maths for practice

Lacs

10 5

Ten

thou

sand

s10

4

Thou

sand

s 1

03

Hun

dred

s 1

02

Tens

101

Uni

ts10

Numbers are collection of certain sym-bols or figures called digits.

The common number system in use isDecimal number system. In this systemwe use ten symbols each representing adigit. These are 0,1,2, 3,4,5,6,7,8 and 9. Acombination of these figures representing anumber is called a numeral. We also haveBinary number system. It uses only 0 and1. There are other number systems too.We would confine our discussion to Deci-mal number system in this topic.

Every digit has a face value which equalsthe value of the digit itself, irrespective of itsplace in the numeral. Each digit in a num-ber or numeral has a place value besides itsface value. For a given number / numeral webegin from the extreme right as unit's place,ten's place, hundred's place, thousand'splace and so on.

Example : The number 795698 is repre-sented as shown below :

7 9 5 6 9 8

We read it as :Seven lac ninty five thousand six hundredand ninty eight.In the above number,The place value of 8 is (8 x 10) = 8The place value of 9 is (9 x 101) = 90The place value of 6 is (6 x 102) = 600 andso on.

The face value of a digit in a number is thevalue of digit itself wherever it may be .Thus, the face value of 7 in the above numeralis 7, the face value of 9 is 9 and so on.Types and Nature of Numbers.Natural numbers : These are also calledcounting numbers as these numbers are theones which we use for counting purpose. It isrepresented by N.eg : N = {1, 2, 3,............................}Whole numbers : It includes all naturalnumbers with zero. We can denote it by W. Eg : W = {0,1,2, ....................}Integers : It includes all whole numbersalong with negative numbers. It is repre-sented by Ieg : I = { -............-2,-1,0,1,2,......}Thus we see that whole numbers are nothingbut positive integers and zero. Similarly, natu-ral numbers consists of positive integers.Even number : A number which is com-pletely divisible by 2 is called an even num-ber. In other words such numbers has 2 asa factor when they are written as a productof different numbers.Eg : 2, 4, 6, 8....................

A number is said to be a factor or submul-tiple of another when it divides the other ex-actly. For example 5 and 3 are factors of 15.Odd number : These numbers are notcompletely divisible by 2. In other words,a number which is not even is an odd number.Eg: 1, 3, 5, 7,.................It may be noted that zero is an exceptionto this even-odd classification.

NUMBER SYSTEM

Bank Probationary OfficerQuantitative Aptitude

Real number : Any measurement car-ried out in the physical world leads to somemeaningful figure or value or number. Thisnumber is called real number. It consistsof two groups :

(i) Rational number : A rational number canalways be represented by a fraction of

the form pq

, , where p and q are integers

and q is not equal to zero. All integersand fractions are rational numbers.

(ii) Irrational number : An irrational num-

ber can't be expressed in the form of

pq

, where q 0

Eg : 3 2 5, ,It gives only an approximate answer in

the form of a fractional or decimal number.The digits after the decimal point are non-ending. The same holds true for which is again irrational. Alternatively wecan say that an infinite non-recurring deci-mal number is an irrational number.

Prime number : A prime number is anumber which has no factors besides it-self and unity, ie, it is divisible only by it-self and 1 and by no other numbers.

Eg : 2, 3, 5, 7, 11, 13, 17, 19, 23,------

The largest prime number known so faris 2 12281 which is a number of about700 digits.

Note :

(1) 2 is the only even number which isprime.

(2) All prime numbers other than 2 are oddnumbers but all odd numbers are notprime numbers, for example, 9 is an odd

number but it is not a prime number as it isdivisible by 3.

Composite number : A composite num-ber is one which has other factors besidesitself and unity. Thus it is a non-prime num-ber.

Eg : 4, 6, 9, 14, 15, ......................

Note : (1) 1 is neither prime nor composite.

(2) A composite number may be even orodd.

The number of ways in which a number Ncan be expressed as the product of twofactors which are relatively prime to eachother. is 2M-1, where M is the number of dif-ferent prime factors of N.

Eg : 540 = 22 x 33 x 5

Prime factors of 540 are 2,3 and 5M Number of ways = 23-1 = 4

ie 20 x 27, 4 x 135, 108 x 5, 540 x 1.

Consecutive Integers : These are se-ries of numbers differing by 1 in ascendingorder.

Eg : 5, 6, 7, 8, ...................

Fractions : A fraction is a number whichrepresents a ratio or divisions of two inte-

gers. It is expressed in the form ab

,

where 'a' and 'b' are integers. 'a' is calledthe numerator and 'b' is called the denomi-nator. A fraction cannot have zero (b 0as a denominator since division by zero isnot defined.

Zero divided by any integer is always zero.A fraction with 1 as the denominator is thesame as its numerator.

Two fractions are equivalent if they repre-sent the same ratio or number. So if we

multiply or divide the numerator and de-nominator of a fraction by the same non-zero integer, the result obtained will beequivalent to the original fraction.

Decimals : A collection of digits (0, 1,2, .........9) after a period ( called the deci-mal point) is called a decimal fraction.Eg : 0.629, 0.013, 10.652, etc.Every decimal number represents a frac-tion. These fractions have denominatorswith power of 10.

eg : . , .0 5 51 0

0 4 5 41 0

51 0 0

A number containing a decimal point iscalled a decimal number.

Mixed number : A mixed number con-sists of a whole number and a fraction.

Eg : 435 is a mixed fraction.

This is equivalent to 4 + 35 and hence

can be written as 435

4 35

Here 4 is the whole number and 35 is

the fraction.

TEST OF DIVISIBILITY1. Divisibility by 2 : A given number is di-

visible by 2, if its unit digit is any one of 0,2, 4, 6 and 8.

2. Divisibility by 3 : A given number is di-visible by 3, if the sum of its digits is divis-ible by 3.

3. Divisibility by 4 : A given number is di-visible by 4, if the number formed by itslast two digits is divisible by 4 or the lasttwo digits are 00.

4. Divisibility by 5 : A given number is di-visible by 5, if its unit digit is either 0 or 5.

5. Divisibility by 6 : A given number is di-visible by 6, if it is divisible by both 2 and3.

6. Divisibility by 8 : A given number is di-visible by 8, if the number formed by itslast three digits is divisible by 8 or the lastthree digits are 000.

7. Divisibility by 9 : A given number isdivisible by 9, if the sum of its digits isdivisible by 9.

8. Divisibility by 10 : A given number isdivisible by 10, if its unit digit is 0.

9. Divisibility by 11 : A given number isdivisible by 11, if the difference of the sumof its digits at odd places and the sum ofits digits at even places, is either 0 or anumber divisible by 11.

10. Divisibility by 12 : A given number isdivisible by 12, if it is divisible by both 3 and4.

11. Divisibility by 14 : A given number is di-visible by 14, if it is divisible by both 2 and7.

12. Divisibility by 15 : A given number isdivisible by 15, if it is divisible by both 3 and5.

13. Divisibility by 16 : A given number isdivisible by 16, if the number formed bythe last 4 digits of the given number isdivisible by 16 or the last four digits are0000.

14. Divisibility by 18 : A given number is divis-ible by 18, if it is divisible by both 2 and 9.

15. Divisibility by 20 : A given number is divisibleby 20, if it is divisible by both 4 and 5.

16. Divisibility by 7, 13, 17 and 19 :

There is no direct method to test thedivisibility by 7, 13, 17 and 19. In all the

above cases, the operator for checking isthe number itself or its prime factors. Buthere the operator is not the number itself orits factors.

In this case each number has a particularoperator. For example, the operator for 7 is2, 13 is 4, 17 is 5 and 19 is 2.

To find the operator for 7; consider themultiples of 7, such that the multiple is verynear to the multiple of 10 (ie 21 =3 x 7 isnear to 20 = 2 x 10). Here 2 ( the numbermultiplied by 10 to get 20) is taken as theoperator. Similarly the multiple of 13 (ie 3x 13 = 39) near to the multiple of 10(ie 40= 4 x 10)is 39. Here 4 (the numbermultiplied by 10 to get 40) is taken as theoperator. Through the same way we candetermine the operator for 17 as 5 and thatfor 19 as 2.

The multiple of 7 and 17 (ie 3 x 7 = 21 and3 x 17 =51) are one more than the multiple of10 (ie 20 = 2x 10 and 50 =5 x10), thereforethe operator for 7 and 17 are called one moreoperator. Also the multiple of 13 and 19 (ie 3x 13 = 39 and 1 x 19 = 19) are one less thanthe multiple of 10 (ie 40 = 4 x 10, and 20 = 2x 10), therefore the operator for 13 and 19are called negative operators.

Divisibility by 7 :

To test the divisibility by 7, multiply thelast digit of the given number by the operatorof 7 (ie 2 ) and subtract the product fromthe number obtained by removing the lastdigit of the given number. Again take thelast digit of the difference and multiply bythe operator, the product is subtracted fromthe number obtained by removing the lastdigit of the just previous difference. Theabove process is repeated. If the differenceobtained is zero or a multiple of 7, then wesay that the number is divisible by 7.

Eg : To test the number 3563 is divisibleby 7.

The operator for 7 is 2 and the last digit ofthe given number is 3. Then its product is3 x 2 = 6.

The number obtained by removing the lastdigit is 356.

Then 356 - 6 = 350

Here the difference is 350 and its last digitis 0. The product with operator is 2 x 0= 0.

The number obtained by removing the lastdigit of the difference is 35

Then 35 - 0 = 35.

Here 35 is a multiple of 7. Therefore thenumber 3563 is divisible by 7.

Divisibility by 13

Multiply the last digit of the given numberby the operator for 13 (ie 4) and add theproduct to the number obtained by remov-ing the last digit of the given number. Againtake the last digit of the sum (previouslyobtained) and multiply by the operator. Addthe product to the number obtained by re-moving the last digit of the sum (just pre-viously obtained). The above process isrepeated. If at last the sum obtained isthe multiple of 13, then we say that thegiven number can be divisible by 13.

Eg : To test the number '6539' is divisibleby 13.

The operator for 13 is 4, and the last digitof the given number is 9; then its productis 4 x 9 = 36.

The number obtained by removing the lastdigit of the given number is 653.

Then, sum = 653 + 36 = 689

Here the sum is 689, its last digit is 9 andits product with the operator is 9 x 4 = 36.

The number obtained by removing the lastdigit of the sum is 68.

Then sum, 68 + 36 = 104

Again 4 x 4 = 16

Then sum = 10 + 16 = 26

Here 26 is a multiple of 13. Ttherefore thenumber 6539 is divisible by 13.

Divisibility by 17 : Same as 7

Divisibility by 19 : Same as 13.

SOLVED EXAMPLES.

1. What is the difference in the place valueand the face value of 8 in 78975?

Ans :

The place value of 8 is 8000

The face value of 8 is 8

the required difference = 8000 - 8 =7992

2. The numbers p, p + 2, p + 4 are all primes.Find the value of p.

Ans : When p = 3, the given numbers are3, 5, 7; which are prime.

3. If 42 k 8 is a multiple of 9, find the value of k

Ans : Sum of digits = (4 + 2 + k +8) = (14 + k).

Now, least value of k for which (14 + k)is divisible by 9 is k = 4.

4. Find the least number of five digits that isexactly divisible by 456.

Ans :

The least five digit number = 10000.when 10000 is divided by 456, the remain-der is 424.

the required number

= 10000 + (456 -424) = 10032

5. When a certain number is multiplied by13, the product consists entirely of fives.

Which is the smallest of such numbers.?

Ans : By hit and trial we find that 555555is divisible by 13.

