A. Aptitude:

I. Aptitude Questions and Answers

1. Problems on Trains

1. km/hr to m/s conversion:

a km/hr = a x5

m/s.18

2. m/s to km/hr conversion:

a m/s = a x18

km/hr.5

3. Formulas for finding Speed, Time and Distance

1. Speed, Time and Distance:

Speed =Distanc

e , Time =Distanc

e , Distance = (Speed x Time).

Time Speed2. km/hr to m/sec conversion:

x km/hr = x x5

m/sec.18

3. m/sec to km/hr conversion:

x m/sec = x x18

km/hr.5

4. If the ratio of the speeds of A and B is a : b, then the ratio of the

the times taken by then to cover the same distance is1

:1

or b : a.a b

5. Suppose a man covers a certain distance at x km/hr and an equal distance at ykm/hr. Then,

the average speed during the whole journey is2xy

km/hr.x + y

4. Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.

5. Time taken by a train of length l metres to pass a stationery object of length bmetres is the time taken by the train to cover (l + b) metres.

6. Suppose two trains or two objects bodies are moving in the same direction at um/s and v m/s, where u > v, then their relative speed is = (u - v) m/s.

7. Suppose two trains or two objects bodies are moving in opposite directions at um/s and v m/s, then their relative speed is = (u + v) m/s.

8. If two trains of length a metres and b metres are moving in opposite directions atu m/s and v m/s, then:

The time taken by the trains to cross each other =(a + b)

sec.(u + v)

9. If two trains of length a metres and b metres are moving in the same direction atu m/s and v m/s, then:

The time taken by the faster train to cross the slower train =(a + b)

sec.(u - v)

10. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:(A's speed) : (B's speed) = (b : a)

2. Time and Distance

Time and Distance1. Speed, Time and Distance:

Speed =Distanc

e , Time =Distanc

e , Distance = (Speed x Time).

Time Speed2. km/hr to m/sec conversion:

x km/hr = x x5

m/sec.18

3. m/sec to km/hr conversion:

x m/sec = x x18

km/hr.5

4. If the ratio of the speeds of A and B is a : b, then the ratio of the

the times taken by then to cover the same distance is1

:1

or b : a.a b

5. Suppose a man covers a certain distance at x km/hr and an equal distance at ykm/hr. Then,

the average speed during the whole journey is2xy

km/hr.x + y

3. Height and Distance

4. Time and Work

1. Work from Days:

If A can do a piece of work in n days, then A's 1 day's work =1

.n

2. Days from Work:

If A's 1 day's work =1

, then A can finish the work in n days.n

3. Ratio:If A is thrice as good a workman as B, then:Ratio of work done by A and B = 3 : 1.Ratio of times taken by A and B to finish a work = 1 : 3.

5. Simple Interest

Simple Interest1. Principal:

The money borrowed or lent out for a certain period is called the principal or thesum.2. Interest:

Extra money paid for using other's money is called interest.3. Simple Interest (S.I.):

If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then

(i). Simple Intereest =P x R x T

100

(ii). P =100 x S.I.

; R =100 x S.I.

and T =100 x S.I.

.R x T P x T P x R

6. Compound Interest

7. Profit and Loss

Profit and Loss

IMPORTANT FACTSCost Price:The price, at which an article is purchased, is called its cost price, abbreviated as C.P.Selling Price:The price, at which an article is sold, is called its selling prices, abbreviated as S.P.Profit or Gain:If S.P. is greater than C.P., the seller is said to have a profit or gain.Loss:If S.P. is less than C.P., the seller is said to have incurred a loss.IMPORTANT FORMULAE

1. Gain = (S.P.) - (C.P.)2. Loss = (C.P.) - (S.P.)3. Loss or gain is always reckoned on C.P.4. Gain Percentage: (Gain %)

    Gain % =Gain x 100

C.P.

5. Loss Percentage: (Loss %)

    Loss % =Loss x 100

C.P.

6. Selling Price: (S.P.)

    SP =(100 + Gain %)

x C.P100

7. Selling Price: (S.P.)

    SP =(100 - Loss %)

x C.P.100

8. Cost Price: (C.P.)

    C.P. =100

x S.P.(100 + Gain %)

9. Cost Price: (C.P.)

    C.P. =100

x S.P.(100 - Loss %)

10. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.11. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.12. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller

always incurs a loss given by:

    Loss % =Common Loss and Gain % 2

=x 2

.10 10

13. If a trader professes to sell his goods at cost price, but uses false weights, then

    Gain % =Error

x 100%.

(True Value) - (Error)

8. Partnership

Partnership1. Partnership:

When two or more than two persons run a business jointly, they are calledpartners and the deal is known as partnership.

2. Ratio of Divisions of Gains:

I. When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year:(A's share of profit) : (B's share of profit) = x : y.

II. When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time). Now gain or loss is divided in the ratio of these capitals.Suppose A invests Rs. x for p months and B invests Rs. y for q months then,(A's share of profit) : (B's share of profit)= xp : yq.

3. Working and Sleeping Partners:A partner who manages the the business is known as a working partner and the one who simply invests the money is a sleeping partner.

9. Percentage

Percentage1. Concept of Percentage:

By a certain percent, we mean that many hundredths.Thus, x percent means x hundredths, written as x%.

To express x% as a fraction: We have, x% =x

.100

    Thus, 20% =20

=1

.100 5

To express

aas a percent: We have,

a=

ax 100

%.b b b

    Thus,1

=1

x 100%

= 25%.4 4

2. Percentage Increase/Decrease:If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:

Rx 100

%(100 + R)

If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:

Rx 100

%(100 - R)

3. Results on Population:Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:

1. Population after n years = P 1 +R n

100

2. Population n years ago =

P

1 +R n

100

4. Results on Depreciation:Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:

1. Value of the machine after n years = P 1 -R n

100

2. Value of the machine n years ago =P

1 - R n

100

3. If A is R% more than B, then B is less than A byR

x 100%.

(100 + R)

4. If A is R% less than B, then B is more than A byR

x 100%.

(100 - R)

10. Problems on Ages

Problems on Ages

1. Odd Days:We are supposed to find the day of the week on a given date.For this, we use the concept of 'odd days'.In a given period, the number of days more than the complete weeks are calledodd days.

2. Leap Year:(i). Every year divisible by 4 is a leap year, if it is not a century.(ii). Every 4th century is a leap year and no other century is a leap year.Note: A leap year has 366 days.Examples:

i. Each of the years 1948, 2004, 1676 etc. is a leap year.ii. Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.iii. None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

3. Ordinary Year:The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

4. Counting of Odd Days:1. 1 ordinary year = 365 days = (52 weeks + 1 day.)

 1 ordinary year has 1 odd day.2. 1 leap year = 366 days = (52 weeks + 2 days)

 1 leap year has 2 odd days.3. 100 years = 76 ordinary years + 24 leap years

  = (76 x 1 + 24 x 2) odd days = 124 odd days.  = (17 weeks + days)   5 odd days.

 Number of odd days in 100 years = 5.Number of odd days in 200 years = (5 x 2)   3 odd days.Number of odd days in 300 years = (5 x 3)   1 odd day.Number of odd days in 400 years = (5 x 4 + 1)   0 odd day.Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

Day of the Week Related to Odd Days:

No. of days:

0 1 2 3 4 5 6

Day:Sun.

Mon.

Tues.

Wed.

Thurs.

Fri.

Sat.

11. Calendar

Calendar

1. Odd Days:We are supposed to find the day of the week on a given date.For this, we use the concept of 'odd days'.In a given period, the number of days more than the complete weeks are calledodd days.

2. Leap Year:(i). Every year divisible by 4 is a leap year, if it is not a century.(ii). Every 4th century is a leap year and no other century is a leap year.Note: A leap year has 366 days.Examples:

i. Each of the years 1948, 2004, 1676 etc. is a leap year.ii. Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.iii. None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

3. Ordinary Year:The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

4. Counting of Odd Days:1. 1 ordinary year = 365 days = (52 weeks + 1 day.)

 1 ordinary year has 1 odd day.2. 1 leap year = 366 days = (52 weeks + 2 days)

 1 leap year has 2 odd days.3. 100 years = 76 ordinary years + 24 leap years

  = (76 x 1 + 24 x 2) odd days = 124 odd days.  = (17 weeks + days)   5 odd days.

 Number of odd days in 100 years = 5.Number of odd days in 200 years = (5 x 2)   3 odd days.Number of odd days in 300 years = (5 x 3)   1 odd day.Number of odd days in 400 years = (5 x 4 + 1)   0 odd day.Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

Day of the Week Related to Odd Days:

No. of days:

0 1 2 3 4 5 6

Day:Sun.

Mon.

Tues.

Wed.

Thurs.

Fri.

Sat.

12. Clock

Clock1. Minute Spaces:

The face or dial of watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.Hour Hand and Minute Hand:A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called minute hand or long hand.

2.i. In 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand.ii. In every hour, both the hands coincide once.iii. The hands are in the same straight line when they are coincident or opposite to each other.iv. When the two hands are at right angles, they are 15 minute spaces apart.v. When the hands are in opposite directions, they are 30 minute spaces apart.vi. Angle traced by hour hand in 12 hrs = 360°vii. Angle traced by minute hand in 60 min. = 360°.viii. If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast.

On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow.

13. Average

Average1. Average:

Average =

Sum of observations

Number of observations

2. Average Speed:Suppose a man covers a certain distance at x kmph and an equal distance at ykmph.

Then, the average speed druing the whole journey is2xy

kmph

14. Area

AreaFUNDAMENTAL CONCEPTS

1. Results on Triangles:i. Sum of the angles of a triangle is 180°.ii. The sum of any two sides of a triangle is greater than the third side.

iii. Pythagoras Theorem:In a right-angled triangle, (Hypotenuse)2 = (Base)2 + (Height)2.

iv. The line joining the mid-point of a side of a triangle to the positive vertex is called the median.v. The point where the three medians of a triangle meet, is called centroid.The centroid divided each

of the medians in the ratio 2 : 1.vi. In an isosceles triangle, the altitude from the vertex bisects the base.vii. The median of a triangle divides it into two triangles of the same area.viii. The area of the triangle formed by joining the mid-points of the sides of a given triangle is one-fourth

of the area of the given triangle.2. Results on Quadrilaterals:

i. The diagonals of a parallelogram bisect each other.ii. Each diagonal of a parallelogram divides it into triangles of the same area.iii. The diagonals of a rectangle are equal and bisect each other.iv. The diagonals of a square are equal and bisect each other at right angles.v. The diagonals of a rhombus are unequal and bisect each other at right angles.vi. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.vii. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

IMPORTANT FORMULAEI. 1. Area of a rectangle = (Length x Breadth).

