 12 = 0.

Practice Exercise 1

1. Solve

1. -4, -3 2. -3, -5 3. -8, -3 4. -2, -4

2. Find the quadratic equation whose roots are and

1. 2.

3. 4.

3. Solve:

1. 2. 3. 4.

157

4. Solve the equation

1. 2. 3. 4.

5. Solve the equation

1. 2. 3. 4.

6. Solve the equation

1. 2. 3. 4.

7. A number consists of two digits such that the square of the digit in the tens place exceeds the digit

in the unit place by 11. If the number is Five times the sum of the digits, then the number is

1. 45 2. 72 3. 18 4. 81

8. If 16 is divided into two parts such that twice the square of one part exceeds the square of the

other part by 164, then one of the parts must be

1. 8 2. 6 3. 11 4. 12

9. The difference between the squares of two consecutive numbers is 59. Find the numbers.

1. 29, 30 2. 27, 28 3. 32, 93 4. 39, 40

10. Jim sold a watch costing $250 to Joe at a profit of z%. Joe sold the watch to Jack at a loss of z%,

losing $60 on the whole deal. What is the value of z?

1. 15 2. 20 3. 25 4.30

11. Ramesh and Mahesh solve an equation. In solving Ramesh commits a mistake in constant term

and finds the root 8 and 2. Mahesh commits a mistake in the coefficient of x and finds the roots –9

and –1. Find the correct roots.

1. 7, 2 2. 9, 1 3. 6, 1 4. 7, 1

12. Two candidates attempt to solve a quadratic of the form x2 + px + q = 0. One starts with a wrong

value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the

roots to be 2, -9. Find the correct roots.

1. -3, -4 2. -2, -3 3. -4, -5 4. -2, -4

13. The coefficient of x in the quadratic equation x2 + px + q = 0 was taken as 17 in place of 13, its

roots were found to be –2 and –15. Find the roots of the original equation.

1. -9, -3 2. -10, -3 3. -3, -6 4. -6, -10

14. If and are the roots of the equation, 2x2-3x-6=0, find the equation whose roots are 2+2, 2+2.

158

1. 2.

3. 4.

15. Mr. Lime bought a basket of oranges for $32. He sold the oranges at $0.25 each and made a profit

equal to the cost price of 40 oranges. How many oranges did Mr. Lime buy?

1. 14 2. 120 3. 160 4. 180

Practice Exercise 2

1. If and are the roots of , find the equation whose roots are and

1. 2.

3. 4.

2. The bending moment M at a point distance x from one end of a beam is given by

M = 20x - x2. Find the points where M = 75.

1. x = 5 or 15 2. x = 10 or 15 3. x = 5 or 10 4. x = 15 or 25

3. A cylinder, which is closed at both ends, has a metal surface area of 200cm2. If its height is 5 cm,

what is the diameter?

(Hint: The surface area of the cylinder s = 2rh + 2r2 where = 227

)

1. d = 5.89 2. d = 6.89 3. d = 4.89 4. d = 7.89

4. Find two consecutive even numbers such that the sum of their squares is 100.

159

1. 4 and 6 2. 8 and 10 3. 6 and 8 4. None of these

5. If a cyclist had traveled 3 km per hour faster he would have taken 1 hour and 20 minutes less to

ride 80 km. What time did he take?

1. 6 hrs 40 min 2. 4 hrs 40 min 3. 5 hrs 40 min 4. 7 hrs 40 min

6. Solve 3 8x = x 2.

1. x = 4 or 6 2. x = 2 or 4 3. x = -3 or 6 4. x = 3 or 4

7. Solve: 9x2+15x-14 = 0

1. 2. 3. 4.

8. Solve (x+5) (x-5) = 39

1. 2. 3. 4.

9. Solve:

1. 2. 3. 4.

10. Solve

1. 2. 3. 4.

11. The sum of the squares of two consecutive natural numbers is 25. Find the numbers.

1. 3 and 4 2. 2 and 7 3. 3 and 6 4. 5 and 9

12. The sum of a number and its reciprocal is 5.2. Find the number.

1. 4 2. 6 3. 3 4. 5

13. Divide 25 into two parts such that the sum of their reciprocals is 1/6.

1. 12 and 9 2. 10 and 15 3. 8 and 15 4. 12 and 10

14. The sum of two numbers is 45 and the product is 500. Find the numbers.

1. 10 and 20 2. 15 and 25 3. 20 and 25 4. 12 & 25

15. A tradesman find that by selling a cycle for Rs.75 which he had bought for Rs. x he gains x%. Find

the value of x.