555555

1342735

the required number = 42735

6. A number when divided by 779 gives 47as remainder. On dividing the same num-ber by 19, what would be the remainder ?

Ans :

When the number is divided by 779, letthe quotient be k. number = 779k + 47

= 19 x 41k + 19 x 2 + 9 = 19 (41k + 2) + 9

So, when the number is divided by 19, thequotient is (41k + 2) and the remainder is 9

7. A positive number when decreased by 4is equal to 21 times the reciprocal of thenumber. Find the number.

Ans : Let the number be x

Then x - 4 = 21

xie x 2 - 4x -21 = 0

(x - 7) (x + 3) = 0 x = 7

8. A certain number consists of two digitswhose sum is 9. If the order of the digits isreversed, the new number is 9 less thanthe original number. The original numberis :

Ans :

Let ten's digit be x and unit's digit be y.

Then x + y = 9

(10x + y) - (10y + x ) = 9 x y 1

Solving x + y = 9 & x- y =1,

we get x = 5 & y = 4

the required number = 54

9. A boy was asked to multiply a certain num-ber by 25. He multiplied it by 52 and gothis answer more than the correct one by324. The number to be multiplied was :

Ans : Let the required number be x

Then 52x - 25x = 324

x 32427

12

10. A number whose fifth part increased by 5is equal to its fourth part diminished by 5.Find the number.

Ans : Let the required number be x

Then x x5

54

5

ie x x x4 5

10 200

PRACTICE TEST

1.34 th of

13

45

rd of thof a number is 80,

find that number1) 300 2) 803) 14 4) 400

2.45 of a number exceeds its

23 rd by 20.

Find the number

1) 150 2) 32

3) 160 4) 300

3. Sum of two numbers is 13 rd of

15 th of

195 and their product is 16 th of

14 th of

960. Find 13 rd of the difference between

them.1) 1 2) 93) 3 4) 27

4. In an examination, a student was asked to find

45 th of a number. By mistake, he found

54

th of it and his answer was 180 more thanthe correct answer. Find the given number.1) 81 2) 8903) 400 4) 500

5. A sum is divided between Vishnu and

Ananthan such that 35 th of the amount of

Vishnu is equal to 57 th of the amount of

Ananthan. If Vishnu gets Rs. 750, howmuch does Ananthan get?1) Rs. 450 2) Rs. 6003) Rs. 630 4) Rs. 730

6.45 th of a number exceeds its

23 rd of

910 th

by 120. Find the number.1) 600 2) 1403) 800 4) 660

7. At an election, a candidate who gets34 th

of the total votes, is elected by a majorityof 2000 votes. The total number of votespolled and the number of votes securedby the candidate who was elected, are re-spectively,1) 4000, 3000 2) 8000, 60003) 4500, 2500 4) 5000, 3000

8. One-third of the total marks are requiredto pass an examination. A candidate whogets 178 marks, fails by 22 marks. Thetotal marks in the examination are1) 550 2) 6003) 535 4) 660

9. A sum is divided between Raju and Biju

such that 56 th of the amount of Raju is

equal to 45 th the amount of Biju. If Raju

gets Rs. 240, find the sum.1) Rs. 500 2) Rs. 2503) Rs. 490 4) Rs. 400

10. The number of students in a school are

1125 more than 14 th of it. Find the total

strength1) 175 2) 15003) 2000 4) 1225

11. A water tank having 1300 litres of capac-ity was filled by adding 75 litres of waterand as many buckets of water as eachbucket had a capacity. What was thecapacity of each bucket in litres ?1) 32 2) 353) 48 4) 45

12. The amount that Satheesh had was Rs.

675 more than 35 th of

19 th of Rs. 450.

Find the amount.1) Rs. 700 2) Rs. 7053) Rs. 625 4) Rs. 605

13.58 th of 120% of a number is 235 more

than 245. Find that number.1) 523 2) 6843) 576 4) 640

14. A basket contains 44 more than 15 th of

total number of apples. How many applesare there in the basket ?1) 66 2) 553) 65 4) 33

15. In a library, 12 of the books are story books,

23 rd of the remaining books are reference

books and the remaining 250 books are

encyclopaedias. Find the total number ofbooks in the library.1) 1,200 2) 1,5003) 2,000 3) 750

16. In a village, 59 th of the population are

adults. 12 of the adults are male,

35 th of

the adult females are illiterate. If 800 adultfemales are illiterate, then the populationof the village is1) 4,000 2) 4,8003) 9,000 4) 5,600

17. In an examination, a student was asked

to find 57 of a number. By mistake, he

found 75 of it. His answer was 96 more

than the correct answer. The given number is1) 196 2) 1283) 156 4) 140

18.78 th of a number exceeds its

716 th by

84. Find 34 th of the number..

1) 144 2) 1923) 138 4) 204

19. The sum of three consecutive even inte-gers is 132. Find the difference between

3 times the least and 12 of the greatest.

1) 44 2) 893) 144 4) 103

20. The sum of four consecutive numbers is90. Find the difference between five timesthe third number and 4 times the fourthnumber.1) 21 2) 293) 24 4) 19

21. 24 is divided into two parts such that 6times the first part exceeds 4 times thesecond part by 4. Find the first part.1) 14 2) 163) 10 4) 12

22. The sum of 11 results is 55. The sum ofthe first 6 results is 30 and that of the last6 is 48. Find the sixth result.1) 20 2) 243) 13 4) 23

23. The total age of 9 boys is 128. The totalage of the first 4 boys is 62 and that of thenext 4 boys is 48. Find the age of the 9th boy.1) 16 2) 183) 12 4) 17

24. 340 soldiers are arranged in a parade suchthat the number of soldiers in each col-umn is the same as the number of col-umns. Find how many soldiers are ex-cluded from the group to make such anarrangement possible.1) 16 2) 213) 14 4) 24

25. The smallest number which must be sub-tracted from 1300 to make it a perfectsquare is1) 2 2) 33) 4 4) 6

26. Three books and a pen cost Rs. 70 and 2pens and a book cost Rs. 40. How muchwill a book and a pen cost ?1) Rs. 10 2) Rs. 203) Rs. 30 4) Rs. 40

27. Each student in a class contributed asmany paise as the number of students in

the class. The teacher contributed Rs.13 to make the total collection of Rs.49.How many students are there in the class?1) 70 2) 503) 60 4) 65

28. The total marks secured by A, B and C are540. A's marks were three times that of B'sand B's marks were twice that of 3) A'smarks are how much more than those of C ?1) 154 2) 3003) 265 4) 286

29. When 6 is added to a number and the sumis multiplied by 8, the result is same aswhen 26 is multiplied by 10 and 12 isadded to the product. The number is1) 28 2) 233) 26 4) 25

30. Two pens and a pencil cost Rs. 18 andtwo pencils and a pen cost Rs. 12. Findthe cost of a pen1) 8 2) 23) 16 4) 6

31. If the square of a number of two digits issubtracted from the square of the numberformed by interchanging the digits, thelargest number by which the result is al-ways divisible is1) 9 2) 103) 11 4) 99

32. The smallest number, which when sub-tracted from the sum of squares of 11 and13 gives a perfect square, is1) 1 2) 43) 5 4) 9

ANSWERS TO PRACTICE TEST

1. (4) 2. (1) 3. (1) 4. (3) 5. (3) 6. (1) 7. (1)8. (2) 9. (3) 10. (2) 11. (2) 12. (2) 13. (4) 14. (2)

15. (2) 16. (2) 17. (4) 18. (1) 19. (4) 20. (4) 21. (3)22. (4) 23. (2) 24. (1) 25. (3) 26. (3) 27. (3) 28. (2)29. (1) 30. (1) 31. (4) 32. (1)

ADDITION

(i). Positive no. + Positive no. = Positive no.

Eg : +2 + + 5 = + 7(ii) Negative no. + Negative no = Negative no.

Eg : -4 + -6 = -10

(iii) Negative no. + Positive no. = Differece ofnos. and sign of greater no.

Eg : + 9 + -11 = -2, -10 + + 4 = -6

QUICKER & SHORT CUT METHODSFOR ADDITION

In the Bank PO examination, there willbe a lot of situations in which you will haveto add various numbers. But the mostimportant thing is to add numbers in light-ning speed. You should develop the habitof seeing the numbers and adding theminstantly. You will not have the time towrite down the numbers with a pen on apiece of paper and calculate in the usualmanner.

1. The moment you see9 + 5, the number 14 should flash in yourmind.

2. As soon as you see

7 + 4 + 9, the number 20 should come.

Remember, you should not even read thenumbers as seven plus four plus nine.Reading is time consuming. You just seeand calculate. Your eye can recognisethese numbers as 7 + 4 + 9 and instantlythe mind can come out with the answer 20.

3. When you see a number, understand andrepresent it with shorter possible words.

(a) 748 should be understood as sevenforty eight, not as seven hundred and fortyeight.

(b) 1098 should be understood as tenninety eight, not as one thousand andninety eight.

(c) 89876 should be eighty nine eightseven six.

4. Double column addition will enable youto add numbers quicker and faster.

(a) Take the example of 78 + 65. Themoment you see the numbers visualisein your mind that they are 78 + 60 + 5.This way you can straight away get theanswer 143.

(b) 84 + 43 + 16 should be visualised as(84 + 40 + 10 + 9) = 143.

(c) 6328 + 4233 + 2495. Here try thedouble column addition

63 2842 3324 95130 56

Ist double column is 28 + 33 + 95 =28 + 30 + 90 + 8 = 156.

Write 56, and 1 is carried.

2nd double column is 63 + 42 + 24 + 1= 63 + 40 + 20 + 7 = 130

Once you master double column addition,you can easily visualise numbers in theaddable form and add numbers horizon-

BANK PROBATIONARY OFFICERQUANTITATIVE APTITUDE

BASIC ARITHMETIC OPERATIONS

tally, as it will provide you lightning speedin addition.

5. For addition of numbers containing deci-mals, the same procedure of double col-umn addition can be used.

Eg. 369.003 + 9.63 + 0.02 + .0003 + 948= 1326.6533

6. In the case of problems involving both op-erations addition and subtraction, subtractthe sum of all the negative terms from thesum of all the positive terms.

Eg. Find the value of 571 - 412 + 173 - 65- 78 + 300

Sum of positive terms = 571 + 173 + 300= 1044

Sum of negative terms = 412 + 65 + 78 =555

Required value = 1044 - 555 = 489

QUICKER & SHORT CUTS :SUBTRACTION

Subtraction can be done through addi-tion easily.Eg : 9687 - 4363 - 2401 = ?

To find the answer, add all the unit's placedigits of the negative integers. (ie) 3 + 1 =4. Now find the number that should beadded to 4 to get 7 of 9687. It is 3 andwrite 3 as the unit's place digit of the an-swer.

Now add all the ten's place digit of thenegative numbers. (ie) 0 + 6 = 6. Thenumber that should be added to 6 to get 8of 9687 will be the ten's place digit of theanswer. It is 2.

Now add all the hundred's place digits ofthe negative numbers. (ie) 3 + 4 = 7. Now

find the number that should be added tothis 7 to get 6 of 9687. But it is not pos-sible to get such a positive number. Sotreat 6 as 16 and this 1 is carried out forthe next step. Here 9 should be added to7 to get 16. Write 9 as the hundredthplace digit of the answer.

In the next step, 4 + 2 + 1 = 7, the 1 is gotfrom the previous step. Here 2 should beadded to this 7 to get 9 of 9687.

9687 - 4363 - 2401 = 2923.

Similarly,

6884 - 2361 - 1592 = 29314328 - 325 - 659 = 33448203 - 3987 - 1697 = 2519.