        Length =Area

and Breadth =

Area.Breadt

hLengt

hII. 2.   Perimeter of a rectangle = 2(Length + Breadth).

III. Area of a square = (side)2 =  (diagonal)2.IV. Area of 4 walls of a room = 2 (Length + Breadth) x Height.

V. 1.   Area of a triangle =   x Base x Height.2.   Area of a triangle = s(s-a)(s-b)(s-c) 

      where a, b, c are the sides of the triangle and s =  (a + b + c).3.   Area of an equilateral triangle = 3 x (side)2.

4

4.   Radius of incircle of an equilateral triangle of side a =a

.23

5.   Radius of circumcircle of an equilateral triangle of side a =a

.3

6.   Radius of incircle of a triangle of area   and semi-perimeter s =s

VI. 1.   Area of parallelogram = (Base x Height).

2.   Area of a rhombus =   x (Product of diagonals).

3.   Area of a trapezium =   x (sum of parallel sides) x distance between them.VII. 1.   Area of a circle =  R2, where R is the radius.

2.   Circumference of a circle = 2 R.

3.   Length of an arc = 2 R , where   is the central angle.360

4.   Area of a sector =1

(arc x R) =R2

.2 360

VIII. 1.   Circumference of a semi-circle =  R.

2.   Area of semi-circle = R2

.2

15. Volume and Surface Area

Volume and Surface Area1. CUBOID

Let length = l, breadth = b and height = h units. Theni. Volume = (l x b x h) cubic units.ii. Surface area = 2(lb + bh + lh) sq. units.iii. Diagonal = l2 + b2 + h2 units.

2. CUBELet each edge of a cube be of length a. Then,

i. Volume = a3 cubic units.ii. Surface area = 6a2 sq. units.iii. Diagonal = 3a units.

3. CYLINDERLet radius of base = r and Height (or length) = h. Then,

i. Volume = ( r2h) cubic units.ii. Curved surface area = (2 rh) sq. units.iii. Total surface area = 2 r(h + r) sq. units.

4. CONELet radius of base = r and Height = h. Then,

i. Slant height, l = h2 + r2 units.

ii. Volume =  r2h  cubic units.iii. Curved surface area = ( rl) sq. units.iv. Total surface area = ( rl +  r2) sq. units.

5. SPHERELet the radius of the sphere be r. Then,

i. Volume =  r3  cubic units.ii. Surface area = (4 r2) sq. units.

6. HEMISPHERELet the radius of a hemisphere be r. Then,

i. Volume =  r3  cubic units.ii. Curved surface area = (2 r2) sq. units.iii. Total surface area = (3 r2) sq. units.

Note: 1 litre = 1000 cm3.

16. Permutation and Combination

Permutation and Combination

1. Factorial Notation:Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.Examples:

i. We define 0! = 1.ii. 4! = (4 x 3 x 2 x 1) = 24.iii. 5! = (5 x 4 x 3 x 2 x 1) = 120.

2. Permutations:The different arrangements of a given number of things by taking some or all at a time, are called permutations.Examples:

i. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

ii. All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

3. Number of Permutations:Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) =n!

(n - r)!Examples:

i. 6P2 = (6 x 5) = 30.ii. 7P3 = (7 x 6 x 5) = 210.iii. Cor. number of all permutations of n things, taken all at a time = n!.

4. An Important Result:If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind, such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is =n!

(p1!).(p2)!.....(pr!)5. Combinations:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.Examples:

1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.Note: AB and BA represent the same selection.

2. All the combinations formed by a, b, c taking ab, bc, ca.3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.4. Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.5. Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations:

The number of all combinations of n things, taken r at a time is:nCr =

n!=

n(n - 1)(n - 2) ... to r factors.

(r!)(n - r!) r!Note:

i. nCn = 1 and nC0 = 1.ii. nCr = nC(n - r)

Examples:

i.   11C4 =(11 x 10 x 9 x 8)

= 330.(4 x 3 x 2 x 1)

ii.   16C13 = 16C(16 - 13) = 16C3 =16 x 15 x 14

=16 x 15 x 14

= 560.3! 3 x 2 x 1

17. Numbers

NumbersSome Basic Formulae:

i. (a + b)(a - b) = (a2 - b2)ii. (a + b)2 = (a2 + b2 + 2ab)iii. (a - b)2 = (a2 + b2 - 2ab)iv. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)v. (a3 + b3) = (a + b)(a2 - ab + b2)vi. (a3 - b3) = (a - b)(a2 + ab + b2)vii. (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)viii. When a + b + c = 0, then a3 + b3 + c3 = 3abc.

18. Problems on Numbers

Problems on NumbersSome Basic Formulae:

i. (a + b)(a - b) = (a2 - b2)ii. (a + b)2 = (a2 + b2 + 2ab)iii. (a - b)2 = (a2 + b2 - 2ab)iv. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)v. (a3 + b3) = (a + b)(a2 - ab + b2)vi. (a3 - b3) = (a - b)(a2 + ab + b2)vii. (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)viii. When a + b + c = 0, then a3 + b3 + c3 = 3abc.

19. Problems on H.C.F and L.C.M

Problems on H.C.F and L.C.M

1. Factors and Multiples:If number a divided another number b exactly, we say that a is a factor of b.In this case, b is called a multiple of a.

2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):The H.C.F. of two or more than two numbers is the greatest number that divided each of them exactly.There are two methods of finding the H.C.F. of a given set of numbers:

I. Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

II. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.Similarly, the H.C.F. of more than three numbers may be obtained.

3. Least Common Multiple (L.C.M.):The least number which is exactly divisible by each one of the given numbers is called their L.C.M.There are two methods of finding the L.C.M. of a given set of numbers:

I. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

II. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

4. Product of two numbers = Product of their H.C.F. and L.C.M.5. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.6. H.C.F. and L.C.M. of Fractions:

    1. H.C.F. =H.C.F. of Numerators

L.C.M. of Denominators

    2. L.C.M. =L.C.M. of Numerators

H.C.F. of Denominators8. H.C.F. and L.C.M. of Decimal Fractions:

In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

9. Comparison of Fractions:Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

20. Decimal Fraction

Decimal Fraction1. Decimal Fractions:

Fractions in which denominators are powers of 10 are known as decimal fractions.

Thus,1

= 1 tenth = .1;        1

= 1 hundredth = .01;10 100

99= 99 hundredths = .99;  

7= 7 thousandths = .007, etc.;

100 10002. Conversion of a Decimal into Vulgar Fraction:

Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.

Thus, 0.25 =25

=1

;       2.008 =2008

=251

.100 4 1000 125

3. Annexing Zeros and Removing Decimal Signs:Annexing zeros to the extreme right of a decimal fraction does not change its value. Thus, 0.8 = 0.80 = 0.800, etc.If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.

Thus,1.84

=184

=8

.2.99 299 13

4. Operations on Decimal Fractions:i. Addition and Subtraction of Decimal Fractions: The given numbers are so placed under each

other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way.

ii. Multiplication of a Decimal Fraction By a Power of 10: Shift the decimal point to the right by as many places as is the power of 10.Thus, 5.9632 x 100 = 596.32;   0.073 x 10000 = 730.

iii. Multiplication of Decimal Fractions: Multiply the given numbers considering them without decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers.Suppose we have to find the product (.2 x 0.02 x .002).Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) = 6.

 .2 x .02 x .002 = .000008iv. Dividing a Decimal Fraction By a Counting Number: Divide the given number without

considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend.Suppose we have to find the quotient (0.0204 ÷ 17). Now, 204 ÷ 17 = 12.Dividend contains 4 places of decimal. So, 0.0204 ÷ 17 = 0.0012

v. Dividing a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number.Now, proceed as above.

Thus,0.00066

=0.00066 x 100

=0.066

= .0060.11 0.11 x 100 11

5. Comparison of Fractions:Suppose some fractions are to be arranged in ascending or descending order of magnitude, then convert each one of the given fractions in the decimal form, and arrange them accordingly.

Let us to arrange the fractions3

,6 an

d7

in descending order.5 7 9

Now,3

= 0.6,  6

= 0.857,  7

= 0.777...5 7 9

Since, 0.857 > 0.777... > 0.6. So,6

>7

>3

.7 9 5

6. Recurring Decimal:If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal.n a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set.

Thus,1

= 0.333... = 0.3;22

= 3.142857142857.... = 3.142857.3 7

Pure Recurring Decimal: A decimal fraction, in which all the figures after the decimal point are repeated, is called a pure recurring decimal.Converting a Pure Recurring Decimal into Vulgar Fraction: Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.

Thus, 0.5 =5

; 0.53 =53

; 0.067 =67

, etc.9 99 999

Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal.Eg. 0.1733333.. = 0.173.Converting a Mixed Recurring Decimal Into Vulgar Fraction: In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated. In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits.

Thus, 0.16 =16 - 1

=15

=1

;   0.2273 =2273 - 22

=2251

.90 90 6 9900 9900

7. Some Basic Formulae:i. (a + b)(a - b) = (a2 + b2)

ii. (a + b)2 = (a2 + b2 + 2ab)iii. (a - b)2 = (a2 + b2 - 2ab)iv. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)v. (a3 + b3) = (a + b)(a2 - ab + b2)vi. (a3 - b3) = (a - b)(a2 + ab + b2)vii. (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)viii. When a + b + c = 0, then a3 + b3 + c3 = 3abc.

21. Simplification

Simplification1. 'BODMAS' Rule:

This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of given expression.Here B - Bracket, O - of, D - Division, M - Multiplication, A - Addition and S - SubtractionThus, in simplifying an expression, first of all the brackets must be removed, strictly in the order (), {} and ||.After removing the brackets, we must use the following operations strictly in the order:(i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.

2. Modulus of a Real Number:Modulus of a real number a is defined as

|a| =a, if a > 0

-a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.3. Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the 'BODMAS' rule, we simplify the expression under the Virnaculum.

22. Square Root and Cube Root

Square Root and Cube Root

1. Square Root:If x2 = y, we say that the square root of y is x and we write y = x.Thus, 4 = 2, 9 = 3, 196 = 14.

2. Cube Root:The cube root of a given number x is the number whose cube is x.We, denote the cube root of x by x.Thus, 8 = 2 x 2 x 2 = 2, 343 = 7 x 7 x 7 = 7 etc.Note:1. xy = x x y

2.x

y=

x=

xx

y=

xy.

y y y y

23. Surds and Indices

Surds and Indices1. Laws of Indices:

i. am x an = am + n

ii.am

= am - n

an

iii. (am)n = amn

iv. (ab)n = anbn

v.a n

=an

b bn

vi. a0 = 12. Surds:

Let a be rational number and n be a positive integer such that a(1/n) = aThen, a is called a surd of order n.

3. Laws of Surds:i. a = a(1/n)

ii. ab = a x biii.