1. 60 2. 70 3. 40 4. 50

160

Answer Keys for Practice Exercise 1

1 3 6 4 11 22 1 7 1 12 13 3 8 2 13 24 1 9 1 14 35 2 10 2 15 3

Answer Keys for Practice Exercise 2

1 2 6 4 11 12 1 7 3 12 43 2 8 3 13 24 3 9 2 14 35 1 10 1 15 4

CHAPTER-13

ProgressionsEvery Progression is a sequence of numbers. We shall discuss three most important progressions here.

Arithmetic Progression (A.P.)A sequence of numbers in which the difference between any two consecutive terms is

constant is an Arithmetic Progression.

The general A.P. is given by a, a+ d, a + 2d, a + 3d......

Where “a” is the first term and “d” is the constant difference (called common difference) are the

components of any A.P.

Clearly the nth term of an A.P, Tn is given by Tn = a + (n 1)d.

Sum of n terms of an A.P.

Let a, a + d, a + 2d......... be the given A.P.

Then, the Sum of n terms of this A.P denoted by Sn is given by,

Sn =

And if the first and the last term of A.P. are known then Sum of n terms is given by,

161

Sn = , where a is the first term and l is the last term of the A.P.

Geometric Progression (G.P.)It is a sequence in which the ratio between any two consecutive terms is a constant.

The common ratio (CR) is obtained by dividing any term by the preceding one.

For example (i) 1, 3, 9, 27 ....... (ii) a, ar, ar2 ....... are all in G.P.

In (i) CR = 3 (ii) CR = r

If a, b, c, d ....... are in G.P., then ba

cb

dc

.

We shall denote the first term of the G.P. by ‘a’ and the common ratio by ‘r’.

If a = 2, and r = 2, then the GP is 2, 4, 8, 16........

The n th term of a G.P.

Let a be the first term and r the common ratio of a G.P.

Then by definition, 2nd term = ar = ar21, 3rd term = ar2 = ar31,

Hence,

nth term of the G.P. = arn1

If l is the last term of the G.P. with n terms, then l = arn1

Corollary: When any two terms are given, the G.P. is completely determined.

To find the sum of ‘n’ terms of a G.P.

Let Sn = a + ar + ar2 + .....arn-1

Multiplying by r we get, r.Sn = ar + ar2 + ..........arn-1 + arn

By subtraction, Sn rSn = a arn or Sn(1 r) = a(1 rn)

Sn = a11

rr

n

=a r

r

n( )

11

, where r 1

If r > 1, we write Sn =

If r < 1, we write Sn =

162

To find the sum to infinite terms of a G.P.

Consider the infinite G.P.

a + ar + ar2 + ......+ arn1 +........ Sn = a r

r

n( )11

If |r| < 1 (i.e. 1 < r < 1) then rn becomes very small, whenever ‘n’ assumes a very large value.

Hence, Sn becomes almost equal to

a or

ar

( )11 1

This is the sum to infinity of the series and is denoted by S

S = whenever |r| < 1

Harmonic Progression (H.P.)A set of numbers a1, a2, a3,........ are said to be in H.P. if their reciprocals i.e.

are in A.P.

For example, 1, 12

13

14

, , ,........ are in H.P. because 1, 2, 3, are in A.P.

The general H.P. is .......

Mean1. If x, a1, a2,........,an, y are in A.P. then the numbers a1, a2, ........an are called the arithmetic means

between x and y.

2. If x, a1, a2,.......,an, y are in G.P, then a1, a2, ........an are called the geometric means between x and

y.

3. If x, a1, a2,........,an, y are in H.P. a1, a2, ........an are the harmonic means between x and y.

Arithmetic Mean (A.M.), Geometric (G.M.) and Harmonic Mean (H.M.) between Two Numbers

1. If a, b, c are in A.P., b is the AM between a and c

163

b a = c b or 2b = a + c i.e. b = a c

2.