MULTIPLICATION

(i) Positive no. x Positive no. = Positive no.Eg : 8 x 3 = 24

(ii) Negative no. x Negative no. = Positive no.

Eg : -12 x -10 = 120

(iii) Positive no. x Negative no. = Negative no.

Eg : 3 x -15 = -45

(iv) Negative no. x Positive no. = Negative no.

Eg : -15 x 3 = -45

DIVISION

(i) Positive No. Positive No = PositiveNo. Eg : 12 3 = 4

(ii) Negative No.Negative No.= Positive No.Eg: -12 -3 = 4

(iii) Positive No Negative No = Negative No.Eg : +12 (-3) = -4

(iv) Negative No.Positive No. = Negative No.Eg : -12 3 = -4

PRACTICE TEST

1. 3543 + 6413 + 5438 = ?

(1) 14294 (2) 15394(3) 15864 (4) 15495

2. 92431 + 64273 + 10428 = ?

(1) 177232 (2) 176134(3) 167132 (4) 168282

3. 98854 -64321 - 12512 = ?

(1) 22021 (2) 20223(3) 20032 (4) 13403

4. 8000.3 + 990.59 + 1885.8 = ?

(1) 10877.69 (2) 10876.69(3) 10886.69 (4) 10866.69

5. 8888 + 888 + 88 + 8 = ?

(1) 9872 (2) 10072(3) 8962 (4) 9962

6. 832.9 + 6.73 + 11.8393 = ?

(1) 851.4693 (2) 861.4793(3) 851.4793 (4) 851.4893

7. 4628 - 954 - 1253 = ?

(1) 2421 (2) 3513(3) 4232 (4) 1029

8. 1286 + 655 - 423 + 638 = ?

(1) 1146 (2) 2656(3) 2056 (4) 2156

9. 1352 + 4352 + ? = 9827

(1) 4213 (2) 4123(3) 3215 (4) 5324

10. 81038 - ? = 61038

(1) 19099 (2) 20100(3) 19999 (4) 20000

11. 7329 - 2564 = 3256 + ?

(1) 1509 (2) 1699(3) 1599 (4) 1409

12. 549 x 99 = ?

(1) 55451 (2) 53451(3) 54351 (4) 54361

13. 15.04 - 0.065 = ?

(1) 15.795 (2) 14.875(3) 14.957 (4) 14.975

14. 2589.47 + 3009.59 + 5099.09 = ?

(1) 11609.85 (2) 10698.12(3) 19808.15 (4) 16989.05

15. 282828280 14 = ?

(1) 202020 (2) 20202(3) 20202020 (4) 2020220

16.6 21 2436 7 15

x xx x

?

(1) 85 (2)

35

(3) 72 (4) 5

2

17. 6142 + ? = 5139 + 8136(1) 7313 (2) 7133(3) 6863 (4) 7033

18. 91 x 33 - 33 = ?

(1) 0 (2) 3003(3) 91 (4) 2970

19. 64640 160 + 120 = ?

(1) 52.40 (2) 89.77(3) 524 (4) 64.40

20. 879 x 37 x 8 = ?

(1) 32523 (2) 292707(3) 260184 (4) 257224

21. 21932 + 67 + 98232 + 100 = ?

(1) 121331 (2) 120331(3) 120231 (4) 100331

22. 2740 x 27 = ?

(1) 73980 (2) 74890(3) 73990 (4) 72880

23.12 48 3 939 38 8

. ..

?

(1) 8 (2) 16(3) 24 (4) 48

24. 5099.09 + 3009.59 + 2589.47 = ?(1) 19608.15 (2) 10698.15(3) 11609.85 (4) 16089.05

25. 4935 x 101 = ?

(1) 498435 (2) 488345(3) 489345 (4) 589425

26. 43488 - 34567 + 14368 = ?

(1) 23289 (2) 24289(3) 13289 (4) 22289

27. 756 18+ 6 = ?

(1) 48 (2) 31.5(3) 46 (4) 50

28. 13025 + 1019 - ? = 7622

(1) 6412 (2) 7422(3) 6422 (4) 7412

29. 182 x 14 - 14 = ?

(1) 2698 (2) 2534(3) 2674 (4) 0

30.32 4 7 227 26 5

. ..

?

(1) 90 (2) 9(3) 2.25 (4) 22.5

31. 6945 + 977 + ? + 59 = 8435

(1) 1044 (2) 575(3) 765 (4) 454

32. 5584 x 51 = ?

(1) 284784 (2) 235439(3) 278484 (4) 327484

33.0 0076

19.

.?

(1) 0.0004 (2) 4.004(3) 0.04 (4) 0.004

34.9 72362x

?

(1) 14 (2)

13

(3) 12 (4)

15

35. 0.9329 - 0.7321 + 4.329 + 0.002 = ?

(1) 4.5318 (2) 45.3180(3) 0.45318 (4) 4.2514

36. 3 x 0.3 x 0.03 x 0.003 = ?

(1) 0.81 (2) 0.000081(3) 0.081 (4) 0.00081

37. 64 0.008 = ?

(1) 80 (2) 800(3) 8000 (4) 0.8

38. 0.999999 0.011 = ?

(1) 90.908 (2) 909.08(3) 9.0908 (4) 90.909

39. 173 x 240 = 48 x ?

(1) 495 (2) 545(3) 685 (4) 865

40. 219 + ? = 7608 - 5719

(1) 2108 (2) 1670(3) 1680 (4) 1570

41.140 20 8 75

11 50x x

x

?

(1) 152 (2) 2(3) 75 (4) 4

42. 1265 x 998 = ?

(1) 1262470 (2) 1263470(3) 226470 (4) 122670

43. 32 3232 x 100 = ?

(1) 1000101 (2)

100101

(3) 1001001 (4)

101100

44. 1.113 - 0.8321 = ?

(1) 0.2809 (2) 0.3809(3) 0.2819 (4) 0.2009

45.480 8 160

160 5x

x

?

(1) 15 (2) 20(3) 1.5 (4) 5

46. 608 x 8 - 48 = ?

(1) 2424 (2) 24240(3) 4816 (4) 4800

47. 625 x 54 = ?

(1) 33750 (2) 34750(3) 43750 (4) 33850

48. 0.023 x 0.5 x 30 = ?

(1) 0.00345 (2) 0.0345(3) 0.345 (4) 3.45

49. 89.467 - 45.971 + 9.991 = ?

(1) 42.505 (2) 32.515(3) 43.404 (4) 53.487

50. 13284 81 = ?

(1) 165 (2) 164(3) 184 (4) 124


1. (2) 2. (3) 3. (1) 4. (2) 5. (1) 6. (1) 7. (1)

8. (4) 9. (2) 10. (4) 11. (1) 12. (3) 13. (4) 14. (2)

15. (3) 16. (4) 17. (2) 18. (4) 19. (3) 20. (3) 21. (2)

22. (1) 23. (2) 24. (2) 25. (1) 26. (1) 27. (1) 28. (3)

29. (2) 30. (2) 31. (4) 32. (1) 33. (4) 34. (3) 35. (1)

36. (2) 37. (3) 38. (4) 39. (4) 40. (2) 41. (4) 42. (1)

43. (2) 44. (1) 45. (4) 46. (3) 47. (1) 48. (3) 49. (4)

50. (2)

VBODMAS'VBODMAS' (Vinculum - Bracket - Of - Division- Multiplication - Addition - Subtraction) ruleshould be applied for solving problems involvingone or more mathematical operations like mul-tiplication, division, addition, subtraction etc.Such problems are solved in the order of vincu-lum, bracket, of, division, multiplication, addi-tion and subtraction. Remember 'Of' inVBODMAS means multiplication.

Solved Examples

1. 45 - 4 x 6 - 5 + 14 7 = ?45 - 4 x 6 - 5 + 14 7 = 45- 24- 5 + 2 = 18

2. 21 3 (10 - 3) - 20 + 1 = ?= 21 3 x 7 - 20 + 1= 7 x 7 - 20 + 1= 49 - 20 + 1 = 30

3. 3 of 45

45

16

?

3 of 45

45

16

125

45

16

125

54

16

3 16

3 16

x

4.

3 8 5 4 2 2 81 3

( ) ( ) ?

3 3 2

3 41 3

3 3 2

1 33 4

x

3 3 137

3 3 1713

x

3 1 33 1 7

1 31 7

xx

5.( ) ?4 4 4 46 6 6 6

( )4 4 4 46 6 6 6

12 4

6 6 13

13

SIMPLIFICATION USING IDENTITIES

1. a x (b + c) = a x b + a x c

2. (a + b)2 = a2 + 2ab + b2

3. (a - b)2 = a2 - 2ab + b2

4. (a + b)2 = (a - b)2 + 4ab

5. (a - b)2 = (a + b)2 - 4ab6. (a - b) (a + b) = a2 - b2

7. (a + b)2 = a3 + 3ab (a + b) + b3

8. (a - b)3 = a3 - 3ab (a - b) - b3

9. a3 + b3 = (a + b) (a2 - ab + b2)

10. a3 - b3 = (a - b) (a2 + ab + b2)

11.a b

a ab ba b

3 3

2 2

12.a b

a ab ba b

3 3

2 2

13.a ab b

a b a b

2 2

3 31

14.a ab b

a b a b

2 2

3 31

15. (a + b)2 + (a - b)2 = 2 (a2 + b2)16. (a + b)2 - (a - b)2 = 4ab

SIMPLIF ICATIONBank ProbationaryQuantitative Aptitude

CLASSIFICATION OF FRACTIONS1 . Proper Fraction : A Proper fraction is onewhose numerator is less than its denomina-tor .

eg. ,13

49

2 . Improper Fraction : An improper fractionis one whose numerator is equal to or greaterthan its denominator

eg. ,65

44

3 . Mixed Fraction: A mixed fraction is aquantity consisting of two parts, one a wholenumber and other a proper fraction.

eg. ,4 18

9 34

A mixed fraction can always be expressed asan improper fraction.

eg x. ( )5 23

5 23

5 3 23

173

Similarly an improper fraction can always beexpressed as a mixed fraction. For that dividethe numerator by the denominator and writethe quotient as the whole number part of themixed fraction, the remainder as the numera-tor and the divisor as the denominator.

eg. ;195

3 45

267

3 57

Basic Property of Fractions1 . The value of a fraction is not altered bymultiplying the numerator and denominator bythe same number.

ie. ab

axcbxc

acbc

2 . The value of a fraction is not altered by

dividing the numerator and the denominator bythe same number.

ie ab

a cb c

.

Reduction of a fraction to its lowest termsTo change a fraction to its lowest terms,

divide its numerator and denominator by theH.C.F. of the numbers.

eg. Reduce 1236 to its lowest terms.

1236

12 1236 12

13

(Since H.C.F. of 12 and 36 is 12)Reducing fractions to their commondenominators

To reduce fractions to their commondenominators, change the denominators intotheir L.C.M.

eg. 34

45

,

L.C.M. of 4 and 5 = 20

To convert the denominator of 34 into 20,

multiply it by 5. To convert the denominator of45 into 20, multiply it by 4.

(ie) 3 54 5

4 45 4

1520

1620

xx

xx

ie; ( ) ;

Comparing Fractions

Let ac and

bc , be two fractions with same

denominator c.