=a

b

iv. (a)n = a

v.vi. (a)m = am

24. Ratio and Proportion

Ratio and Proportion1. Ratio:

The ratio of two quantities a and b in the same units, is the fraction   and we write it as a : b.In the ratio a : b, we call a as the first term or antecedent and b, the second term or consequent.

Eg. The ratio 5 : 9 represents5

with antecedent = 5, consequent = 9.9

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.Eg. 4 : 5 = 8 : 10 = 12 : 15. Also, 4 : 6 = 2 : 3.

2. Proportion:The equality of two ratios is called proportion.If a : b = c : d, we write a : b :: c : d and we say that a, b, c, d are in proportion.Here a and d are called extremes, while b and c are called mean terms.Product of means = Product of extremes.Thus, a : b :: c : d   (b x c) = (a x d).

3. Fourth Proportional:If a : b = c : d, then d is called the fourth proportional to a, b, c.Third Proportional:a : b = c : d, then c is called the third proportion to a and b.Mean Proportional:Mean proportional between a and b is ab.

4. Comparison of Ratios:

We say that (a : b) > (c : d)      a

>c.

b d5. Compounded Ratio:

6. The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).7. Duplicate Ratios:

Duplicate ratio of (a : b) is (a2 : b2).Sub-duplicate ratio of (a : b) is (a : b).Triplicate ratio of (a : b) is (a3 : b3).Sub-triplicate ratio of (a : b) is (a1/3 : b1/3).

Ifa

=c

, thena + b

=c + d

.     [componendo and dividendo]b d a - b c - d

8. Variations:We say that x is directly proportional to y, if x = ky for some constant k and we write, x   y.We say that x is inversely proportional to y, if xy = k for some constant k and

we write, x 1

.y

25. Chain Rule

Chain Rule1. Direct Proportion:

Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same extent.Eg. Cost is directly proportional to the number of articles.       (More Articles, More Cost)

2. Indirect Proportion:Two quantities are said to be indirectly proportional, if on the increase of the one, the orther decreases to the same extent and vice-versa.Eg. The time taken by a car is covering a certain distance is inversely proportional to the speed of the car. (More speed, Less is the time taken to cover a distance.)Note: In solving problems by chain rule, we compare every item with the term to be found out.

26. Pipes and Cistern

Pipes and Cistern1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.Outlet:A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

2. If a pipe can fill a tank in x hours, then:

part filled in 1 hour =1

.x

3. If a pipe can empty a tank in y hours, then:

part emptied in 1 hour =1

.y

4. If a pipe can fill a tank in x hours and another pipe can empty the full tank in yhours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =1

-1

.x y

5. If a pipe can fill a tank in x hours and another pipe can empty the full tank in yhours (where y > x), then on opening both the pipes, then

the net part emptied in 1 hour =1

-1

.y x

27. Boats and Streams

Boats and Streams

1. Downstream/Upstream:In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.

2. If the speed of a boat in still water is u km/hr and the speed of the stream is vkm/hr, then:Speed downstream = (u + v) km/hr.Speed upstream = (u - v) km/hr.

3. If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water =1

(a + b) km/hr.2

Rate of stream =1

(a - b) km/hr.2

28. Alligation or Mixture

29. Logarithm

Logarithm1. Logarithm:

If a is a positive real number, other than 1 and am = x, then we write:m = logax and we say that the value of log x to the base a is m.

Examples:(i). 103 1000       log10 1000 = 3.(ii). 34 = 81       log3 81 = 4.

(iii). 2-3 =1

      log21

= -3.8 8

(iv). (.1)2 = .01       log(.1) .01 = 2.2. Properties of Logarithms:

1. loga (xy) = loga x + loga y

2. loga

x= loga x - loga yy

3. logx x = 14. loga 1 = 05. loga (xn) = n(loga x)

6. loga x =1

logx a

7. loga x =logb x

=log x

.logb a log a

3. Common Logarithms:Logarithms to the base 10 are known as common logarithms.

4. The logarithm of a number contains two parts, namely 'characteristic' and 'mantissa'.Characteristic: The internal part of the logarithm of a number is called itscharacteristic.Case I: When the number is greater than 1.In this case, the characteristic is one less than the number of digits in the left of the decimal point in the given number.Case II: When the number is less than 1.In this case, the characteristic is one more than the number of zeros between the decimal point and the first significant digit of the number and it is negative.Instead of -1, -2 etc. we write 1 (one bar), 2 (two bar), etc.Examples:-

Number

Characteristic

Number

Characteristic

654.24

20.6453

1

26.649

10.06134

2

8.3547

00.00123

3

Mantissa:The decimal part of the logarithm of a number is known is its mantissa. For mantissa, we look through log table.

30. Races and Games

Races and Games1. Races: A contest of speed in running, riding, driving, sailing or rowing is called a race.2. Race Course: The ground or path on which contests are made is called a race course.3. Starting Point: The point from which a race begins is known as a starting point.4. Winning Point or Goal: The point set to bound a race is called a winning point or a goal.5. Winner: The person who first reaches the winning point is called a winner.6. Dead Heat Race: If all the persons contesting a race reach the goal exactly at the same time, the race is

said to be dead heat race.

7. Start: Suppose A and B are two contestants in a race. If before the start of the race, A is at the starting point and B is ahead of A by 12 metres, then we say that 'A gives B, a start of 12 metres'.To cover a race of 100 metres in this case, A will have to cover 100 metres while B will have to cover only (100 - 12) = 88 metres.In a 100 race, 'A can give B 12 m' or 'A can give B a start of 12 m' or 'A beats B by 12 m' means that while A runs 100 m, B runs (100 - 12) = 88 m.

8. Games: 'A game of 100, means that the person among the contestants who scores 100 points first is the winner'.If A scores 100 points while B scores only 80 points, then we say that 'A can give B 20 points'.

31. Stocks and Shares

Stocks and Shares1. Stock Capital:

The total amount of money needed to run the company is called the stock capital.2. Shares or Stock:

The whole capital is divided into small units, called shares or stock.For each investment, the company issues a 'share-certificate', showing the value of each share and the number of shares held by a person.The person who subscribes in shares or stock is called a share holder or stock holder.

3. Dividend:The annual profit distributed among share holders is called dividend.Dividend is paid annually as per share or as a percentage.

4. Face Value:The value of a share or stock printed on the share-certificate is called its Face Value or Nominal Value or Par Value.

5. Market Value:The stock of different companies are sold and bought in the open market through brokers at stock-exchanges. A share or stock is said to be:

i. At premium or Above par, if its market value is more than its face value.ii. At par, if its market value is the same as its face value.iii. At discount or Below par, if its market value is less than its face value.

Thus, if a Rs. 100 stock is quoted at premium of 16, then market value of the stock = Rs.(100 + 16) = Rs. 116.Likewise, if a Rs. 100 stock is quoted at a discount of 7, then market value of the stock = Rs. (100 -7) = 93.

6. Brokerage:The broker's charge is called brokerage.(i)  When stock is purchased, brokerage is added to the cost price.(ii) When stock is sold, brokerage is subtracted from the selling price.Remember:

i. The face value of a share always remains the same.ii. The market value of a share changes from time to time.iii. Dividend is always paid on the face value of a share.iv. Number of shares held by a person

= Total Investment = Total Income =Total Face

Value .

Investment in 1 share Income from 1 share Face of 1 share7. Thus, by a Rs. 100, 9% stock at 120, we mean that:

i. Face Value of stock = Rs. 100.ii. Market Value (M.V) of stock = Rs. 120.iii. Annual dividend on 1 share = 9% of face value = 9% of Rs. 100 = Rs. 9.iv. An investment of Rs. 120 gives an annual income of Rs. 9.v. Rate of interest p.a = Annual income from an investment of Rs. 100

=9

x 100 % = 71

%.120 2

32. Probability

Probability1. Experiment:

An operation which can produce some well-defined outcomes is called an experiment.2. Random Experiment:

An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.Examples:

i. Rolling an unbiased dice.ii. Tossing a fair coin.iii. Drawing a card from a pack of well-shuffled cards.iv. Picking up a ball of certain colour from a bag containing balls of different colours.

Details:v. When we throw a coin, then either a Head (H) or a Tail (T) appears.vi. A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the

outcome is the number that appears on its upper face.vii. A pack of cards has 52 cards.

It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.Cards of spades and clubs are black cards.Cards of hearts and diamonds are red cards.There are 4 honours of each unit.There are Kings, Queens and Jacks. These are all called face cards.

3. Sample Space:When we perform an experiment, then the set S of all possible outcomes is called the sample space.Examples:

1. In tossing a coin, S = {H, T}2. If two coins are tossed, the S = {HH, HT, TH, TT}.3. In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.

Event:Any subset of a sample space is called an event.

Probability of Occurrence of an Event:Let S be the sample and let E be an event.Then, E   S.

 P(E) =n(E)

.n(S)

Results on Probability:i. P(S) = 1ii. 0   P (E)   1iii. P( ) = 0iv. For any events A and B we have : P(A   B) = P(A) + P(B) - P(A   B)v. If A denotes (not-A), then P(A) = 1 - P(A).

33. True Discount

True DiscountIMPORTANT CONCEPTSSuppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to R. 156 in 4 years. So, the payment of Rs. now will clear off the debt of Rs. 156 due 4 years hence. We say that:Sum due = Rs. 156 due 4 years hence;Present Worth (P.W.) = Rs. 100;True Discount (T.D.) = Rs. (156 - 100) = Rs. 56 = (Sum due) - (P.W.)We define: T.D. = Interest on P.W.;     Amount = (P.W.) + (T.D.)Interest is reckoned on P.W. and true discount is reckoned on the amount.

IMPORTANT FORMULAELet rate = R% per annum and Time = T years. Then,

1.   P.W. =100 x Amount

=100 x T.D.

100 + (R x T) R x T

2.   T.D. =(P.W.) x R x T

=Amount x R x T

100 100 + (R x T)

3.   Sum =(S.I.) x (T.D.)

(S.I.) - (T.D.)4. (S.I.) - (T.D.) = S.I. on T.D.

5.   When the sum is put at compound interest, then P.W. =

Amount

1 +R T

100

34. Banker's Discount

Banker's Discount

IMPORTANT CONCEPTSBanker's Discount:Suppose a merchant A buys goods worth, say Rs. 10,000 from another merchant B at a credit of say 5 months. Then, B prepares a bill, called the bill of exchange. A signs this bill and allows B to withdraw the amount from his bank account after exactly 5 months.The date exactly after 5 months is called nominally due date. Three days (known as grace days) are added to it get a date, known as legally due date.Suppose B wants to have the money before the legally due date. Then he can have the money from the banker or a broker, who deducts S.I. on the face vale (i.e., Rs. 10,000 in this case) for the period from the date on which the bill was discounted (i.e., paid by the banker) and the legally due date. This amount is know as Banker's Discount (B.D.).Thus, B.D. is the S.I. on the face value for the period from the date on which the bill was discounted and the legally due date.Banker's Gain (B.G.) = (B.D.) - (T.D.) for the unexpired time.Note: When the date of the bill is not given, grace days are not to be added.IMPORTANT FORMULAE1. B.D. = S.I. on bill for unexpired time.