Thus the A.M. between a and c is a c

2

2. If a, b, c are in G.P. then b is the GM between a and c.

ba

cb

or b2 = ac i.e. b = ac

Hence the GM between a and c is ac .

3. If a, b, c are in H.P. then b is the HM between a and c.

Hence 1 1 1 1b a c b or

2 1 1b a c

i.e. 2b

a cac

b = 2aca c

Thus the HM between a and c is 2aca c

Note:

If A, G, H are the AM, GM & HM between two positive numbers a and b (a > 2), then

(i) G2 = AH (ii) a > A > G > H > b

Arithmetic-Geometric Series (A.G.):A sequence in which each term is the product of the corresponding terms of an A.P and

a G.P is known as an Arithmetic-geometric sequence.

A typical A.P is a, a+d, a+2d…………………..

A typical geometric progression is 1, r, r2……………..

A typical arithmetic –Geometric sequence: a, (a+4.r), (a+24.r2)…………

The nth term of this sequence is , Tn= [a + (n-1)d] r(n-1)

Example:

1. 1+2x+3x2+4x3 ………………, is an arithmetic-geometric series whose corresponding A.P and G.P

are 1,2,3……… and 1,x, x2……. respectively.

2. is an arithmetic- geometric series whose corresponding A.P and G.P are

1,3,5,7…. And respectively.

To find the sum of n terms of an Arithmetic-geometric series

Let the arithmetic-geometric series be,

164

and let be the sum of n terms of this series . Then,

On subtracting these 2 series, we get

[By summing the G.P]

Sum of the Infinite A.G. series

Let . Then and , when

, which is the sum to infinity of the A.G. series

Worked Examples1. Find the 12th term of the following A.P.

1, 4, 7, 10.......

Solution:

Here a = 1, d = 3, n = 12.

Tn = a + (n1)d

T12 = 1 + (12 1)3 = 1 + 33 = 34.

2. Which term of the A.P. 7, 10, 13,....... is (1) 283 (2) 148.

Solution:

(1) Here a = 7, d = 3, Tn = 283.

Tn = a + (n 1)d

283 = 7 + (n 1)3

165

283 = 7 + 3n 3 = 4 + 3n

283 4 = 3n.

Therefore, n = 279

3 = 93

Therefore 283 is the 93rd term.

(2) a = 7, d = 3, Tn =148

Tn = a + (n 1)d

148 = 7 + (n 1) 3

148 = 7 + 3n 3

3n = 144 or n = 144

3 = 48. Therefore 148 is the 48th term.

3. Find the sum of the arithmetic series 2 + 5 + 8 + ...... to 10 terms.

Solution:

Sn = n2

[2a + (n 1)d].

Here a = 2, n = 10, d = 3.

Sn = 102

[22 + (10 1)3] = 5[4 + 27] = 5 31 = 155.

4. A man saved Rs.16,500 in ten years. In each year after the first, he saved Rs.100 more than he

did in the preceding year. How much did he save in the first year?

Solution:

Here Sn = 16,500, n = 10 and d = 100

Sn = n

a n d2

2 1[ ( ) ]

16500 = 102

[2a + (10 1)100] = 5(2a + 900)

= 10a + 4500 10a = 12000 or a = 1200.

Hence he saved Rs.1200 in the first year.

5. How many terms of the series 1, 5, 9 ......... must be taken so that their sum is equal to 190.

Solution:

Here a = 1, d = 4, Sn = 190

Applying the formula Sn = n2

[2a + (n 1)d]

166

190 =

or 380 = n[2 + 4n 4] = n(4n 2)

or 4n2 2n 380 =0

2n2 n 190 = 0

2n2 20n + 19n 190 = 0

2n(n 10) + 19(n 10) = 0

Therefore n = 10. Hence 10 terms must be taken.

6. Find the sum of the series 1 + 3 + 9 + 27 + ..... to 10 terms

Solution:

Here a = 1, r = 3, n = 10, r > 1

We know that Sn .

7. Find the sum to n terms of 1,4,16.......

Solution:

Here a = 1, r =4, nth term = n, r>1

.