FRACTIONSBank Probationary

Quantitative Aptitude

Thenac

bc

if a>b eg.45

35

ac

bc

if a

fraction together, perform separate multiplica-tion and then add the results.

eg x x x. ( )1 8 5 23

18 5 1 8 23

90 12 120

2 . To divide a mixed fraction by a wholenumber divide the whole number part of themixed fraction by the divisor (let the quotientbe a). Reduce the remainder to a single frac-tion and divide this single fraction by the divi-sor. (Let the quotient be b). Now the requiredresult is a+b.

eg. 2123

4

4 2123

5) (

20

123

53

Now 53

4 53

14

512

x

2123

4 5 512

5 512

More Solved Examples1 . There are 40 students in a class. One day

only 7

10 th of total students were present.

Find the number of absentees on that day.Number of absentees= Fraction of absentees x Total number

= 17

1040 12

x students

2 . A man spends 25 of his salary on food,

310 of his salary on house rent and

18 of the

salary on clothes. He still has Rs. 1,400 leftwith him. Find his total salary.

Totally he spends 25

310

18

of his to-

tal salary.

He saves 125

310

18

part of his sal-

ary.

13340 x total salary = 1400

(ie) 7

40x total salary = 1400

total salary = 1400407

8000x Rs .

3 . In an examination, a studnet was asked

to find 3

14 of a certain number. By mistake,

he found 34 of it. His answer was 150 more

than the correct answer. Find the given num-ber.

Let the given number be x, then34

312

150x x

x x

34

314

1501528

150

x = 150 28

15280x

4 . By how much is 45 of 70 less than

57 of

11 2?57

112 45

70 5 16 4 14 24x x x x

5 .5

12 part of what amount will be equal to

3 34 part of Rs. 100.

Let the amount be Rs.y

512

3 34

100of y of

5

12154

100y x

y x x15 1004

125

y 900

Decimal FractionsFractions that have powers of 10 in the

denominators are called decimal fractions.

(ie) Fractions whose denominators are 10, 102,103, 104 ......... are called decimal fractions.

eg. 0.5, 0.063, 8.98 etc.Here

0.5 =5

100 063 63

10008 98 898

100; . ; .

Annexing zeros to the extreme right ofdecimal fraction does not change its value. 0.47= 0.470 = 0.4700 etc.

Addition

For adding a decimal number with anotherdecimal number or with another whole num-ber write the given number in such a way thatthe number of decimal places are equal for allthe numbers.

eg. 2+0.63 + 0.712

Here maximum number of decimal place= 3

Convert all the numbers to 3 decimal places.

2+ 0.63 + 0.712 =

2.000 + 0.630 + 0.712 = 3.342

SubtractionIn subtraction also, the given numbers are

to be written in such a way that the number ofdecimal places become equal for all numbers.

eg. 5 - 0.473Maximum number of decimal place= 3 (in 0.473)ie. 5-0.473=5.000 - 0.473 = 4.527

Multiplication1 . Multiplication of a Decimal Fraction by apower of 10:

Shift the decimal point to the right by asmany places of decimal as the power of 10.eg. 4.5291 x 100 = 452.912 . Multiplication of two or more decimalfractions :0.002 x 0.08 x 0.5 = ?Step 1: Multiply the given numbers as if theyare without any decimal point.

ie. 2x8x5 = 80Step 2 : Add the total number of decimalplaces in the given numbers

ie 3+2+1 = 6Step 3 : Write the result of step 1 and con-vert it to a number whose number of decimalplaces is same as the number obtained in step2 by shifting the decimal point to the left.

0.002x0.08x0.5=0.000080 = 0.00008Division1 . While dividing a decimal fraction by pow-ers of 10, the result is obtained by shifting thedecimal point to the left by as many places ofdecimal as is the power of 10.

eg. 3.45 10 = 0.345961.1 100 = 9.611

2 . While dividing a decimal fraction by a natu-ral number, divide the given fraction withoutthe decimal point by the given natural number.In the answer thus got, place the decimal point

to the left as many places of decimal as arethere in the dividend.

eg. 12525. ?

First step is 12525

5

12525

0 05. .

3 . While dividing a decimal fraction by a deci-mal fraction, shift the decimal point to the rightof the dividend and the divisor both by equalnumber of digits such that the divisor is con-verted into a whole number.

eg. ..

. .3153 5

31535

0 9

28 60143

28600143

200..

Expressing a decimal into a vulgar fractionPut 1 in the denominator under the deci-

mal point and annex with it as many zeros as isthe number of digits after the decimal point.Remove the decimal point and reduce the frac-tion to its lowest terms.

Thus 01251000

1251000

18

..

If numerator and denominator of a frac-tion contain the same number of decimalplaces, then we may remove the decimal sign.

eg. ..

8 869 25

886925

To multiply a decimal by any multiple often, move the decimal point as many places tothe right as is the number of zeros in the multi-plier. To divide a decimal by any multiple of tenmove the decimal point as many places to theleft as is the number of zeros in the divisor.

When a divisor as well as dividend is adecimal, we multiply both the dividend and thedivisor by suitable multiple of 10 to make thedivisor a whole number and then proceed divi-sion.

SOLVED EXAMPLES1 . Evaluate 237x237+ 363x363+2x237x363

Given expression is of the forma2 + b2 + 2ab = (a+b)2

= ( ) ( )237 363 600 3600002 2

2 . 221 2202 2 ?

Given expression isa2 - b2 =(a+b) (a-b)= (221 + 220) (221 - 220)= 441 x 1 = 21

3 .0 45 0 45 0 45 0 21 0 21 0 21

0 45 0 45 0 45 0 21 0 21 0 21. . . . . .

. . . . . .x x x xx x x

Given expression is of the form

a b3 3

a ab b2 2

4 .4 7 6 5 5 3 6 513 7 9 13 6 9. . . .. . . .

?x xx x

Given expression is a bcy dy

a bc d y

x x x

( )( )

( . . ) .( . . ) .

..

4 7 5 3 6 57 9 6 9 13

10 6 51 13

50

x

x

5 .0 75 0 75 0 74 0 74

149. . . .

.?x x

Given expression is a ba b

a b2 2

a b 0 45 0 21 0 24 . . .

= 0.75 - 0.74 = 0.01

6 .6 4 6 4 2 6 4 3 6 3 6 3 6

6 4 3 62 2. . . . . .

( . ) ( . )?x x x x

The given expression is a ab b

a b

2 2

2 22

=( )

( )( )( . . )( . . )

a ba b a b

a ba b

2 6 4 3 66 4 3 6

=102 8

10028

257

3 47.

7 . 0.7 x 0.7 x 0.7 - 0.3 x 0.3 x 0.3- 3x0.7x0.3x0.4 = ?The given expression isa3 - b3 - 3ab (a-b)= (a-b)3=(0.7-0.3)3=(0.4)3

= 0.0648 . Simplify

7 12

2 14

1 14

12

1 12

13

16

7 12

2 14

1 14

12

1 12

13

16

= 712

2 14

1 14

12

1

x

= 712

2 14

34

= 712

94

43

7 12

3 4 12

x

9 . Find the value of 4 5

1 1

3 1

2 14

4 5

1 1

3 1

2 14

=

4 5

1 1

3 49

= 45

1 93 1

=

= 4318

32 318

18

10 . Find the value of

1 12

1 12

1 13

1 13

1 14

1 14

112

1 12

1 13

1 13

1 14

1 14

= 114

1 19

1 116

=34

89

1516

58

x x

11 . Find the value of

2 2 12 2

12 2

2 2 12 2

12 2

=

2 22 2 2 2

2 2 2 2

= 2 22 22 4

= 2 2 2 2

6 . 2 of 34

34

14

?

1)49

2)32 3) 2 4)

2 14

7 .2 3 5 2 3 5 3 5 3 5

9 6 9 6 2 9 6 8 6 8 6 8 6. . . .

. . . . . .?x x

x x x x

1) 5 4 0 2) 2 7 3) 5 4 4) 6 7 08 . 32.5x32.5-2x32.5x2.5+2.5x2.5= ?

1) 9 0 0 2) 3 0 3) 5 0 0 4) 12 25

9 .8 9 8 9 8 9 14 14 148 9 8 9 8 9 14 14 14

. . . . . .. . . . . .

?x x x xx x x

1) 7 5 2) 10. 3 3) 14. 5 4) 7 . 5

10 .24 4 24 4 2 24 4 5 6 5 6 5 6

24 4 24 4 5 6 5 6. . . . . .

. . . .?x x x x

x x

1)1880 3.. 2)

18830 3)

4775 4)

1625

11.0.7x0.7x0.7+0.3x0.3x0.3+3x0.7x0.3=?1) 4 2) 1 3) 1 0 4) 1 6

12 .( . ) . . ( . )( . ) . . ( . )

?0 356 2 0 356 0106 01060 632 2 0 632 0 368 0 368

2 2

2 2x x x

x x

1) 0. 625 2) 0. 06253) 0. 0345 4) 0. 345

13 . 0637 0637 2 0637 0395 0395 03950242

. . . . . ..

?x x x x

1) 1. 132 2) 0. 2423) 1. 422 4) 1

14 .475 475 475 125 125 125475 475 125 125 475 125

. . . . . .. . . . . .

?x x x xx x x

1) 5 .25 2) 3 . 5 3) 0 4) 6

15 .7 7 5 7 7 5 2 2 5 2 2 5 7 7 5 2 2 57 7 5 7 7 5 7 7 5 2 2 5 2 2 5 2 2 5

x x xx x x x

?

1) 10 00 2) 0 .013) 0. 001 4) 0.0001

12 . If xy

34 then find the value of

67

yy

xx

67

67

1

1

y xy x

xyxy

= 67

1 34

1 34

=67

1474

67

17

1

PRACTICE TEST

1 . 20 - [9-{7+(2x3)} +5] = ?1) 2 0 2) 1 5 3) 1 7 4) 1 9

2 . 6+[2+{4x(8-3) - (2x6)-1}+2]= ?1) 1 7 2) 1 3 3) 1 9 4) 1 5

3 .15

of 35+4 (9-3) = ?

1) 3 1 2) 1 5 3) 2 4 4) 4 2

4 .12

13

14

112

x

1)13

2)45 3)

12 4)

16

5 .( ) ?7 7 7 73 3 3 3

1)3

11 2)3

13 3)57 4)

37 +

16 .( . . ) ( . . )

. .?

0 337 0 126 0 337 0 1260 337 0 126

2 2

x1) 0. 211 2) 0. 4633) 4 4) 2 .11

17 .( ) ( )

( ) ( )?6 9 5 3 45 6 9 5 3 45

6 95 3 45

2 2

2 2

1)2 2) 3 4 5 3) 6 9 5 4) 4

18 .( . . ) ( . . )

. . . .?4 621 2 954 4 621 2 954

4 621 4 621 2 954 2 954

2 2

x x

1) 4 2) 2 3) 0 4) 1

19 . 6 4 85 2 6 48 5 2 24 8 52 2 4 85 26 4 85 2 2 48 5 2

x x

?

1) 20000 2) 800003) 30000 4) 40000

20. 126.5x126.5-2x126.5x6.5+6.5x6.5=?1) 12000 2) 144003) 17689 4) 14 40

21 . 0 52 0 52 0 4 0 4 2 0 52 0 40 52 0 4

. . . . . .. .

?x x x x

1) 1 . 2 2) 0 .923) 0 .48 4) 0 .12

22 .( . ) .

( . ) . .?4 8 0 0 27

4 8 1 4 4 0 0 9

3

2

1) 4 . 5 2) 0 .45 3) 5 . 1 4) 2 .20

23 .1

1 1

1 12

?

1) 3 2)53

3) 1 4)35

24 . 534

2 12

0 5 16

17

. ?

1) 11984 2)

2 6184 3)

2 2384 4)

2 4784

25 .

3 14

45

56

4 13

15

31 0

2 1 15

of?

1) 16 2)

2 712

3) 1512 4)

21 12

26 .12

12

34

12

78

34

?