2.   B.G. = (B.D.) - (T.D.) = S.I. on T.D. =(T.D.)2

P.W.3.   T.D. P.W. x B.G.

4.   B.D. =Amount x Rate x Time

100

5.   T.D. =Amount x Rate x Time

100 + (Rate x Time)

6.   Amount =B.D. x T.D.

B.D. - T.D.

7.   T.D. =B.G. x 100

Rate x Time

35. Odd Man Out and Series

II. Data Interpretation Questions and Answers

1. Table Charts

Directions to SolveStudy the following table and answer the questions based on it.

Expenditures of a Company (in Lakh Rupees) per Annum Over the given Years.

YearItem of Expenditure

Salary Fuel and Transport Bonus Interest on Loans Taxes

1998 288 98 3.00 23.4 83

1999 342 112 2.52 32.5 108

2000 324 101 3.84 41.6 74

2001 336 133 3.68 36.4 88

2002 420 142 3.96 49.4 98

1.  What is the average amount of interest per year which the company had to pay during this period?

A. Rs. 32.43 lakhs B. Rs. 33.72 lakhs

C. Rs. 34.18 lakhs D. Rs. 36.66 lakhs

Answer & Explanation

Answer: Option DExplanation:Average amount of interest paid by the Company during the given period

    = Rs.23.4 + 32.5 + 41.6 + 36.4 + 49.4

lakhs

5

    = Rs.183.3 lakh

s5

    = Rs. 36.66 lakhs.View Answer Workspace Report Discuss in Forum

2. Bar Charts

Directions to SolveThe bar graph given below shows the sales of books (in thousand number) from six branches of a publishing company during two consecutive years 2000 and 2001.

Sales of Books (in thousand numbers) from Six Branches - B1, B2, B3, B4, B5 and B6 of a publishing Company in 2000 and 2001.

1.  What is the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years?

A. 2:3 B. 3:5

C. 4:5 D. 7:9

Answer & Explanation

Answer: Option DExplanation:Required ratio

=(75 + 65)

=140

=7

.(85 + 95) 180 9

View Answer Workspace Report Discuss in Forum

3. Pie Charts

Directions to SolveThe following pie-chart shows the percentage distribution of the expenditure incurred in publishing a book. Study the pie-chart and the answer the questions based on it.

Various Expenditures (in percentage) Incurred in Publishing a Book

1.  If for a certain quantity of books, the publisher has to pay Rs. 30,600 as printing cost, then what will be amount of royalty to be paid for these books?

A. Rs. 19,450 B. Rs. 21,200

C. Rs. 22,950 D. Rs. 26,150

Answer & Explanation

Answer: Option CExplanation:

Let the amount of Royalty to be paid for these books be Rs. r.

Then, 20 : 15 = 30600 : r           r = Rs.30600 x 15

= Rs. 22,950.20

View Answer Workspace Report Discuss in Forum

4. Line Charts

Directions to SolveStudy the following line graph and answer the questions.

Exports from Three Companies Over the Years (in Rs. crore)

1.  For which of the following pairs of years the total exports from the three Companies together are equal?

A. 1995 and 1998 B. 1996 and 1998

C. 1997 and 1998 D. 1995 and 1996

Answer & Explanation

Answer: Option DExplanation:Total exports of the three Companies X, Y and Z together, during various years are:In 1993 = Rs. (30 + 80 + 60) crores = Rs. 170 crores.In 1994 = Rs. (60 + 40 + 90) crores = Rs. 190 crores.In 1995 = Rs. (40 + 60 + 120) crores = Rs. 220 crores.In 1996 = Rs. (70 + 60 + 90) crores = Rs. 220 crores.In 1997 = Rs. (100 + 80 + 60) crores = Rs. 240 crores.In 1998 = Rs. (50 + 100 + 80) crores = Rs. 230 crores.In 1999 = Rs. (120 + 140 + 100) crores = Rs. 360 crores.Clearly, the total exports of the three Companies X, Y and Z together are same during the years 1995 and 1996.View Answer Workspace Report Discuss in Forum

B. Reasoning:

i. Logical Reasoning Questions and Answers

1. Number Series

2. Letter and Symbol Series

3. Verbal Classification

4. Essential Part

5. Analogies

6. Artificial Language

7. Matching Definitions

8. Making Judgments

9. Verbal Reasoning

10. Logical Problems

11. Logical Games

12. Analyzing Arguments

13. Statement and Assumption

14. Course of Action

15. Statement and Conclusion

16. Theme Detection

17. Cause and Effect

18. Statement and Argument

19. Logical Deduction

IntroductionIn Logic, any categorical statement is termed as the Proposition.A Proposition (or a categorical statement) is a statement that asserts that either a part of, or the whole of, one set of objects - the set identified by the subject term in the sentence expressing that statement - either is included in, or is excluded from, another set - the set identified by the predicate term in that sentence.The standard form of a proposition is :Quantifier + Subject + Copula + PredicateThus, the proposition consists of four parts :1. Quantifier: The words 'all', 'no' and 'some' are called quantifiers because they specify a quantity 'All' and 'no' are universal quantifiers because they refer to every object in a certain set, while the quantifier 'some' is a particular quantifier because it refers to at least one existing object in a certain set.2. Subject (denoted by 'S'): The subject is that about which something is said.

3. Predicate (denoted by 'P'): The predicate is the part of the proposition denoting that which is affirmed or denied about the subject.4. Copula : The copula is that part of the proposition which denotes the relation between the subject and the predicate.Examples:

Four-Fold Classification of Propositions :A proposition is said to have a universal quantity if it begins with a universal quantifier, and a particular quantity if it begins with a particular quantifier. Besides, propositions which assert something about the inclusion of the whole or a part of one set in the other are said to have affirmative quality, while those which deny the inclusion of the whole or a part of one set in the other are said to have a negative quality. Also, a term is distributed in a proposition if it refers to all members of the set of objects denoted by that term. Otherwise, it is said to be undistributed. Based on the above facts, propositions can be classified into four types :1. Universal Affirmative Proposition (denoted by A): It distributes only the subject i.e. the predicate is not interchangeable with the subject while maintaining the validity of the proposition.e.g., All snakes are reptiles. This is proposition A since we cannot say 'All reptiles are snakes'.2. Universal Negative Proposition (denoted by E): It distributes both the subject and the predicate i.e. an entire class of predicate term is denied to the entire class of the subject term, as in the proposition.e.g., No boy is intelligent.3.Particular Affirmative Proposition (denoted by I): It distributes neither the subject nor the predicate.e.g.,Some men are foolish. Here, the subject term 'men' is used not for all but only for some men and similarly the predicate term 'foolish' is affirmed for a part of subject class. So, both are undistributed.

4. Particular Negative Proposition (denoted by O): It distributes only the predicate. e.g., Some animals are not wild. Here, the subject term 'animals' is used only for a part of its class and hence is undistributed while the predicate term 'wild' is denied in entirety to the subject term and hence is distributed. These facts can be summarized as follows :

Statement Form Quantity Quality Distributed

(A): All S is P. Universal Affirmative S only

(E): No S is P. Universal Negative Both S and P

(I): Some S is P. Particular Affirmative Neither S nor P

(O): Some S is not P Particular Negative P only

Logical Deduction - Important Formulas

Logical Deduction:

The phenomenon of deriving a conclusion from a single proposition or a set of given propositions, is known as logical deduction. The given propositions are also referred to as the premises.Two Inferential Processes of Deduction :I. Immediate Deductive Inference :Here, conclusion is deduced from one of the given propositions, by any of the three ways -conversion, obversion and contraposition.1. Conversion: The Conversion proceeds with interchanging the subject term and the predicate term i.e. the subject term of the premise becomes the predicate term of the conclusion and the predicate term of the premise becomes the subject of the conclusion. The given proposition is called convertend, whereas the conclusion drawn from it is called its converse.Table of Valid Conversions

Convertend Converse

A: All S is PEx. All pins are tops.

I: Some P is SSome tops are pins.

E: No S is P.Ex. No fish is whale.

E: No P is S.No whale is fish.

I: Some S is P.Ex. Some boys are poets.

I: Some P is S.Some poets are boys.

O: Some S is not P. No valid conversion

Note that in a conversion, the quality remains the same and the quantity may change.2. Obversion: In obversion, we change the quality of the proposition and replace the predicate term by its complement.Table of Valid Obversions

Obvertend Obverse

A: All birds are mammals. E: No birds are non-mammals.

E: No poets are singers. A: All poets are non-singers.

I: Some nurses are doctors. O: Some nurses are not non-doctors.

O: some politicians are not statesmen. I: Some politicians are non-statesmen.

3. Contraposition: To obtain the contrapositive of a statement, we first replace the subject and predicate terms in the proposition and then exchange both these terms with their complements.Table of Valid Contrapositions

Proposition Contrapositive

A: All birds are mammals. A: All non-mammals are non-birds.

I: Some birds are mammals. I: Some non-mammals are non-birds.

Note: The valid converse, obverse or contrapositive of a given proposition always logically follows from the proposition.II. Mediate Deductive Inference (SYLLOGISM): First introduced by Aristotle, a Syllogism is a deductive argument in which conclusion has to be drawn from two propositions referred to as the premises.Example:1. All lotus are flowers.2. All flowers are beautiful.3. All lotus are beautiful.Clearly, the propositions 1 and 2 are the premises and the proposition 3, which follows from the first two propositions, is called the conclusion.Term : In Logic, a term is a word or a combination of words, which by itself can be used as a subject or predicate of a proposition.Syllogism is concerned with three terms :1. Major Term : It is the predicate of the conclusion and is denoted by P (first letter of 'Predicate').2. Minor Term: It is the subject of the conclusion and is denoted by S (first letter of 'Subject').3. Middle Term: It is the term common to both the premises and is denoted by M (first letter of 'Middle').

Example:Premises:1. All dogs are animals.2. All tigers are dogs.Conclusion :All tigers are animals.Here 'animals' is the predicate of the conclusion and so,.it is the major term. P.'Tigers' is the subject of the conclusion and so, it is the minor term, S.'Dogs' is the term common to both the premises and so, it is the middle term, M.Major And Minor Premises : Of the two premises, the major premise is that in which the middle term is the subject and the minor premise is that in which the middle term is the predicate.RULES FOR DERIVING CONCLUSION FROM TWO GIVEN PREMISES:1. The conclusion does not contain the middle term.Example.Statements :1. All men are girls.2. Some girls are students.Conclusions :1. All girls are men.2. Some girls are not students.Since both the conclusions 1 and 2 contain the middle term 'girls', so neither of them can follow.