8. Find the sum of n terms of = 7 + 77 +777 +......

Solution:

7+77 +777 +....to n terms = 7(1+11+111+....to n terms)

=79

(9+99+999+.....to n terms)

= 79

10 1 10 1 10 12 3

...... to n terms

= 79

(101 + 102 + 103 + ....... to n terms) (1 + 1 + 1 + ....... to n terms)

= 79

9. Find the sum to 6 terms of the series 1, 4, 16,.......

Solution:

The first term a = 1, the common ratio r = 4, Tn = arn1, T6 = 1 46 = 1 45 = 1024.

167

Sn = a rr

n( )

1

14 14 1

40953

6 = 1365.

10. Find the 3rd term of the H.P. given that its 6th term is 161

and the 10th term is 1

105.

Solution:

Let a and d be the first term and the common difference of the corresponding A.P.

Then T6 = a + 5d =61 .......(1)

T10 = a + 9d = 105 .......(2)

Solving (1) and (2) a = 6, d = 11

T3 = a + 2d = 6 + 2(11) = 6 + 22 = 28

Hence the 3rd term of the H.P. is 128

.

Note:

If a, b, c are any three distinct positive real numbers then A = a b c

3 G = abc3 then

A > G a b c

abc

3

3 i.e. (a + b + c)3 > 27abc

11. Insert 5 AM’s between 5 and 15.

Solution:

let a1, a2, a3, a4 and a5 be the five A.M’s

Therefore 5, a1, a2, a3, a4, a5, 15 are in A.P.

a = 5 and T7 = a + 6d = 15.

i.e. 5 + 6d = 15 or 6d = 10 d = 53

.

Hence a1 = 553

203

a2 = 203

53

253

; a3 = 253

53

= 10; a4 = 10 + 53

353

and a5 = 353

53

403

The required means are 203

253

10353

403

, , , and

168

12. The H.M. between two numbers is 1137

and their A.M. is 14. Find them.

Solution:

Let the numbers be a and b.

2

1137

807

aba b

ab

a b

407

40(a + 2. = 7ab .....(1)

also a b

2

14 a + b = 28 ......(2)

Solving (1)and (2) a = 8, b = 20.

Hence the required numbers are 8 and 20.

13. Sum the series to n terms

Solution:

The given series is an A.G series , in which

The A.P is,

The G.P is,

term of the A.P

term of the G.P

By subtracting these two equations, we get

[Taking the sum of the G.P ]

14. Find the sum infinity of the series,

Solution:

169

Let denote the sum to infinity of the given series. Then,

By subtracting these two equation,we get

[Taking the sum of the infinite G.P]

15. If the sum to infinity of the series is , find r

Solution:

The given series is an A.G series in which the

A.P is with a=3,d=2 and G.P is with common ratio is r

or

But and So,

170

Practice Exercise 11. Find the 80th term of the progression 2, 5, 8, 11, ......

1. 239 2. 249 3. 349 4. 149

2. Given Tn = 2n2 + 1, find the first four terms of the sequence (Tn)

1. 3, 9, 12, 15 2. 3, 9, 19, 33 3. 2, 4, 8, 10 4. 3, 7, 11, 15

3. Find the nth term of A.P. 1, 4, 9, 14......

1. -6n+3 2. 3n-6 3. -5n+6 4. 5n-6

4. Find the sum of all natural numbers between 200 and 300 which are multiples of 3.

1. 8262 2. 8217 3. 6939 4. 8989

5. Find the sum of the series 1 + 4 + 7+..... to 40 terms.

1. 2380 2. 3240 3. 4280 4. 2390

6. The angles of a triangle are in AP. The smaller angle is 40. Find the angles of the triangle. 1.

1. 2. 3. 4.

7. The sum of the three numbers which are in AP is 12 and the sum of their cubes is 408. Find the

numbers.

1. 1, 4, 7 and 7, 4, 1 2. 1, 3, 8 and 8, 3, 1

3. 1, 3, 5 and 5, 3, 1 4. 1, 4, 6 and 6, 4, 1

8. A man 50 years old has 8 sons born at equal intervals. The sum of the ages of the father and eight

sons is known to be 186 years. Calculate the age of the eldest son if the youngest one is 3 years

old.