1)2716 2)

2732

3)2764 4)

107112

27 .15

15

15

15

15

15

of

of?

1) 1 2) 5 3)15 4) 2 5

28 . 79 24 11 6

5 9 13 12

x

x ( )?

1) 87

20 2)554

3)54

4)120

29 . If ab

78 , then

1423

22

b ab a

is equal

t o

1)5

14 2)59 3)

523 4)

592

30 . 213

2 35

2 57

2 79

2 997999

..... ?

1)5

999 2)7

1000

3) 10007

4)1001

3

31 . If ab

43

, then the value of 6 46 5a ba b

is

1) -1 2) 3 3) 4 4) 5

32 . If 1 1

213

14

1 12

13x

, then the

value of x is

1)14 2)

1322 3) 2 4) 4

33 . 1

2 1

2 1

2 12

?

1)8

19 2)198 3)

78 4)

89

34 . 1 1

1 1

1 19

?

1) 159 2)

11019 3)

1019 4)

1910

35 . If (a-2) is 6 more than (c+4) and (a+2)is 3 less than (c-4), then (a-3) is1) 0 . 5 2) 1 . 0 3) 1 . 5 4) 2 . 0

36 . The expression(7.98 x 7.98+7.98x x + 0.02 x 0.02) willbe a perfect square for x equal to1) 4 . 0 2) 0 . 43) 0 .04 4) 0. 004

37 . The sum of the smallest six digit numberand the greatest five digit number is1) 199999 2) 2011103) 211110 4) 1099999

38 . The sum of two numbers is 22 and theirdifference is 14. Find the product of thenumbers.1) 7 0 2) 7 5 3) 7 2 4) 8 2

39 . The sum of squares of two numbers is80 and the square of their difference is36. The product of the two numbers is1) 2 2 2) 4 4 3) 5 8 4) 1 1 6

40 . The product of two numbers is 120. Thesum of their squares is 289. The sum ofthe two numbers is1) 2 0 2) 2 3 3) 1 6 9 4) 1 5 0

41 . The difference of two numbers is 11 andone-fifth of their sum is 9. The numbersare:1) 31 ,20 2) 30 ,193) 29 ,18 4) 28 ,17

42 . The sum of two numbers is 10 and theirproduct is 20. The sum of their recipro-cals is

1)14 2)

12 3) 1 4)

1 14

43 . If the sum of two numbers exceeds theirdifference by 10, then the smaller of thetwo numbers is1) 3 2) 5 3) 8 4) 1 3

44 . The simplification of

3 1

3 13

5 1

5 15

92 5

95

of yields

1) 0 2) 1 3)81

125 4) 2

74125

45 . Given that 5625

83 9 93 40 7178

. ..

x then

5 6258 39 9 34

.. .x is equal to

1) 717. 8 2) 71 .783) 0. 07178 4) 0. 7178


1 1

1 1

1 13

1 47

yields

1) 113 2)

1 14

3) 117 4) 1


34

214

23

12

13

12

13

313

56

of x yields

1)7

18 2)4954

3)23 4)

16


1 1 1 1 113

yields

1) 113 2)

1 47

3) 118 4)

123


1. (4) 2. (1) 3. (1) 4. (3) 5. (4) 6. (4) 7. (1) 8. (1)

9. (4) 10. (3) 11. (2) 12. (2) 13. (2) 14. (4) 15. (3) 16. (3)

17. (1) 18. (2) 19. (4) 20. (2) 21.(4) 22. (1) 23. (4) 24. (3)

25. (3) 26. (2) 27. (4) 28. (2) 29.(3) 30. (4) 31. (3) 32.(2)

33. (1) 34. (2) 35. (3) 36 (3) 37. (1) 38. (3) 39. (1) 40.(2)

41. (4) 42 (2) 43 (2) 44. (1) 45. (3) 46. (4) 47. (3) 48.(2)

219961 = 4692 . Find the square root of 59.1361

7.697 59 .1361

4 91 4 6 10 13

87 615 29 13761

13761 0

59.1361 = 7.69Properties of a perfect square

No perfect square ends with 2,3,7,8 No perfect square ends with an odd

number of zeros. The perfect square consisting of (n-1)

or n digits will have n2 digits in theirroot

The square of a number other than unityis either a multiple of 4 or exceeds amultiple of 4 by 1.

CUBE ROOTThe cube root of a number is one of three

equal factors which if multiplied gives thatnumber. Cube root of a number can be foundout by using the following steps.1 . Write down all the prime factors of thegiven numbers.2 . Write the prime factors in the index no-tation, ie, in an form.3 . Divide the index by 3, then the result willbe the cube root of the given number.

Square : If a number is multiplied by itself thenthe product is the square of the number. Thusthe square of 5 is 5x5 = 25

eg.12

12

12

14

2

x

23

23

23

49

2

x

Square root: The square root of a number isone of two equal factors which is multipliedtogether gives that number.

eg. 121 = 11x11 = 1110000 = 100 x 100 = 100

Finding Square root by means of factorisationWhen a given number is a perfect square,

we resolve it into prime factors and take theproduct of prime factors, choosing one out ofeach pair.

eg. Find the square root of 1156Factors of 1156 is 2x2x17x171156= 2x2x17x17= 2 2x1 7 2

1156 = 22 x 172 = 2x17 = 34General method to find the square root

In the given number mark off the digitsin pairs, from right and then find the squareroot as shown in the example below.eg. 1. Find the square root of 219961

4694 21,99,61

1 68 6 599

5169 2 9 8361

8361 0

BANK PROBATIONARY OFFICERQUANTITATIVE APTITUDE

SQUARE & SQUARE ROOTS

eg. 1. Find the cube root of 512512=(2x2x2) x (2x2x2) x (2x2x2)

5123 = ( )2 2 89 1 3 3 2. Find the cube root of 0.000027

3.

0000027 27

1000000

3

1003

13 313

3 13.

= 3

1000 03 .

Learn by heart the following square rootsSQUARES AND CUBES

No. Square Cube No Square1 1 1 1 6 2 5 62 4 8 1 7 2 8 93 9 2 7 1 8 3 2 44 1 6 6 4 1 9 3 6 15 2 5 1 2 5 2 0 4 0 06 3 6 2 1 6 2 1 4 4 17 4 9 3 4 3 2 2 4 8 48 6 4 5 1 2 2 3 5 2 99 8 1 7 2 9 2 4 5 7 6

1 0 1 0 0 10 00 2 5 6 2 51 1 1 2 1 13 31 2 6 6 7 61 2 1 4 4 17 28 2 7 7 2 91 3 1 6 9 21 97 2 8 7 8 41 4 1 9 6 27 44 2 9 8 4 11 5 2 2 5 33 75 3 0 9 0 0

2= 1.41421 6 =2.449493= 1.73205 7 = 2.645755= 2 . 2360 10 = 3.1622

PRACTICE TEST

1 . If x = 3018 + 36+169, the valueof x is

1) 4 4 2) 5 5 3) 6 9 4) 4 3

2 . If x = 169 0 913 0 13. .. .

xx

, the value of x is

1) 3 2) 1 3 3) 3 9 4) 0 .39

3 . If x

4947

, the value of x is

1) 7 2) 4 9 3) 1 6 4) 4

4 . If x = 1967

18324

65169

the value

of x is1) 1 2 2) 1 0 3) 8 4) 1 5

5 .48411

3249

0 4x x . ?

1)58 2)

85 3)

95 4)

105

6 . If 169 169

x . , the value of x is

1) 1 0 0 2) 10 003) 10000 4) 1,00,000

7 .1446

3246

104169

x ?

1) 2 6 2) 1 43) 101. 82 4) 3 6

8 . 8 ? + 44 = 25% of 4001) 3 2) 3 6 3) 4 9 4) 1 6

9 . If x = 81

0 09., the value of x is

1) 3 2) 3 0 3) 3 0 0 4) 0 . 310 . The largest of four digit numbers whichis a perfect squre is

1) 98 01 2) 99 043) 98 04 4) 98 09

11 . A certain number of people collected Rs.125. If each person contributed as many fivepaise as they are in number, the number ofperson were

1) 2 5 2) 5 0 3) 1 0 0 4) 1 2 5

12 . A gardener plants an orchard with 5776trees. In each row there were as many treesas the number of rows. Find the number ofrows.

1) 7 6 2) 9 6 3) 6 6 4) 1 8 613 . Each student in a class contributed asmany rupees as the number of studnets in theclass for a picnic. The school contributed Rs.150 per teacher who led the trip. If the totalamount collected was Rs. 1350 and the num-ber of teachers who led the trip was 3, howmany students were there in that class?

1) 3 6 2) 3 5 3) 3 4 4) 3 014 . Some persons contributed Rs. 1089. Eachperson gave as many rupees as they are innumber. Find their number.

1) 3 3 2) 6 6 3) 4 5 4) 2 3

15 . If 1 1441312

x

then x is equal to.

1) 0 2) 1 2 3) 1 3 4) 2 5

16 .169225

49

81121

?

1)9

55 2)1145 3)

4 111 4)

4511

17 .7 57 5 is equal to

1) 6 35 2) 6 353) 2 4) 1

18 . If 0 04 0 4. .x xa = 0.004 x 0.4 x b ,

then ab

is

1) 16x10 - 4 2) 16x10 - 3

3) 16x10 - 5 4) 16x10 - 6

19 . The value of 1

2 11

3 21

2 3

is

(1) -1 2) 0

3) 1 4) 13

12 3

20 . The value of 4 0 0 1 6 8 1 is

1) 4 1 2) 2 1 3) 3 1 4) 5 1

21 . if 5 2 24 . and 6 2 45 . , the

value of 23

56

is

1) 1 .37 2) 1 .57 3) 1 .73 4) 1 .75

22 . If 15625 125 then the value of

1 5 6 2 5 1 5 6 2 5 15 6 2 5 . . is1) 1. 3875 2) 13 .8753) 138. 75 4) 156. 25

23 . If 4 096 = 64 to then the value of

40 96 0 4096 0004096 0 00004096. . . . is

1) 7 .09 2) 7. 10143) 7. 1104 4) 7 .12

24 . The expression

2 21

2 21

2 2

equals

1) 2 2) 2 23) 2 2 4) 2 2

25 . 3 15 5 4 27. is equal to

1) 12 3 2) 9 33) 3 3 4) 3

26 . If 5 2 24 . then the value of

3 5

2 5 0 48 . is

1) 0. 168 2) 1 .683) 16. 8 4) 168. 0

27 . Suppose you know that 15 is approxi-mately 3.88 which of the following is the best

approximation to 53

?

1) 0 .43 2) 1 .893) 1 .29 4) 1 .63

28 . If 13 69 0 6025 37 25. . . x then xis equal to

1) 1 0 2) 1 0 0 3) 10 00 4) 11 25

29 .4413

2255

64256

x x ?

1) 2 1 2) 2 8 3) 4 8 4) 8 4

30 . 4 24 81 273205

3 3x x equals

1) 8 2) 814

3) 823 4)

8 34

31 . If 6 2 45 . then the value of 3 23 2

correct upto three places of decimals is1) 1. 000 2) 0. 9903) 0. 100 4) 0. 101

32 . If 86 when approximated to one place

of decimal is 9.3 then the value of 4 32 upto

one place of decimal is1) 18. 6 2) 4 .633) 4 . 6 4) 4. 636

33 . If 18225 135 , then the value of

18225 182 25 18225 0 018225 . . . is

1) 1. 49985 2) 14 .99853) 149.985 4) 1499. 85

34 . If 6 2 45 . then the value of

3 2 33 2 3

equals

1) 0 .41 2) 0 .423) 4 .10 4) 4 .20

35 . The square root of

0 324 0 081 4 624

15625 0 0289 72 9 64. . .