2. No term can be distributed in the conclusion unless it is distributed in the premises.Example.Statements :1. Some dogs are goats.2. All goats are cows.Conclusions :1. All cows are goats.2. Some dogs are cows.Statement 1 is an I-type proposition which distributes neither the subject nor the predicate.Statement 2 is an A type proposition which distributes the subject i.e. 'goats' only.Conclusion 1 is an A-type proposition which distributes the subject 'cow' only Since the term 'cows' is distributed in conclusion 1 without being distributed in the premises, so conclusion 1 cannot follow.

3. The middle term (M) should he distributed at least once in the premises. Otherwise, the conclusion cannot follow.For the middle term to be distributed in a premise.(i) M must be the subject if premise is an A proposition.(ii) M must be subject or predicate if premise is an E proposition.(iii) M must be predicate if premise is an O proposition.Note that in an I proposition, which distributes neither the subject nor the predicate, the middle term cannot be distributed.Example.Statements :1. All fans are watches.2. Some watches are black.Conclusions :1. All watches are fans.2. Some fans are black.In the premises, the middle term is 'watches'. Clearly, it is not distributed in the first premise which is an A proposition as it does not form its subject. Also, it is not distributed in the second premise which is an I proposition. Since the middle term is not distributed even once in the premises, so no conclusion follows.

4. No conclusion follows(a) if both the premises are particularExample.

Statements :1. Some books are pens.2. Some pens are erasers.Conclusions:1. All books are erasers.2. Some erasers are books.Since both the premises are particular, so no definite conclusion follows.(b) if both the premises are negative.Example.Statements :1. No flower is mango.2. No mango is cherry.Conclusions :1. No flower is cherry.2. Some cherries are mangoes. Since both the premises are negative, neither conclusion follows.(c) if the major premise is particular and the minor premise is negative.Example.Statements:1. Some dogs are bulls.2. No tigers are dogs.Conclusions:1. No dogs are tigers.2. Some bulls are tigers.Here, the first premise containing the middle term 'dogs' as the subject is the major premise and the second premise containing the middle term 'dogs' as the predicate is the minor premise. Since the major premise is particular and the minor premise is negative, so no conclusion follows.

5. If the middle term is distributed twice, the conclusion cannot be universal.Example.Statements :1. All fans are chairs.2. No tables are fans.Conclusions:1. No tables are chairs.2. Some tables are chairs.Here, the first premise is an A proposition and so, the middle term 'fans' forming the subject is distributed. The second premise is an E proposition and so, the middle term 'fans' forming the predicate is distributed. Since the middle term is distributed twice, so the conclusion cannot be universal.

6. If one premise is negative, the conclusion must be negative.Example.Statements:1. All grasses are trees.2. No tree is shrub.Conclusions:1. No grasses are shrubs.2. Some shrubs are grasses.Since one premise is negative, the conclusion must be negative. So, conclusion 2 cannot follow.

7. If one premise is particular, the conclusion must be particular.Example.Statements:1. Some boys are thieves.2. All thieves are dacoits.Conclusions :1. Some boys are dacoits.2. All dacoits are boys.

Since one premise is particular, the conclusion must be particular. So, conclusion 2 cannot follow.

8. If both the premises are affirmative, the conclusion must be affirmative.Example.Statements :1. All women are mothers.2. All mothers are sisters.Conclusions :1. All women are sisters.2. Some women are not sisters.Since both the premises are affirmative, the conclusion must be affirmative. So, conclusion 2 cannot follow.

9. If both the premises are universal, the conclusion must be universal.Complementary pair:A pair of contradictory statements i.e. a pair of statements such that if one is true, the other is false and when no definite conclusion can be drawn, either of them is bound to follow, is called a complementary pair. E and I-type propositions together form a complementary pair and usually either of them follows, in a case where we cannot arrive at a definite conclusion, using the rules of syllogism.Let us study the various possible cases and draw all possible inferences in each case, along with verification through Venn diagrams.Case 1: All men are boys. All boys are students.Immediate Deductive Inferences:The converse of first premise i.e. 'Some boys are men' and the converse of second premise i.e. 'Some students are boys' both hold.Mediate Deductive Inferences:Since both the premises are universal and affirmative, the conclusion must be universal affirmative. Also, the conclusion should not contain the middle term. So, it follows that 'All men are students'. The converse of this conclusion i.e. 'Some students are men' also holds.

Case 2: All birds are animals. All fishes are animals.Immediate Deductive Inferences:The converse of first premise i.e. 'Some animals are birds' and the converse of second premise i.e. 'Some animals are fishes' both hold.Mediate Deductive Inferences:Both, being A-type propositions, distribute subject only. Thus, the middle term 'animals' is not distributed even once in the premises. So, no definite conclusion follows.

Case 3: All puppets are dolls. Some dolls are rattles.Immediate Deductive Inferences:The converse of the first premise i.e. 'Some dolls are puppets' and the converse of the second premise i.e. 'Some rattles are dolls', both hold.Mediate Deductive Inferences:First premise, being an A-type proposition, distributes the subject only while the second premise, being an I-type proposition, distributes neither subject nor predicate. Since the middle term 'dolls' is not distributed even once in the premises, so no definite conclusion can be drawn.

Case 4: Some writers are players. All players are musicians.Immediate Deductive Inferences :The converse of the first premise i.e. 'Some players are writers' and the converse of the second premise i.e. 'Some musicians are players', both hold.Mediate Deductive Inferences:Since one premise is particular, the conclusion must be particular and should not contain the middle term. So, it follows that 'Some writers are musicians'. The converse of this conclusion i.e. 'Some musicians are writers' also holds.

Case 5: All boxes are toys. Some boxes are clips.Immediate Deductive Inferences :The converse of the first premise i.e. 'Some toys are boxes' and the converse of the second premise i.e. 'Some clips are boxes', both hold.Mediate Deductive Inferences:Since one premise is particular, the conclusion must be particular and should not contain the middle term. So, it follows that 'Some toys are clips'. The converse of this conclusion i.e. 'Some clips are toys' also holds.

Case 6: All buses are vans. Some cycles are vans.Immediate Deductive Inferences:The converse of the first premise i.e. 'Some vans are buses' and the converse of the second premise i.e. 'Some vans are cycles', both hold.Mediate Deductive Inferences:First premise, being an A-type proposition, distributes subject only and the second premise, being an I-type proposition, distributes neither subject nor predicate. So, the middle term 'vans' is not distributed even once in the premises. Hence, no definite conclusion can be drawn.

Case 7: Some radios are cameras. Some cameras are statues.Immediate Deductive Inferences:The converse of the first premise i.e. 'Some cameras are radios' and the converse of the second premise i.e. 'Some statues are cameras', both hold.Mediate Deductive Inferences :Since both premises are particular, no definite conclusion follows.

Case 8: All cakes are candies. No candy is pastry.Immediate Deductive Inferences:The converse of the first premise i.e. 'Some candies are cakes' and the converse of the second premise i.e. 'No pastry is candy', both hold.Mediate Deductive Inferences:Since both premises are universal, the conclusion must be universal. Since one premise is negative, the conclusion must be negative. So, it follows that 'No cake is pastry'. The converse of this conclusion i.e. 'No pastry is cake' also holds.

Case 9: No coin is ring. All rings are bangles.Immediate Deductive Inferences :The converse of the first premise i.e. 'No ring is coin' and the converse of the second premise i.e.'Some bangles are rings', both hold.Mediate Deductive Inferences:First premise, being an E-type proposition, distributes both the subject and the predicate.Second premise, being an A-type proposition, distributes the subject. Thus, the middle term 'ring' is distributed twice in the premises. So, the conclusion cannot be universal. Also, since one premise is negative, the conclusion must be negative. Thus, the conclusion must be particular negative i.e. O-type. So, it follows that 'some bangles are not coins'.

Case 10: Some lamps are candles. No candle is bulb.Immediate Deductive Inferences :The converse of the first premise i.e. 'Some candles are lamps' and the converse of the second premise i.e. 'No bulb is candle', both hold.Mediate Deductive Inferences:Since one premise is particular and the other negative, the conclusion must be particular negative i.e. O-type, So, it follows that 'Some lamps are not bulbs'.

Important Points To Remember:While deriving logical conclusions, always remember that the following conclusions hold :1. The converse of each of the given premises;2. The conclusion that directly follows from the given premises in accordance with the rules of syllogism;3. The converse of the derived conclusions.

ii. Verbal Reasoning Questions and Answers

1. Logical Sequence of Words

In this type of question, some words are given. You have to arrange these words in a meaningful order. The order may be according to age, size and need etc.Example 1:Arrange the following words in a meaningful order.

1. Death2. Marriage

3. Education

4. Birth 5. Funeral  

(A) 5, 1, 2, 3, 4(B) 4, 2, 3, 1, 5(C) 4, 3, 2, 5, 1(D) 4, 3, 2, 1, 5Answer is Option DExplanation:First of all a man is born then he takes education; after this he is married. Then after sometimes he dies. After death the order is of Funeral. So the correct order is 4, 3, 2, 1 and 5.

Example 2:Arrange the following words in a logical sequence.1. Grass 2. Curd 3. Milk

4. Cow 5. Butter  

(A) 1, 2, 3, 4, 5(B) 2, 3, 4, 5, 1(C) 4, 1, 3, 2, 5(D) 5, 4, 3, 2, 1Answer is Option CExplanation:We know that cow eats grass and then gives milk. With the milk, curd is made and then from curd, butter is made.Hence logical sequence is Cow, Grass, Milk, Curd, Butter.

2. Blood Relation Test

Introduction:The questions which are asked in this section depend upon Relation. You should have a sound knowledge of the blood relation in order to solve the questions.To remember easily the relations may be divided into two sides as given below:1. Relations of Paternal side:

1. Father's father → Grandfather2. Father's mother → Grandmother3. Father's brother → Uncle4. Father's sister → Aunt5. Children of uncle → Cousin6. Wife of uncle → Aunt7. Children of aunt → Cousin8. Husband of aunt → Uncle

2. Relations of Maternal side:

1. Mother's father → Maternal grandfather2. Mother's mother → Maternal grandmother3. Mother's brother Maternal uncle4. Mother's sister → Aunt5. Children of maternal uncle → Cousin6. Wife of maternal uncle → Maternal aunt

Relations from one generation to next:

Differenct types of questions with explanation:Type 1:If A + B means A is the mother of B; A x B means A is the father of B; A $ B means A is the brother of B and A @ B means A is the sister of B then which of the following means P is the son of Q?(A) Q + R @ P @ N       (B) Q + R * P @ N(C) Q x R $ P @ N        (D) Q x R $ P $ NSolution: (D)Q x R = Q is the mother of R [-Q, ±R]R $ P = R is the brother of P [+ R, ±P]P $ N = P is the brother of N [+ P, ±N]Therefore P is the son of Q.