1. 25 years 2. 30 years 3. 40 years 4. 31years

9. Find the sum of the series 2 + 4 + 8 +..... +512.

1. 1055 2. 1099 3. 1022 4. 1056

10. The sum of 3 numbers in GP is 26 and their product is 216. Find the numbers.

171

1. 3, 4, 5 or 5, 4, 6 2. 2, 6, 18 or 18, 6, 2

3. 2, 4, 8 or 8, 4, 2 4. 4, 8, 12 or 12, 8, 4

11. Sum to n terms: 5 + 55 + 555 + .......

1. 2.

3.

4.

12. A man borrows Rs.1000 from a friend. There is no interest charge on the loan, but it has to be paid

back in monthly installments, starting with Rs. 64 for the first month and decreasing by Rs.2

successively each following month. In what time will the loan be repaid?

1. n = 30 months 2. n = 40months 3. n = 25 months 4. n = 35 months

13. How many terms of the series 2+4+8+.....make the sum equal to 1022.

1. n = 7 2. n = 5 3. n = 8 4. n = 9

14. Find the nth term of the A.G series 1.

1. 2. 3. 4.

15. Find the sum to infinity of the series

1. 5 2. 4 3. 6 4. 9

172

Practice Exercise 21. Given Tn = 3n + 4. Find the first three terms of the sequence (Tn).

1. 7, 10 and 13 2. 7, 11 and 13 3. 7, 10 and 12 4. 7, 11 and 12

2. If Tn = 3n 2. Find the value of T1 + T2 + T3

1. 13 2. 14 3. 12 4. 10

3. What is the 21st term of the A.P. whose 4th term is 17 and 9th term is 42.

1. 100 2. 101 3. 104 4. 102

4. A man borrows Rs.1000 and agrees to repay it with a total interest of Rs.140 in 12 installments

each installment being less than the one preceding it by Rs.10. What should be his first

installment?

1. 170 2. 150 3. 120 4. 110

5. The first term of an AP is 5, the last 45, and the sum 400. Find the number of terms and the

common difference.

1. n=13, d= 2. n=16, d= 3. n=15, d= 4. n=16, d=

6. Find the sum of the series 72 + 70 + 69 +.......+40

1. 950 2. 951 3. 952 4. 953

7. If you save 1p today, 2p the next day, 3p the succeeding day and so on, what will be your total

savings in 365 days.

1. Rs.657.30 2. Rs.640.60 3. Rs.667.95 4. Rs.659.50

8. Find the 10th 15th and 20th term of the GP whose first term is 4 and CR is 2.

1. 4.29, 4.214, 4.219 2. 4.29, 4.211, 4.210 3. 4.29, 4.214, 4.214 4. 4.29, 4.212, 4.219

9. Find the sum of the series 1 + 13

1

3

1

32 1

............... ........

n

1. 2. 3. 4.

173

10. In a G.P the ratio of the sum of the first eight terms to the sum of the first four terms is 97 : 81.

Find the common ratio of the GP.

1. r= 2. r= 3. r= 4. r=

11. The 3rd and 7th terms of an H.P. are 6

1929

and respectively. Find the H.P.

1. 2. 3. 4.

12. If pth term of a H.P. is q and the qth term is p, what is the (pq)th term ?

1. 0 2. 1 3. 2 4.3

13. Find the sum to n terms of the series

1. 2. 3. 4.

14. Sum the series to n terms

1. 2. 3. 4.

15. If the sum of infinity of the series is , find d

1. 5 2. 6 3. 2 4. 7

Answer Keys: Practice Exercise 1

1 1 6 4 11 12 2 7 1 12 33 3 8 4 13 44 2 9 3 14 15 1 10 2 15 2

Answer Keys: Practice Exercise 2

1 1 6 3 11 12 3 7 4 12 23 4 8 1 13 14 2 9 2 14 25 1 10 3 15 4

174

CHAPTER-14

LogarithmsNeed of Logarithms

1. The operations of multiplication and division can be replaced by those of addition and subtraction

using logarithms which makes computation much simpler.

2. If ‘a’ is a positive real number other than 1 such ax = m then ‘x’ is called the logarithm of ‘m’ to the

base ‘a’ symbolically written as x = logam a>0, (a 1)

For example 25 = 32 log232 = 5

In other words, the logarithm of a number to a given base is the power to which the base

must be raised in order to get that number.

loga1 = 0 a0 = 1 Here a > 0, a 1

logaa = 1 a1 = a.

loga0 =

3. Logarithms to base 10 are called common logarithms and those to the base e are natural or

Naprierian logarithms [e is an irrational number, 2 < e < 3].