. . .x x

x x x is

1) 24. 0 2) 2 . 43) 0 .24 4) 0. 024


1 . (2) 2 . (1) 3 . (3) 4 . (3) 5 . (2) 6 . (3) 7 . (1) 8 . (3)

9 . (2) 10 . (1) 11 . (2) 12 . (1) 13. (4) 14 . (1) 15 . (4) 16.(2)

17 . (1) 18 . (3) 19 . (3) 20 . (2) 21.(3) 22 . (3) 23 . (3) 24 . (1)

25 . (3) 26 . (2) 27 . (3) 28 . (4) 29.(4) 30 . (3) 31 . (3) 32.(3)

33 . (3) 34 . (2) 35 . (4)

iii). a am n mniv). ab a bm m m .

v). a 1

vi). aa

mm

1

SOLVED EXAMPLES :

1 . 8 16 21 3 1 4 2x x ?

2 2 23 1 3 4 1 4 2x x

2 2 2 2 2 11 1 2 1 1 2x x

2 . 2 2 22 3 3 2 6x ?

2 22

2 6 46 6

66x

3 . 3 3 2 72 2 2 4 4x

3 3

3

3 33

33

14 8

3 4

4

12

8 12

12x x

4 . Evaluate

3 3 27 29

252 2 3 63

1 2

( )

= 1 13 3 2352

3 2 3 6 1 32

12

= 119

3 2 53

2 2

a m

A long product axaxa ....m factors canbe expressed, in short by notation am, whereà' is called the base and `m' the power (orindex or exponent) Thus axaxa.... 10 factors= a10.Definition am: If a is a natural number, then amstands for the product of `m' factors eachequal to à'.

Property 1: am x an = am+n

(eg. a3xa4 = a3+4 = a7)

Property 2 : an=am-n

(eg. a3 = a5-3 = a2)

Property 3 : ( )a am n mn

(eg. ( )a a2 3 6

Meaning of a : Any non-zero number raisedto zero power is equal to 1.

ie. a = 1 where a 0 eg. 5 = 1.

Negative Index : If a 0 then a am

m

1

(eg. aa

33

1)

Fractional Index : a an n1

eg. a1 2 a , a a2 3 23

If à' and `b' are non-zero integers, `m' and`n' are rational numbers, then

i). a xa am n m n

ii).aa

am

nm n

a 5

LAWS OF EXPONENTSBank Probationery Officer


= 119

9 4 53

14 179

15 79

5 . If 5 3 1 2 5a , find the value of 5 3a

5 3 1 2 5 5 5 5a a

a 5

5 5 5 253 5 3 2a

6 . If m and n are whole numbers such that

mn 121 , find the value of ( )m n 1 1

m m nn 121 11 11 22 ,

( ) ( )m n1 11 1 10 100001 2 1 3

PRACTICE TEST

1 . 2525 is divided by 24, the remainder is

1) 2 3 2) 2 2 3) 1 4) 2

2 . ( ) ( )6 4 3 21

24

5

equals

1)1

16 2)18 3)

316 4) 2

3 . If 2 64n , the value of n is

1) 2 2) 4 3) 6 4) 1 24 . If 1.125x10k=0.001125, the value of k

is1) -4 2) -3 3) -2 4) -1

5 . The unit 's digit in the product of

2 4 6 7 3 4 1 2 2 51 5 3 7 2 7 2 1x x is

1) 1 2) 3 3) 5 4) 7

6 . If 21

82 1

3x

x

, the value of x is

1) 0 2) -1 3) -2 4) 2

7 . The value of 3 9

3

12 2 7

5n

n nx is

1)13 2)

913 3)

19 4)

23

8 . If ab

ba

x x

1 3

, x is equal to

1) 1 2)12 3)

72

4) 2

9 . A boy was asked to write the value of

2 95 2x . He wrote it as 2592. The differencebetween the obtained and the actual value is

1) 2 91 2x 2) 2 92 3x

3) 2 93 4x 4) Zero

10 . 5 1251

4 0 25x ( ) . is equal to

1) 5 2) 2 5 3) 5 4) 5 5

11 . If a b ab c3 3 180 and a, b, and c arepositive integers, the value of c is

1) 1 0 2) 1 5 3) 2 54) data inadequate

12 . ( ) ( . ).4 0 50 5 4x is equal to

1) 1 2) 4 3)18 4)

13 2

13 . If2 2 12801 1x x , then the value of xis

1) 7 2) 8 3) 9 4) 1 0


11 2 5

23

gives

1)1

2 5 2)12 5

3) 2 5 4) -25

15 . The units digit in 2 4 6 9 1 5 3 will be

1) 1 2) 3 3) 7 4) 9


1. (3) 2. (1) 3. (4) 4. (2) 5. (3) 6. (4) 7. (3) 8. (4)

9. (4) 10. (3) 11. (4) 12. (3) 13. (3) 14. (3) 15. (4) 16.(1)

17. (4) 18. (3) 19. (2) 20. (3) 21.(4) 22. (4) 23. (1) 24.(4)

16 . 2 4 3 2 4 30 1 2 0 08. .x equals1) 3 2) 9 3) 1 2 4) 2 7

17 . The value of 4 1 6 1 67 4 x is euqal to:

1)1

1 6 2)14 3) 4 4) 1

18 . If 41

64x , the value of x is

1) 3 2) 9 3) -3 4) -9

19 . The value of 7 2 9 7 2 914

11 2x is

1) 3 2) 9 3) 1 2 4) 2 7

20 . The fraction x yx y

2 2

1 1

1

is equal to

1) x y 1 1 2) x y 1 1

3)xy

x y 3)xy

x y

21 . The value of ( ) 1

5 31 1 equals

1)13215 2)

15132

3)152 4)

158

22 . If x xx

x3 2 4

10

xp , p is equal to

1) 2 6 2) 2 3) 1 4) 0

23 . If 5 5 5 5 53 3 2 2x a the value of`a' is

1) 4 2) 5 3) 6 4) 8

24 . 2 2 2

2 22

4

33

n n

n

( ) is equal to

1) 2 1n 2) 2 1

81n

3)98

2 n 4) 1

1 3 2

Factors and Multiples: If a number `m' dividesanother number `n' exactly, then we say that`m' is a factor of `n' and that `n' is a multipleof `m'.

eg. 3 is a factor of 12 and therefore 12is a multiple of 3.Least Common Multiple (L.C.M.)L.C.M. is the least non-zero number in com-mon multiples of two or more numbers.

Multiple of 6 = 6, 12, 18, 24, 30, ........Multiple of 8 = 8, 16, 24, 32, 40, ........Common Multiple of 6 and 8 = 24, 48 ...............Least Common Multiple = 24

Factorisation Method:Find the L.C.M. of 12, 27 and 40Factors of 12= 2x2x3 = 22x3Factors of 27= 3x3x3 = 33

Factors of 40= 2x2x2x5 = 23x5

L C M x x. . . 2 3 5 10 8 03 3

SHORT CUT METHOD(Division Method)

Find the L.C.M. of 12, 27, 40

L C M. . . 2x2x3x9x10 = 1080

2 1 2 3 2 7 2 4 02 6 3 9 2 2 03 3 3 3 2 1 0

1 1 5 51

2 12 , 27 , 4 02 6 , 27 , 2 03 3 , 27 , 1 0

1 , 9 , 1 0

HIGHEST COMMON FACTOR (H.C.F)The highest common factor of two or

more numbers is the greatest number whichdivides each of them exactly.eg. Find the H.C.F. of 24 and 56Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56Common factors of 24 and 56 are 1, 2, 4, 8H.C.F. of 24 and 56 = 8Factorisation Method: H.C.F. can be found byresolving the given numbers into prime fac-tors and then taking the product of least pow-ers of all common factors, that occur in thesenumbers.

Eg. Find H.C.F. of 48, 108, 140

Factors of 48= 2x2x2x2x3= 24x3Factors of 108= 2x2x3x3x3 = 22x33

Factors of 140= 2x2x5x7 = 22 x 5 x 7H.C.F. = 22 = 4Division MethodFind the H.C.F. of 48, 108, 1402 48 , 108 , 1 4 02 24 , 54 , 7 0

12 , 27 , 3 5

H.C.F.=2x2= 4

2 4 8 2 1 0 8 2 1 4 02 2 4 2 5 4 2 7 02 1 2 3 2 7 5 3 52 6 3 9 7

3 3

L.C.M & H.C.F

Bank probationary Officer

Quantitative Aptitutde

QUICKER & SHORT CUT METHODFind the H.C.F. of 777 and 1147 777) 11 47 (1

777 370) 777(2

7 4 0 37) 370 (10

370 0

H.C.F. of 777 and 1147 is 37* The product of two given numbers is equalto the product of their H.C.F. and L.C.M.L.C.M. of two numbers

= Product of numbers

. . .H C F of numbers

L.C.M. of given fractions

= L C M of numerators

H C F of deno ators. . .

. . . min

H.C.F of given fractions

= H C F of numerators

L C M of deno ators. . .

. . . min

The L.C.M of a given set of numberswould be either the highest or higher than thehighest of the given numbers. The H.C.F. of a given set of numberswould be either the lowest or lower than thelowest.Solved Examples1 . Find the L.C.M. of 125,64,8 and 3.Ans : Given numbers are 53, 26, 23 and 3

L.C.M. 53x26x3 = 24,000

2 . Find the L.C.M. of 13

56

59

1027

, , , ?

Ans: L.C.M. of fractions

= L C M of numerators

H C F of deno ators. . .. . . min

L.C.M. of 1, 5 and 10 is 10H.C.F of 3, 6, 9 and 27 is 3

L.C.M. of given fractions = 103

3 . Find the H.C.F. of 12

34

56

78

910

, , , ,

Ans: H.C.F. of fractions

=H C F of numerators

L C M of deno ators. . .

. . . min

H.C.F. of 1, 3, 5, 7 and 9 is 1L.C.M of 2, 4, 6, 8 and 10 is 120

H.C.F. of given fractions = 1

1204 . The L.C.M. of two number is 2310. TheirH.C.F. is 30. If one number is 210, the otheris:Ans: The other number

= L C M xH C Fgivennumber. . . . . .

= 2310 30

210330x

5 . The H.C.F. and L.C.M. of two numbersare 44 and 264 respectively. If the first num-ber is divided by 2, the quotient is 44, Theother number isAns: First number = 2x44 = 88

Second number = 44 264

88132x

6 . The least square number which is divis-ible by 6, 8 and 15 is:Ans: The least number divisible by 6, 8 and 15is their L.C.M. which is 120

Now 120 = 2x2x2x3x5To make it a perfect square, it must be

multiplied by 2x3x5Required Number=120x2x3x5=36007 . The least number of square tilesrequired to pave the ceiling of a room 15m17cm long and 9m. 2cm broad is:Ans: Size of largest square tile

= H.C.F. of 1517 cm and 902 cm= 41 cm. Least number of tiles required

= Areaof theroomAreaof one tile

= 1517 902

41 41814x

x

8. Find the least number which when dividedseparately by 15, 20, 36 and 48 leaves 3 as re-mainder in each case.Ans : Required number

= L.C.M. of (15,20,36 and 48) +3= 720 + 3 = 723

9 . Find the greatest number that will divide197 and 269 and leaves 5 as remainder ineach case.