Type 2:A has 3 children. B is the brother of C and C is the sister of D, E who is the wife of A is the mother of D. There is only one daughter of the husband of E. what is the relation between D and B?Solution: With the chart

Therefore, D is a boy because there is only one daughter of E.Hence, B is the brother of D.

Type 3:Pointing to a photograph, Rekha says to Lalli, "The girl in the photo is the second daughter of the wife of only son of the grandmother of my younger sister." How this girl of photograph is related to Rekha?Solution:First Method - By Generating Charts:

Second method:1. Grandmother of younger sister of Rekha → Grandmother of Rekha2. Wife of only son of grandmother → Mother of Rekha3. Younger daughter of the mother → Younger sister.

Note: While solving the question (+) can be used for male and (-) can be used for female.

3. Syllogism

Introduction:

The questions which are asked in this section contain two or more statements and these statements are followed by two or more conclusions. You have to find out which of the conclusions logically follow from the given statements. The statements have to be taken true even if they seem to be at variance from the commonly known facts.For such questions, you can take the help of Venn Diagrams. On the basis of the given statements, you should draw all the possible diagrams, and then derive the solution from each of these diagrams separately. Finally, the answer common to the all the diagrams is taken.

Example 1:Statements:

1. All dogs are asses.2. All asses are bulls.

Conclusions:1. Some dogs are not bulls.2. Some bulls are dogs.3. All bulls are dogs.4. All dogs are bulls.

Solution:On the basis of both statements, the following one diagram is possible.

From the diagram it is clear that (2) and (4) conclusions logically follow.

Example 2:Statements:

1. Some dogs are asses.2. Some asses are bulls.

Conclusions:1. Some asses are not dogs.2. Some dogs are bulls.

Solution:From these given statements the following diagrams are possible:

From the diagram neither (1) nor (2) conclusions follow.

4. Series Completion

5. Cause and Effect

Introduction:In this type of questions two statements are given. Out of these two statements one may be the cause and other the effect or either these two may be independent causes any effect or independent effects of any cause etc.The following examples will you a clear cut idea to solve this type of problems.Example 1:Statements:

I. Ram's father was ill.II. Ram brought medicine after consulting the doctor.

Answer with Explanation:As Ram's father was ill, he brought medicine on the advice of doctor.Therefore, I statement is the cause while II statement is the effect.

Example 2:Statements:

I. The Central Government has recently declared to finish the rebate on farming.II. The Central Government faces financial loss on account of giving rebate on farming for the last few years.

Answer with Explanation:As the Central Government faced financial loss on accounts of giving rebate on farming for the last few years, therefore, they declared to finish the rebate of farming.Hence statement II is the cause while statement I is the effect.

6. Dice

Introduction:

Dice is a cube. In cube there are 6 faces. Some important points are given below:1. There are 6 faces in the cube - ABCG, GCDE, DEFH, BCDH, AGEF and ABHF.

2. Always four faces are adjacent to one face.3. Opposite of ABCG is DEFH and so on.4. CDEG is the upper face of the cube.5. ABHF is the bottom of the cube.

There are certain rules with the help of these rules question on dice can easily determined.

Rule No. 1:Two opposite faces cannot be adjacent to one another.Example:

Two different positions of a dice are shown below. Which number will appear on the face opposite to the face with number 4?

Solution:Faces with four numbers 6, 2, 5 and 3 are adjacent of to the face with No. 4.Hence the faces with no. 6, 2, 5 and 3 cannot be opposite to the face with no. 4.Therefore the remaining face with no.1 will be the opposite of the face with no. 4.

Rule No. 2:If two different positions of a dice are shown and one of the two common faces is in the same position then of the remaining faces will be opposite to each other.Example:Two different positions of a dice are shown below.

Here in both shown positions two faces 5 and 3 are common.The remaining faces are 2 and 4.Hence the number on the face opposite to the face with number 2 is 4.

Rule No. 3:If in two different positions of dice, the position of a common face be the same, then each of the opposite faces of the remaining faces will be in the same position.Example:

Here in both positions of common (3) is same.Therefore, opposite of 5 is 6 and opposite of 4 is 2.

Rule No. 4:If in two different positions of a dice, the position of the common face be not the same, then opposite face of the common face will be that which is not shown on any face in these two positions. Besides, the opposite faces of the remaining faces will not be the same.Example:

Here in two positions of a dice the face with number 1 is not in the same position.The face with number 6 is not shown.Hence the face opposite to the face with number 1 is 6.Besides the opposite face of 3 will be the face with number 2 and opposite face to face 5 will be the face with number 1.

7. Venn Diagrams

Introduction:The main aim of this section is to test your ability about the relation between some items of a group by diagrams. In these questions some figures of circles and some words are given. You have to choose a figure which represents the given words.Some critical examples are given below:Example 1:If all the words are of different groups, then they will be shown by the diagram as given below.Dog, Cow, Horse

All these three are animals but of different groups, there is no relation between them. Hence they will be represented by three different circles.

Example 2:If the first word is related to second word and second word is related to third word. Then they will be shown by diagram as given below.Unit, Tens, Hundreds

Ten units together make one Tens or in one tens, whole unit is available and ten tens together make one hundreds.

Example 3:If two different items are completly related to third item, they will be shown as below.Pen, Pencil, Stationery

Example 4:If there is some relation between two items and these two items are completely related to a third item they will be shown as given below.Women, Sisters, Mothers

Some sisters may be mothers and vice-versa. Similarly some mothers may not be sisters and vice-versa. But all the sisters and all the mothers belong to women group.

Example 5:Two items are related to a third item to some extent but not completely and first two items totally different.

Students, Boys, Girls

The boys and girls are different items while some boys may be students. Similarly among girls some may be students.

Example 6:All the three items are related to one another but to some extent not completely.Boys, Students, Athletes

Some boys may be students and vice-versa. Similarly some boys may be athletes and vice-versa. Some students may be athletes and vice-versa.

Example 7:Two items are related to each other completely and third item is entirely different from first two.Lions, Carnivorous, Cows

All the lions are carnivorous but no cow is lion or carnivorous.

Example 8:First item is completely related to second and third item is partially related to first and second item.Dogs, Animals, Flesh-eaters

All the dogs are belonging to animals but some dogs are flesh eater but not all.

Example 9:First item is partially related to second but third is entirely different from the first two.Dogs, Flesh-eaters, Cows

Some dogs are flesh-eaters but not all while any dog or any flesh-eater cannot be cow.

8. Cube and Cuboid

Intrducion:

In a cube or a cuboid there are six faces in each. In a cube length, breadth and height are same while in cuboid these are different. In a cube the number of unit cubes = (side)3. In cuboid the number of unit cube = (l x b x h).

Example:A cube of each side 4 cm, has been painted black, red and green on pars of opposite faces. It is then cut into small cubes of each side 1 cm.

The following questions and answers are based on the information give above:1. How many small cubes will be there ?Total no. of cubes = (sides)3 = (4)3 = 642. How many small cubes will have three faces painted ?From the figure it is clear that the small cube having three faces coloured are situated at the corners of the big cube because at these corners only three faces of the big cube meet.Therefore the required number of such cubes is always 8, because there are 8 corners.3. How many small cubes will have only two faces painted ?From the figure it is clear that to each edge of the big cube 4 small cubes are connected and two out of them are situated at the corners of the big cube which have all three faces painted.Thus, to edge two small cubes are left which have two faces painted. As the total no. of edges in a cube are 12.Hence the no. of small cubes with two faces coloured = 12 x 2 = 24(or)No. of small cubes with two faces coloured = (x - 2) x No. of edgeswhere x = (side of big cube / side of small cube)

4. How many small cubes will have only one face painted ?

The cubes which are painted on one face only are the cubes at the centre of each face of the big cube.Since there are 6 faces in the big cube and each of the face of big cube there will be four small cubes.Hence, in all there will be 6 x 4 = 24 such small cubes (or) (x - 2)2 x 6.

5. How many small cubes will have no faces painted ?No. of small cubes will have no faces painted = No. of such small cubes= (x - 2)3 [ Here x = (4/1) = 4 ]= (4 - 2)3

= 8.6. How many small cubes will have only two faces painted in black and green and all other faces unpainted ?There are 4 small cubes in layer II and 4 small cubes in layer III which have two faces painted green and black.Required no. of such small cubes = 4 + 4 = 8.7. How many small cubes will have only two faces painted green and red ?No. of small cubes having two faces painted green and red = 4 + 4 = 8.8. How many small cubes will have only two faces painted black and red ?No. of small cubes having two faces painted black and red = 4 + 4 = 8.9. How many small cubes will have only black painted ?No. of small cubes having only black paint. There will be 8 small cubes which have only black paint. Four cubes will be form one side and 4 from the opposite side.10. How many small cubes will be only red painted ?No. of small cubes having only red paint = 4 + 4 = 8.

11. How many small cubes will be only green painted ?No. of small cubes having only green paint = 4 + 4 = 8.12. How many small cubes will have at least one face painted ?No. of small cubes having at least one face painted = No. of small cubes having 1 face painted + 2 faces painted + 3 faces painted= 24 + 24 + 8

= 56.13. How many small cubes will have at least two faces painted ?No. of small cubes having at least two faces painted = No. of small cubes having two faces painted + 3 faces painted= 24 + 8= 32.

Exmp:

Directions to SolveThe following questions are based on the information given below:

1. A cuboid shaped wooden block has 6 cm length, 4 cm breadth and 1 cm height.2. Two faces measuring 4 cm x 1 cm are coloured in black.3. Two faces measuring 6 cm x 1 cm are coloured in red.4. Two faces measuring 6 cm x 4 cm are coloured in green.5. The block is divided into 6 equal cubes of side 1 cm (from 6 cm side), 4 equal cubes of

side 1 cm(from 4 cm side).1.  How many cubes having red, green and black colours on at least one side of the cube

will be formed ?

A. 16 B. 12

C. 10 D. 4

Answer & Explanation

Answer: Option DExplanation:

Such cubes are related to the corners of the cuboid. Since the number of corners of the cuboid is 4.Hence, the number of such small cubes is 4.View Answer Workspace Report Discuss in Forum

9. Analogy

Introduction:Analogy means similarity. In this type of questions, two objects related in some way are given and third object is also given with four or five alternatives. You have to find out which one of the alternatives bears the same relation with the third objects as first and second objects are related.Example 1:Curd : Milk :: Shoe : ?(A) Leather      (B) Cloth(C) Jute           (D) SilverAnswer: Option AAs curd is made from milk similarly shoe is made from leather.