4. Characteristics and Mantissa: The integral part of a logarithm is called the Characteristic and

the decimal part is called mantissa.

Laws of Logarithms 1. loga (mn) = loga m + loga n

2. logamn

= logam logan

3. loga mn = nloga m

4. loga N = loglog

e

e

Na

175

Properties of Logarithms1. , then

2. etc

3. etc

4. or

Proof:

Let,

putting the value of b, we get

xya= a xy =1or

1x =

y

log a.log b =1ab or 1

log a=b log ba

5. Base change Formula

log a=log a.log ccb b or

log aclog a=b log bc

Proof:

Let xlog a= x, a=bb …..1)

ylog a= y, a= cc …..2)

and ……3)

by 2)and 3),

176

we have ………..4)

Or by 1) and 4)

By using property 4)

In general,

6. and

Proof:

Let

Consider,

[By property1) ]

7. or in particular

177

Proof:

Let

And

…….1)

Particular Case: Putting m=a , in equation1)

By property 2) we have

8. . In particular,

Proof:

Let

……….1)

Let or

………..2)

Particular Case: Putting a=n , in

178

We get,

[since, ]

9.

Proof:

Let

By property 1),We have

Worked Examples1. Find x, if logx 36 =2

Solution:

or .

2. Solve x if log x = log 5 + 2 log 2

Solution:

log x = log 5 + 2 log2 = log 5 + log 22 = log 5 + log4 or log 5 4)

log x = log 20 or x = 20

3. Simplify log(x2 9) log(x + 3)

Solution:

Log(x2 9) log(x + 3) = log xx

2 93

= log( )( )

( )x x

x

3 3

3 =log(x 3).

4. Find the value of log2716

2log37

+ log48

147.

Solution:

We have, log 2716

log 37

2

+ log

48147

= log2716

log949

48147

log

179

= log27 169 49

//

+ log

48147

= log2716

499

48147

3x x

log .

5. Find the value of log6 2163 .

Solution:

Let log6 2163 = x.

6x = 2163 = 6x = 63 1/3= 6x = 61 x = 1.

180

Practice Exercise 11. Give an equivalent exponential form for the following expression: log3 729 = 6

1. 36 = 729 2. 36 3.729 4. 36

2. Give an equivalent logarithm form for the following: 32 = 25

1. 5 = log2 32 2. log2 32 3. 5 = log2 4. log 10

3. Evaluate: log10 100

1. 2 2. 3 3. 4 4. 5

4. Solve: log2 0.25 = x

1. 5 = log2 32 2. x = 2 3. 5 = log2 4. 36

5. If , then find ‘x’.

1. 0 2. 1 3. 2 4. 3

6. If , find x.

1.+1 2.- 1 3. 1 4. 3

7. If , find ‘x’.

1. 0 2. 1 3. 2 4. 3

8. If , then find ‘x’.

1. 6 2. 4 3. 46 4. 64

9. Compute without using table: 2.log10 5 + log10 4 log10 2

1. 0 2. 1 3. 2 4. 3

10. If . Find x.

1. 12 2. 14 3. 16 4. 18

Practice Exercise 2

181

1. Find the value of

1. 1 2. 2 3. 3 4. 4

2. Find the value of

1. 1 2. 2 3. 4.

3. Evaluate:

1. 2. 0 3. 4. 1

4. If and ,then find

1. 1 2. 2 3. 3 4. 4

5. Find the value

1. 0 2. 1 3. 4.

6. Find the value of

1. 2. log 1 3. both 1 & 2 4. None of these

7. Find the value of

1. 0 2. 1 3. both 1 & 2 4.

8. If , then find

1. 2. 1 3. 0 4. None of these

182

9. Solve x if log x = log 5 + 2 log 2

1. 10 2. 15 3. 20 4. 25

10. Simplify log(x2 9) log (x + 3)

1. log (x-2) 2. log (x-3) 3. log (x-4) 4.

Answer Keys for Practice Exercise 1

1 1 6 32 1 7 33 1 8 44 2 9 25 1 10 3

Answer Keys for Practice Exercise 2

1 4 6 12 4 7 43 1 8 44 1 9 35 3 10 2

183