Required number = H.C.F. of [(197-5)and (269-5)]= H.C.F. of (192 and 264) = 8

12 . Five bells begin to toll together and tollrespectively at intervals of 6,7,8,9 and 12 sec-onds. How many times they will toll togetherin one hour, excluding the one at the start?Ans: L.C.M. of 6,7,8,9 and 12

= 2x2x3x7x2x3 = 504ie, The bells will toll together after each 504seconds. In one hour, they will toll together

60 60

5047x t imes

PRACTICE TEST1 . Find the L.C.M of 12, 15, 18 and 27.

1) 1, 080 2) 5 4 03) 2 7 0 4) 7 6 0

2 . Find the H.C.F. of 72, 48 and 30.1) 3 0 2) 1 2 3) 6 4) 3

3 . Find the L.C.M. of 22x33x53 and23x3 2x5.1) 27 ,000 2) 1 8 03) 3 6 4) 13 ,500

4 . Find the L.C.M. of 25

310

, and 4

15

1)1

302) 2

25

3)24

7504)

25

5 . Find the H.C.F. of 45

310

715

, and

1)15

2)845

3)8430

4)1

306 . If the L.C.M of x and y is z, their H.C.F.is.

1)xyz

2) xyz

3)x + y

z4)

zxy

7 . H.C.F of two numbers is 24 and theirL.C.M is 1080. If one of the numbers is 120,find the other.

1) 2 1 6 2) 5 3 2 3) 1 0 8 4) 8 2 08 . L.C.M. of 2.5, 0.5 and 0.175 = ?

1) 2 . 5 2) 0 . 53) 0. 175 4) 17. 5

9 . H.C.F. of two numbers is 24 and theirL.C.M is 1344. If the difference between thenumbers is 80, their sum is:

1) 3 6 8 2) 3 5 63) 3 3 2 3) 3 0 4

10 . Find the greatest number which can di-vide 1354, 1866 and 2762 leaving the sameremainder 10 in each case.

1) 6 4 2) 1 2 4 3) 1 5 6 4) 2 6 011 . Find the least perfect square which is di-

visible by 3, 4, 5, 6 and 8.1) 25 00 2) 12 003) 36 00 4) 9 0 0

12 . The least number which when divided by15, 27, 35 and 42 leaves in each case a re-mainder 7 is:

1) 18 97 2) 19 873) 18 83 4) 20 07

13 . Two containers contain 60 and 165 litresof milk respectively. Find the maximum capac-ity of a container which can measure the milkin each container an exact number of times(in litres)

1) 1 5 2) 3 3) 5 4) 1 014 . Two baskets contain 195 and 250 ba-nanas respectively, which are distributed inequal number among children. Find the largestnumber of bananas that can be given, so that3 bananas are left over from the first basketand 2 from the second.

1) 4 2) 1 8 3) 8 4) 6Qn: (15- 18) :- Write in ascending order

15 .12

25

34

32

, , ,

1)25

12

34

32

, , , 2)34

12

25

32

, , ,

3)32

34

12

25

, , , 4)32

12

25

34

, , ,

16 .53

119

56

712

, , ,

1)119

712

53

56

, , , 2)7

1256

119

53

, , ,

3)56

712

119

53

, , , 4)53

119

56

712

, , ,

17 .56

78

34

13

, , ,

1)78

13

34

56

, , , 2)56

78

34

13

, , ,

3)34

78

13

34

, , , 4)13

34

56

78

, , ,

Qn: 18- 20 Write in descending order

18 .13

25

34

16

, , ,

1)13

25

34

16

, , , 2)16

25

13

34

, , ,

3)25

34

13

16

, , , 4)34

25

13

16

, , ,

19 .56

78

1112

310

, , ,

1)56

78

1112

310

, , , 2)78

56

1112

310

, , ,

3)1112

78

56

310

, , , 4)78

56

1112

310

, , ,

20 .53

119

56

712

, , ,

1)53

119

56

712

, , , 2)119

53

712

56

, , ,

3)53

119

56

712

, , , 4)119

56

53

712

, , ,

Qn 21-23 Find the greatest of the given frac-tions

21 .23

415

35

34

, , ,

1)4

15 2)34

3)35 4)

23

22 .58

611

1322

913

, , ,

1)58 2)

611

3)1322 4)

913

23 .34

57

23

811

, , ,

1)34 2)

57

3)23 4)

811

Qn: (24 - 26) Find the smallest of the givenfraction.

24 .23

57

913

914

74

, , , ,

1)9

14 2)23

3)74 4)

57

25 .1114

1417

1720

2326

2932

, , , ,

1)2932 2)

1114

3)1720 4)

1417

26 .56

34

58

67

, , ,

1)34 2)

67

3)58 4)

56

27 . A heap of stones can be made in groupsof 21 but when made up into groups of 16,20, 25 and 45 there are 3 stones left in each

case, The number of stones in the heap is1) 36 00 2) 36 033) 72 00 4) 72 03

28 . Three measuring rods are 64cm, 80cmand 96 cm in length. The least length of cloth(in metres) that can be measured exact num-ber of times using any of the three rods is

1) 0.96m 2) 9.6m3) 96m 4) 960m

29 . The largest number, which exactly dividesthe product of any three consecutive integersis

1) 2 2) 3 3) 6 4) 1 230 . The L.C.M. of two numbers is 63 and theirH.C.F. is 9. If one of the numbers is 27, theother number will be

1) 9 2) 2 1 3) 1 7 4) 1 8 931 . The HCF of two numbers is 32 and theirproduct is 10240. Find their L.C.M?

1) 6 4 0 2) 3 2 0 3) 3 2 4 4) 2 3 032 . A gardener had a number of shrubs toplant in rows. At first he tried to plant 8, then12 and then 16 in a row but he had always 3shrubs left with him. On trying 7 he had noneleft. Find the total number of shrubs.

1) 1 4 7 2) 1 5 0 3) 1 3 7 4) 1 5 433 . Six bells commencing tolling together andtoll at intervals of 2,4,6,8,10 and 12 secondsrespectively. In 30 minutes, how many timesdo they toll together.

1) 1 7 2) 1 5 3) 1 6 4) 2 034 . In a seminar the number of participantsin Hindi, English and Mathematics are 60, 84and 108 respectively. Find the minimum num-ber of rooms required, where in each roomthe same number of participants are to beseated; and all of them being in the same sub-ject.

1) 2 0 2) 2 2 3) 2 5 4) 2 135 . Find the least number that being increasedby 8 is divisible by 32, 36 and 40

1) 14 30 2) 14 003) 14 32 4) 14 25

36 . Find the least multiple of 11 which whendivided by 8, 12 and 16 leaves 3 as remain-der.

1) 1 0 0 2) 9 0 3) 9 9 4) 8 837 . The least multiple of 7 which leaves aremainder 4 when divided by 6, 9, 15 and 18is

1) 7 4 2) 9 4 3) 1 8 4 4) 3 6 438 . The greatest number that will divide 187,

233 and 279 leaving the same remainder ineach case is

1) 3 0 2) 3 6 3) 4 6 4) 5 639 . The floor of a room of dimensions 6.5mx 4.0 m is to be paved with square marbleslabs. The length of the largest possible slabis:

1) 25cm 2) 75cm3) 50cm 4) 100cm

ANSWERS TO PRACTICE TEST - 6

1. (2) 2. (3) 3. (1) 4. (2) 5. (4) 6. (1) 7. (1) 8.(4)

9. (1) 10. (1) 11. (3) 12. (1) 13. (1) 14. (3) 15. (1) 16.(2)

17. (4) 18. (4) 19. (3) 20. (1) 21.(2) 22. (4) 23. (1) 24. (1)

25. (2) 26. (3) 27. (4) 28.(2) 29.(3) 30. (2) 31. (2) 32.(1)

33. (2) 34. (4) 35. (3) 36. (3) 37. (4) 38. (3) 39. (3)

A fraction with its denominator as `100'is called a percentage. Percentage means perhundred.

So it is a fraction of the form6

1 0 03 7

1 0 01 5 11 0 0

, and and these fractions

can be expressed as 6%, 37% and 151%respectively.

In such a fraction, the numerator is calledrate percent.

To express x% as a fraction or a deci-mal, divide x by 100.

If the price of an item increases by r%,then the reduction in consumption, so that theexpenditure remains the same is

rr

x

1 0 0

1 0 0 %

If the price of the commodity decreasesby r%, the increase in consumption, so thatthe expenditure remains the same is

rr

x1 0 0

1 0 0 %

If the value is first increased by x% andthen by y%, the final increase is

x yxy

100%

If there is a decrease instead of increase, anegative sign is attached to the correspondingrate percent.

If the value of a number is first increasedby x% and later it is decreased by x% then netchange is always a decrease which is equal to

x2

100

%

If pass marks in an examination is x% and

if a student secures y marks and fails by zmarks, then the maximum mark

= 1 0 0 ( )y z

xA candidate scores x% in an examina-

tion fails by `a' marks while another candi-date who scores y% gets `b' marks more thanthe minimum required for a pass, then the

maximum mark = 100 (a b)

y x

If the length of a rectangle is increasedby x% and the breadth is decreased by y%,then the area is increased or decreased by

( )%x y xy 1 0 0

according to the (+) ve or

(-) ve sign obtained.If the present population is P which in-

creases R% annually, then(i) the population after n years

= PR n100

100

(ii) the population n years ago

= n

R100100P

If the present value of a machine is Pwhich depreciates at R% per annum, then(i) the value of the machine after n years

= PR n100

100

(ii) the value of the machine n years ago

= P R

n100

100

P ER C EN T AG E

Bank Probationary Officer

Q uantitative Aptitude

4 In an examination 36% are pass marks.If an examinee gets 17 marks and fails by 10marks, what are the maximum marks?Ans : Pass mark=(17+10)= 27

Let maximum marks be x

Then 36% of x = 27 or 3 6

1 0 02 7x x

x 2 7 1 0 03 6

7 5x

Hence, maximum marks = 75The answer can be arrived quickly by

Maximum marks = 1 0 0 1 7 1 0

3 6( )

=1 0 0 2 7

3 67 5x

5 . Subtracting 40% of a number from thenumber, we get the result as 30. Find the num-ber.Ans: Let the number be x.

x x4 01 0 0

3 0 (ie) x 125

3 0

x 3 0 53

5 0x

6 . If the price of sugar be increased by25%, find by how much percent must its con-sumption be decreased to keep the expendi-ture fixed on sugar?Ans:

Decrease in consumption

= 25

100 25100%

=

2 5 1 0 01 2 5

2 0 %x %

7 . The salary of a worker was first increasedby 10% and thereafter decreased by 5%.What was the effect in his salary?

Ans: % effect = 10 510 5100

x%

His salary is increased by 4.5% (becausethe sign is +ve.)

If x% students failed in a particular sub-ject, y% students failed in another subject, andz% students failed in both subjects, then thepass present = 100+z-(x+y)Fractional Equivalents of important percents

1%1

100

40%25

614

116

813

112

%

%

2%1

50

60%35

1212

18

1623

16

%

%

4%1

25

80%45

25%14

3313

13

%

5%1

20100% 1

3712

38

6623

23

%

%

8% 225

50% 12

83 13

56

%

10% 110

62 12

58

%

20%1

5

75%34

SOLVED EXAMPLES:

1 . Find 3313 % of 600

Ans: 3313

% of 600 = 13

x600 = 200

2 . What percent of 144 is 36?Ans: Let x% of 144 = 36

(ie) x

1 0 01 4 4 3 6x

(ie) x 3 6 1 0 0

1 4 42 5x

3 . 2.5 is 5% of what?Ans : Let the number be x

5% 2 5of x .

5100

2 5 50x x x .