Example 2:Calf : Piglet :: Shed : ?(A) Prison      (B) Nest(C) Pigsty      (D) DenAnswer: Option CCalf is young one of the cow and piglet is the young of Pig. Shed is the dwelling place of cow. Similarly Pigsty is the dwelling place of pig.

Example 3:Malaria : Mosquito :: ? : ?(A) Poison : Death      (B) Cholera : Water(C) Rat : Plague          (D) Medicine : DiseaseAnswer: Option BAs malaria is caused due to mosquito similarly cholera is cause due to water.

Example 4:ABC : ZYX :: CBA : ?(A) XYZ      (B) BCA(C) YZX      (D) ZXYAnswer: Option ACBA is the reverse of ABC similarly XYZ is the reverse of ZYX.

Example 5:4 : 18 :: 6 : ?(A) 32      (B) 38(C) 11      (D) 37Answer: Option BAs, (4)2 + 2 =18Similarly, (6)2 + 2 = 38.

10. Seating Arrangement

Introduction:

In order to solve seating arrangement questions, first of all diagram should be made. By doing so questions are easily and quickly solved.Example 1:

1. 6 Boys are sitting in a circle and facing towards the centre of the circle.2. Rajeev is sitting to the right of mohan but he is not just at the left of Vijay.3. Suresh is between Babu and Vijay.4. Ajay is sitting to the left of Vijay.

Who is sitting to the left of Mohan ?Solution :

Hence, Babu is sitting to the left of Mohan.

Example 2:1. Eleven students A, B, C, D, E, F, G, H, I, J and K are sitting in first line facing to the teacher.2. D who is just to the left of F, is to the right of C at second place.3. A is second to the right of E who is at one end.4. J is the nearest neighbour of A and B and is to the left of G at third place.5. H is next to D to the right and is at the third place to the right of I.

Who is just in the middle ?Solution :

Hence, I is just in the middle.

Example 3:Siva, Sathish, Amar and Praveen are playing cards. Amar isto the right of Sathish, who is to the right of Siva.Who is to the right of Amar ?Solution :

Hence Praveen is to the right of Amar.

Example 4:1. A, B and C are three boys while R, S and T are three girls. They are sitting such that the boys are facing the

girls.2. A and R are diagonally opposite to each other.3. C is not sitting at any of the ends.4. T is left to R but opposite to C.

(A). Who is sitting opposite to B ?(B). Who is sitting diagonally opposite to B ?Solution :

(A). Hence, R is sitting opposite to B.(B). Hence, S is sitting diagonally opposite to B.

11. Character Puzzles

Introduction:

In this type of questions, a figure or a matrix is given in which some numbers are filled according to a rule. A place is left blank. You have to find out a character (a number or a letter) from the given possible answers which may be filled in the blank space.

Some examples are given below.Example 1:Which number will replace the question mark ?

Solution :From fig. a: 6 + 4 + 8 = 1818 + 2 = 20From fig. b: 7 + 9 + 8 = 2424 + 2 = 26From fig. c: 6 + 5 + 12 = 2323 + 2 = 25Hence the number 25 will replace the question mark.

Example 2:Which number will replace the question mark ?

Solution :From fig. a: (3)2 + (2)2 = 13From fig. b: (4)2 + (8)2 = 80From fig. c: ? = (1)2 + (5)2

? = 1 + 25? = 26Hence the number 26 will replace the question mark.

Example 3:Which number will replace the question mark ?

Solution :From fig. a: 7 x 6 + 3 = 45From fig. b: 5 x 4 + 6 = 26From fig. c: 7 x 3 + 8 = 29Hence the number 29 will replace the question mark.

Example 4:Which number will replace the question mark ?

Solution :From fig. a: 92 + 82 + 72 + 62 = 81 + 64 + 49 + 36 = 230From fig. b: 62 + 72 + 32 + 42 = 36 + 49 + 9 + 16 = 110From fig. c: 92 + 62 + 52 + 42 = 81 + 36 + 25 + 16 = 158Hence the number 158 will replace the question mark.

Example 5:Which number will replace the question mark ?

Solution :(4 + 3)2 = (7)2 = 49(8 + 5)2 = (13)2 = 169(11 + 12)2 = (23)2 = 529(10 + 9)2 = (19)2 = 361Hence the number 361 will replace the question mark.

Example 6:Which number will replace the question mark ?

Solution :From column I: (9 x 5) % 5 = 9From column II: (17 x 4) % 4 = 17From column III: (16 x ?) % 8 = 816? = 64? = 4Hence the number 4 will replace the question mark.

Example 7:Which number will replace the question mark ?

Solution :From fig. a: (8 x 5) - (4 x 3) = 28From fig. b: (12 x 7) - (8 x 9) = 12From fig. c: (5 x 3) - (6 x ?) = 2115 - 6? = 216? = -6? = -1Hence the number -1 will replace the question mark.

Example 8:Which number will replace the question mark ?

Solution :In each row there are 'A', 'B' and 'C'In second row 'A' and 'C' are already thereHence in place of ?, there will be 'B'.From first row : 4A x 6C = 24BFrom third row : 9B x 4C = 36AFrom second row : 5A x ? = 45C? = (45C/5A)? = 9BHence the number 9B will replace the question mark.

Example 9:Which number will replace the question mark ?

Solution :(5)2 = 25(6)2 = 36(4)2 = 16(7)2 = 49Hence the number 49 will replace the question mark.

12. Direction Sense Test

Introduction:

There are four main directions - East, West, North and South as shown below:

There are four cardinal directions - North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.2. At the time of sunset the shadow of an object is always in the east.3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of

sunset it will be towards his right.4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.

Main types of questions are given below:Type 1:Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?Solution:

From third position it is clear he is 4 km from his house and is in North direction.

Type 2:Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?Solution:

Type 3:One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli's shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.Now Lalli's shadow falls to the right of the Juhi. Hence Juhi is facing South.

Type 4:Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

13. Classification

Directions to SolveIn each of the following questions, five words have been given out of which four are alike in some manner, while the fifth one is different. Choose the word which is different from the rest.

1.  Choose the word which is different from the rest.

A. Chicken B. Snake

C. Swan D. Crocodile

E. Frog

Answer & Explanation

Answer: Option AExplanation:All except Chicken can live in water.View Answer Workspace Report Discuss in Forum

14. Data Sufficiency

Directions to SolveIn each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements andGive answer

(A) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question

(B) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question

(C) If the data either in statement I alone or in statement II alone are sufficient to answer the question

(D) If the data given in both statements I and II together are not sufficient to answer the question and

(E) If the data in both statements I and II together are necessary to answer the question.

1.  Question: In which year was Rahul born ?Statements:

I. Rahul at present is 25 years younger to his mother.II. Rahul's brother, who was born in 1964, is 35 years younger to his mother.

A. I alone is sufficient while II alone is not sufficient

B. II alone is sufficient while I alone is not sufficient

C. Either I or II is sufficient

D. Neither I nor II is sufficient

E. Both I and II are sufficient

Answer & Explanation

Answer: Option EExplanation:From both I and II, we find that Rahul is (35 - 25) = 10 years older than his brother, who was born in 1964. So, Rahul was born in 1954.View Answer Workspace Report Discuss in Forum

15. Arithmetic Reasoning

1.  The total of the ages of Amar, Akbar and Anthony is 80 years. What was the total of their ages three years ago ?

A. 71 years

B. 72 years

C. 74 years

D. 77 years

Answer & Explanation

Answer: Option AExplanation:Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.View Answer Workspace Report Discuss in Forum

16. Verification of Truth

1.  A train always has

A. Rails B. Driver

C. Guard D. Engine

Answer & Explanation

Answer: Option DExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

iii. Non Verbal Reasoning Questions and Answers

1. Series

Directions to SolveEach of the following questions consists of five figures marked A, B, C, D and E called the Problem Figures followed by five other figures marked 1, 2, 3, 4 and 5 called the Answer Figures. Select a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.

2. Analogy

Directions to SolveEach of the following questions consists of two sets of figures. Figures A, B, C and D constitute the Problem Set while figures 1, 2, 3, 4 and 5 constitute the Answer Set. There is a definite relationship between figures A and B. Establish a similar relationship between figures C and D by selecting a suitable figure from the Answer Set that would replace the question mark (?) in fig. (D).

3. Classification

Classification

Directions to SolveIn each problem, out of the five figures marked (1), (2), (3), (4) and (5), four are similar in a certain manner. However, one figure is not like the other four. Choose the figure which is different from the rest.

4. Analytical Reasoning

1.  Find the number of triangles in the given figure.

A. 8 B. 10

C. 12 D. 14

Answer & Explanation

Answer: Option DExplanation:The figure may be labelled as shown.

The simplest triangles are AHG, AIG, AIB, JFE, CJE and CED i.e. 6 in number.The triangles composed of two components each are ABG, CFE, ACJ and EGI i.e. 4 in number.The triangles composed of three components each are ACE, AGE and CFD i.e. 3 in number.There is only one triangle i.e. AHE composed of four components.Therefore, There are 6 + 4 + 3 + 1 = 14 triangles in the given figure.View Answer Workspace Report Discuss in Forum

5. Mirror Images

Mirror Images

Directions to SolveIn each of the following questions you are given a combination of alphabets and/or numbers followed by four alternatives (1), (2), (3) and (4). Choose the alternative which is closely resembles the mirror image of the given combination.1.  Choose the alternative which is closely resembles the mirror image of the given

combination.

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option BExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

6. Water Images

Directions to SolveIn each of the following questions, you are given a combination of alphabets and/or numbers followed by four alternatives (1), (2), (3) and (4). Choose the alternative which is closely resembles the water-image of the given combination.1.  Choose the alternative which is closely resembles the water-image of the given

combination.

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option DExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

7. Embedded Images

Directions to SolveIn each of the following questions, you are given a figure (X) followed by four alternative figures (1), (2), (3) and (4) such that figure (X) is embedded in one of them. Trace out the alternative figure which contains fig. (X) as its part.

8. Pattern Completion

Directions to SolveIn each of the following questions, select a figure from amongst the four alternatives, which when placed in the blank space of figure (X) would complete the pattern.

9. Figure Matrix

Directions to SolveIn each of the following questions, find out which of the answer figures (1), (2), (3) and (4)

completes the figure matrix ?1.  Select a suitable figure from the four alternatives that would complete the figure matrix.

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option DExplanation:The third figure in each row comprises of parts which are not common to the first two figures.View Answer Workspace Report Discuss in Forum

10. Paper Folding

Directions to SolveIn each of the following problems, a square transparent sheet (X) with a pattern is given. Figure out from amongst the four alternatives as to how the patter would appear when the transparent sheet is folded at the dotted line.1.  Find out from amongst the four alternatives as to how the pattern would appear when

the transparent sheet is folded at the dotted line.

     (X)                     (1)                 (2)                 (3)                 (4)

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option DExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

11. Paper Cutting

Paper Cutting

Directions to SolveEach of the following questions consists of a set of three figures X, Y and Z showing a sequence of folding of apiece of paper. Figure (Z) shows the manner in which the folded paper has been cut. These three figures are followed by four answer figures from which you have to choose a figure

which would most closely resemble the unfolded form of figure (Z).1.  Choose a figure which would most closely resemble the unfolded form of Figure (Z).

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

12. Rule Detection

Directions to SolveIn each of the following questions, choose the set of figures which follows the given rule.

13. Grouping of Images

Directions to SolveIn each of the following questions, group the given figures into three classes using each figure only once.1.  Group the given figures into three classes using each figure only once.

A. 1,4 ; 2,3 ; 5,6

B. 1,5 ; 2,6 ; 4,3

C. 1,6 ; 2,3 ; 4,5

D. 1,2 ; 3,6 ; 4,5

Answer & Explanation

Answer: Option AExplanation:(1, 4), (2, 3) and (5, 6) are three different pairs of identical figures.View Answer Workspace Report Discuss in Forum

14. Dot Situation

Directions to SolveFrom amongst the figures marked (1), (2), (3) and (4), select the figure which satisfies the same conditions of placement of the dots as in figure (X).

1.  Select the figure which satisfies the same conditions of placement of the dots as in Figure-X.

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option DExplanation:In fig. (X), one of the dots lies in the region common to the circle and the square only and the other dot lies in the region common to all the three figures -the circle, the square and the triangle. In each of the alternatives (1), (2) and (3), there is no region common to the square and the circle only. Only fig. (4) consists of both the types of regions.View Answer Workspace Report Discuss in Forum

15. Shape Construction

Directions to SolveIn each of the following questions, a set of five alternative figures 1, 2, 3, 4 and 5 followed by a set of four alternatives (A), (B), (C) and (D) is provided. It is required to select the alternative which represents three out of the five alternative figures which when fitted into each other would form a complete square.1.  Select the alternative which represents three out of the five alternative figures which

when fitted into each other would form a complete square.

A. 145 B. 245

C. 123 D. 234

Answer & Explanation

Answer: Option BExplanation:

View Answer Workspace Report Discuss in Forum

16. Image Analysis

Image Analysis1.  Find out which of the figures (1), (2), (3) and (4) can be formed from the pieces fiven in figure (X).

A. 1 B. 2

C. 3 D. 4

Answer & Explanation

Answer: Option AExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

17. Cubes and Dice

Introduction:Construction of Boxes:The details of the cube formed when a sheet is folded to form a box:

Form I In this case:1 lies opposite 5;2 lies opposite 4;3 lies opposite 6.

Form II

In this case:1 lies opposite 6;2 lies opposite 4;3 lies opposite 5.

Form III

In this case:1 lies opposite 4;2 lies opposite 6;3 lies opposite 5.

Form IV In this case:1 lies opposite 4;2 lies opposite 5;3 lies opposite 6.

Form V

In this case:1 lies opposite 3;2 lies opposite 5;4 lies opposite 6.

Form VIIn this case:

 will be the one of the faces of the cube and it lies opposite 3;2 lies opposite 4;1 lies opposite 5.

Form VII

In this case:

 will be the one of the faces of the cube and it lies opposite 3;2 lies opposite 4;1 lies opposite 5.

Form VIIIIn this case:

 and   are two faces of the cube that lie opposite to each other.1 lies opposite 3;2 lies opposite 4;

Directions to SolveThe sheet of paper shown in the figure (X) given on the left hand side, in each problem, is folded to form a box. Choose from amongst the alternatives (1), (2), (3) and (4), the boxes that are similar to the box that will be formed.1.  Choose the box that is similar to the box formed from the given sheet of paper (X).

A. 1 and 2 only

B. 2 and 3 only

C. 1 and 4 only

D. 1, 2, 3 and 4

Answer & Explanation

Answer: Option DExplanation:The fig. (X) is similar to the Form VI. So, when a cube is formed by folding the sheet

shown in fig. (X), then   is one of the faces of the cube. Clearly, each one of the four cubes shown in figures (1), (2), (3) and (4) can be formed by folding the sheet shown in fig. (X).View Answer Workspace Report Discuss in Forum

C. Verbal ability:

I. Verbal Ability Questions and Answers

1. Spotting Errors

Directions to SolveRead the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'D'. (Ignore the errors of punctuation, if any).

2. Synonyms

Directions to SolveIn the following the questions choose the word which best expresses the meaning of the given word.

3. Antonyms

Directions to SolveIn the following questions choose the word which is the exact OPPOSITE of the given words.

4. Selecting Words

Directions to SolvePick out the most effective word(s) from the given words to fill in the blank to make the sentence meaningfully complete.

5. Spellings

Directions to SolveFind the correctly spelt words.

6. Sentence Formation

Directions to SolveIn each question below a sentence broken into five or six parts. Join these parts to make a meaningful sentence. The correct order of parts is the answer.

7. Ordering of Words

Directions to SolveIn each question below, there is a sentence of which some parts have been jumbled up. Rearranage these parts which are labelled P, Q, R and S to produce the correct sentence. Choose the proper sequence.

8. Sentence Correction

Directions to SolveWhich of phrases given below each sentence should replace the phrase printed in bold type to make the grammatically correct? If the sentence is correct as it is, mark 'E' as the answer.

9. Sentence Improvement

Directions to SolveIn questions given below, a part of the sentence is italicised and underlined. Below are given alternatives to the italicised part which may improve the sentence. Choose the correct alternative. In case no improvement is needed, option 'D' is the answer.

10. Completing Statements

Directions to SolveIn each question, an incomplete statement (Stem) followed by fillers is given. Pick out the best one which can complete incomplete stem correctly and meaningfully.

11. Ordering of Sentences

Directions to SolveIn questions below, each passage consist of six sentences. The first and sixth sentence are given in the begining. The middle four sentences in each have been removed and jumbled up. These are labelled as P, Q, R and S. Find out the proper order for the four sentences.

12. Paragraph Formation

Directions to SolveRearrange the following five sentences in proper sequence so as to for a meaningful paragraph, then answer the questions given below them.

1. After Examining him, the doctor smiled at him mischievously and took out a syringe.2. Thinking that he was really sick, his father summoned the family doctor.3. That day, Mintu wanted to take a day off from school4. Immediately, Mintu jumped up from his bed and swore the he was fine5. Therefor; he pretended to be sick and remained in bed.

1.  Which sentence should come third in the paragraph?

A. 1 B. 2

C. 3 D. 4

E. 5

Answer & Explanation

Answer: Option BExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

2.  Which sentence should come last in the paragraph?

A. 1 B. 2

C. 3 D. 4

E. 5

Answer & Explanation

Answer: Option DExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

3.  Which sentence should come fourth in the paragraph?

A. 1 B. 2

C. 3 D. 4

E. 5

Answer & Explanation

Answer: Option AExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

4.  Which sentence should come second in the paragraph?

A. 1 B. 2

C. 3 D. 4

E. 5

Answer & Explanation

Answer: Option EExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

5.  Which sentence should come first in the paragraph?

A. 1 B. 2

C. 3 D. 4

E. 5

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

13. Closet Test

Directions to SolveToday most businessmen are very worried. To begin with, they are not used to competition.In the past they sold whatever ...(1)... produced at whatever prices they chose. But ...(2)... increasing competition, customers began to ...(3)... and choose. Imports suddenly became ...(4)... available and that too at cheaper ...(5)...1.  (solve as per the direction given above)

A. it B. he

C. they D. we

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

2.  (solve as per the direction given above)

A. with B. by

C. after D. from

Answer & Explanation

Answer: Option AExplanation:

No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

3.  (solve as per the direction given above)

A. buy B. take

C. pick D. want

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

4.  (solve as per the direction given above)

A. hardly B. easily

C. frequently D. conveniently

Answer & Explanation

Answer: Option BExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

5.  (solve as per the direction given above)

A. costs B. returns

C. dividend D. prices

Answer & Explanation

Answer: Option DExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

14. Comprehension

Directions to SolveI felt the wall of the tunnel shiver. The master alarm squealed through my earphones. Almost simultaneously, Jack yelled down to me that there was a warning light on. Fleeting but spectacular sights snapped into ans out of view, the snow, the shower of debris, the moon, looming close and big, the dazzling sunshine for once unfiltered by layers of air. The last twelve hours before re-entry were particular bone-chilling. During this period, I had to go up in to command module. Even after the fiery re-entry splashing down in 81o water in south pacific, we could still see our frosty breath inside the command module.1.  The word 'Command Module' used twice in the given passage indicates perhaps that it

deals with

A. an alarming journey B. a commanding situation

C. a journey into outer space D. a frightful battle.

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

2.  Which one of the following reasons would one consider as more as possible for the warning lights to be on?

A. There was a shower of debris.

B. Jack was yelling.

C. A catastrophe was imminent.

D. The moon was looming close and big.

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

3.  The statement that the dazzling sunshine was "for once unfiltered by layers of air" means

A. that the sun was very hot B. that there was no strong wind

C. that the air was unpolluted D. none of above

Answer & Explanation

Answer: Option DExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

15. One Word Substitutes

Directions to SolveIn questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.1.  Extreme old age when a man behaves like a fool

A. Imbecility B. Senility

C. Dotage D. Superannuation

Answer & Explanation

Answer: Option CExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

16. Idioms and Phrases

Directions to SolveSome proverbs/idioms are given below together with their meanings. Choose the correct meaning

of proverb/idiom, If there is no correct meaning given, E (i.e.) 'None of these' will be the answer.

17. Change of Voice

Directions to SolveIn the questions below the sentences have been given in Active/Passive voice. From the given alternatives, choose the one which best expresses the given sentence in Passive/Active voice.1.  After driving professor Kumar to the museum she dropped him at his hotel.

A. After being driven to the museum, Professor Kumar was dropped at his hotel.

B. Professor Kumar was being driven dropped at his hotel.

C.After she had driven Professor Kumar to the museum she had dropped him at his hotel.

D.After she was driven Professor Kumar to the museum she had dropped him at his hotel.

Answer & Explanation

Answer: Option AExplanation:No answer description available for this question. Let us discuss.View Answer Workspace Report Discuss in Forum

18. Change of Speech

Directions to SolveIn the questions below the sentences have been given in Direct/Indirect speech. From the given alternatives, choose the one which best expresses the given sentence in Indirect/Direct speech.

19. Verbal Analogies

Directions to SolveEach question consist of two words which have a certain relationship to each other followed by four pairs of related words, Select the pair which has the same relationship.