8712

78

%

2 8

133 13

43

%

8 . The value of a machine depriciates at therate of 10% per annum. If its present value isRs. 81,000 what will be its worth after 2year s?Ans: The value of the machine after

2 years = Rs. 81,000x 11 0

1 00

2

= Rs. 81000 x 9

1 09

1 06 5 6 1 0x Rs . ,

9 . Due to fall of 10% in the rate of sugar,500 gm more sugar can be purchased for Rs.140. Find the original rate and reduced rate.Ans : Money spent originally=Rs. 140

Less Money to be spent now= 10% of 140= Rs. 14

Rs. 14 now yield 500gm sugar

Present rate of sugar = Rs. 28 per kg.If the present value is Rs. 90, the original value= Rs. 100If the present value is Rs. 28 the original value

= Rs. 1 0 09 0

2 8x

= Rs. 31.1110 . In an examination, 42% students failedin History and 52% failed in Geography. If 17%students failed in both subjects, find the per-centage of those students who passed in boththe subjects.Ans:- Pass percent=100+17-(42+52)

= 117 - 94= 23

PRACTICE TEST

1 . 65% of 7+35% of 3 = ?% of 561) 1 2) 1 0 3) 5 0 4) 1 0 0

2 . What is 20% of a number whose 200%is 360?

1) 7 2 2) 3 6 3) 5 2 4) 1 4 4

3 . What percent of 47

235

is ?

1) 2 . 5% 2) 10 00 %3) 2 5 % 4) 1 0 %

4 . The total income of A and B is Rs. 6000.A spends 60% of his income and B spends80% of his income. If their savings are equal,then the income of A is,

1) Rs. 3500 2) Rs. 20003) Rs. 4000 4) Rs. 3000

5 . With an increase of Rs. 2,000, Vishnu'smonthly salary became Rs. 12,000. What isthe percent increase in his salary?

1) 2 0 2) 2 5 3) 4 0 4) 8 06 . if 75% of the students in a school areboys and the number of girls is 420, the num-ber of boys is

1) 11 76 2) 13 503) 12 60 4) 11 25

7 . The salary of a worker is first increasedby 10% and therafter it was reduced by 10%.What was the change in his salary?

1) 1% increase 2) 5% increase3) no change 4) 1% decrease

8 . A water tank contains 5% salt byweight. x litres of fresh water is added to 40litres of tank water, so that the solution con-tains 2% salt. The value of x is

1) 4 0 2) 5 0 3) 5 5 4) 6 09 . The population of a town increases 5%annually. If it is 15,435 now, what was it 2years ago?

1) 14 ,000 2) 13 ,4733) 12 ,345 4) 10 ,145

10 . Navin spends 15% of his salary on cloths,30% on food and 10% on transport. After thisif he is left with Rs. 900/- what is his salary?

1) Rs. 1,500 2) Rs. 20003) Rs. 1,635 4) Rs. 2500

11 . When the price of an article was reducedby 15% the sale of the article is increased by

20%. What was the effect on the sales?1) 2% increase 2) 1% increase3) 2% decrease 4) 1% decrease

12 . In an election between two candidates,the one gets 35% of the votes polled is de-feated by 15000 votes. The number of votescasted by the winning candidate is

1) 15 ,000 2) 1,75,0003) 32 ,500 4) 52 ,500

13 . In an examination, 70% students passedin English and 75% in Hindi while 20% failedin both the subjects. If 260 students passed inboth the subjects, the total number of studentsis

1) 4 0 0 2) 5 0 03) 3 4 0 4) 4 6 0

14 . If the radius of a circle is diminished by10%, the area is diminished by

1) 3 6 % 2) 2 0 %3) 1 9 % 4) 1 0 %

15 . The price of an article is cut by 10%. Inorder to restore it to its former value, the newprice must be increased by

1) 1013

% 2) 1 1 %

3) 1119

% 4) 1219

%

16 . The breadth of a rectangular field is 60%of its length. If the perimeter of the field is800m, What is the area of the field?

1) 37,500 sq.m. 2) 4,800 sq.m3) 18,750 sq.m 4) 40,000 sq.m

17 . In a factory, 60% of the employees aremales. Among them 20% are matriculates andthe remaining are graduates. Among the fe-males 40% are matriculates and the remain-ing are graduates. If the total number of fe-male employees in the factory is 640, howmany graduates are there in the factory?

1) 1024 2) 8963) 1,152 4) 10 00

18 . In an employment exchange, 40% of thejob seekers are graduates, 20% are post-graduates and remaining 6000 are non-gradu-ates. How many post-graduate job seekers arethere?

1) 3, 000 2) 6, 0003) 9, 000 4) 12 ,000

19 . A company hired a salesman on a monthlysalary of Rs. 3,000. In addition to it, the sales-man was entitled for 20% commission on themonthly sale. How much sale the salesmanshould do if he wants his monthly income asRs. 10,000?

1) Rs. 50,000 2) Rs. 15,0003) Rs. 35,000 4) Rs. 21,000

20 . In a public sector company, 30% employ-ees opted for pension and 50% employeesopted for provident fund. The remaining em-ployees were uncertain. If the difference be-tween those who opted for provident fund andthose who were uncertain was 1440, howmany employees were there in the company?

1) 7, 200 2) 2, 4003) 2, 880 4) 4, 800

21 . Prasanna spends 25% of her monthly in-

come on petrol for her car,23

rd of the remain-

ing income on house hold items, rent, etc. Ifshe is left with Rs. 1,800 with her at the endof the month how much does she spend onpetrol?

1) Rs. 1,800 2) Rs. 7203) Rs. 2,500 4) Rs. 1,440

22 . Rajesh earns Rs. 2,300 per month. Hespends Rs. 1,200 on food, Rs. 630 on convey-ance, 10% of his monthly income on other in-cidentals and saves the remaining amount. Howmuch money will he save in one year?

1) Rs. 2300 2) Rs. 28803) Rs. 2600 4) Rs. 2400

23 . In an examination, Hari got 8 marks lessthan 80% of the full marks and Ravi got 5marks more than 70% of the full marks. Hari

beats Ravi by 2 marks. The marks scored byRavi is

1) 9 0 2) 1 1 0 3) 1 3 0 4) 1 4 024 . A candidate secured 20% marks in a testand failed by 10 marks. Another candidatesecured 42% and got one mark more than thebare minimum to pass. The maximum mark is

1) 5 0 2) 6 0 3) 6 5 4) 7 025 . A's salary is 20% less than B's salary andB's salary is 20% more than C's salary. If thesum of the salaries of A and B is Rs. 5,400then C's salary is

1) Rs. 3,000 2) Rs, 2,8803) Rs. 2,500 4) Rs. 2,700

26 . The price of some commodity was re-duced by 20%. To bring the price of that com-modity to the original level, by how much per-centage of the increase in the price of thatcommodity will have to be made?

1) 12. 5% 2) 2 0 %3) 2 5 % 4) 37. 5%

27 . In a college election between two candi-dates, the candidate who got 62% of thevotes, won by 144 votes. The total number ofvotes is

1) 6 0 0 2) 8 0 03) 9 2 5 4) 12 00

28 . In a class, 30% of the boys play football,40% of the remaining play cricket and theremaining 21 boys play different other games.

How many boys are there in the class?1) 5 0 2) 1 0 0 3) 4 8 4) 9 6

29 . In an examination 40% of the studentsfailed in English, 60% passed in Mathematics.If 10% of the students failed in both the sub-jects, what is the pass percent?

1) 3313 2) 3 0

3) 3623 4) 5 0

30 . In a library, 30% of books are on com-puters, 5% on English, 35% on Science andremaining 900 are on various other fields. Howmany books on English are there in the library?

1) 30 00 2) 3 0 03) 1 5 0 4) 2 0 0

31 . Ramu spends 40% of his income on food,1/3 rd of the remaining on transport and 10%of the remaining on books. If he spends Rs.250 for rent of his house, what is his salary?

1) Rs. 6,000 2) Rs. 6253) Rs. 62,500 4) Cannot be

determined32 . In an examination, A got 10% marks lessthan B, B got 25% more than C, and C got20% less than D. If A got 360 out of 500, thepercentage marks obtained by D was

1) 7 0 2) 7 53) 8 0 4) 8 5

ANSWERS TO PRACTICE

1 . (2) 2 . (2) 3 . (4) 4 . (2) 5 . (1) 6 . (3) 7 . (4) 8.(4)

9 . (1) 10 . (2) 11 . (1) 12 . (3) 13. (1) 14 . (3) 15 . (3) 16.(1)

17 . (3) 18 . (1) 19 . (3) 20 . (4) 21.(1) 22 . (2) 23 . (2) 24 . (1)

25 . (3) 26 . (3) 27 . (1) 28.(1) 29.(2) 30 . (3) 31 . (4) 32.(3)

Cost Price: The price for which an article ispurchased is called the Cost Price (C.P.)Selling price : The price at which an article issold is called the Selling Price (S.P.)Profit (Gain) : The difference between the sell-ing price and the cost price (when S.P. is morethan C.P) is called the Profit.Loss:The difference between the cost priceand selling price (when C.P. is more than S.P.)is called the Loss.Points to remember:1 . Gain = (S.P) - (C.P); Loss = (C.P) - (S.P)

2 . Gain%=Gainx

C P1 0 0

. .=

SP CP

CP

x 100

Loss%=Lossx

C P1 0 0

. . = C S

CPx

P P

100

3 . When the cost price and gain percent aregiven,

S.P. = C.P. x 100

100

Gain%

4 . When the cost price and loss percent aregiven

S.P. = C.P. x100

100

Loss%

5 . When the selling price and gain percentare given

C.P. = S.P. x 100

100

Gain%

6 . When the selling price and loss percentare given

C.P. = S.P. x 1 0 0

1 0 0( % ) Loss

7 . The discount percent is calculated on themarked price.

Discount percent

= Discount

M arked pricex 1 0 0

8 . If there are two successive profits of x%and y% in a transaction then the resultantprofit percent is

x x y y1 0 0

9 . If there is a profit of x% and loss of y% ina transaction, then the resultant profit and losspercent is

x x y y1 0 0 according to the (+)ve and

the (-)ve signs respectively.10 . If cost price of x articles is equal to theselling price of y articles, then profit percent.

=x y

yx

100

SOLVED EXAMPLES

1 . A man buys a toy for Rs. 25 and sells itfor Rs. 30. Find his gain percent.Ans: Gain = 30 - 25 = Rs. 5

Gain % = 52 5

1 0 0 2 0 %x Rs .

2 . By selling a watch for Rs. 144 a manloses 10%. At what price should he sell it inorder to gain 10%?Ans: S.P. = Rs. 144; Loss = 10%

C P Rs x. .100

100 10144

PROFIT & LOSSBank Probationery Officer


= Rs. 1 0 09 0

1 4 4 1 6 0x Rs .

Gain required = 15%

S P Rs x Rs. . . .1 1 51 0 0

1 6 0 1 8 4

Short cut:

Required SP = 1 4 49 0

1 1 5 1 8 4x Rs .

3 . I sold a book at a profit of 16%. Had Isold it for Rs. 18 more, 20% would have beengained. Find the cost price.Ans: Here 120% of C.P._ 116% of C.P.

= Rs. 18

4% 18of t Rscos .

C P x Rs. . .18 1004

450

Formula :

C.P = M o re gain x

D if f in percen t age p rof it1 0 0

.

= 1 8 1 0 02 0 1 6

4 5 0x Rs

.

4 . A shopkeeper mixes two varieties of tea,one costing Rs. 35 per kg. and another atRs. 45 per kg. in the ratio 3:2. If he sells themixed variety at Rs

maths for practice

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