tnp - venkata subbu.b

Training & Placement

Pondicherry Engineering College

Puducherry

Preparatory Material

1. Numerical Reasoning

2. Analytical Reasoning

3. Logical Reasoning

4. Verbal Reasoning


Pondicherry Engineering College 1

1. Numerical Reasoning

1.1 Problems on Numbers

1.1.1. Concept: Numbers are broadly classified into Real and Imaginary Numbers. The Real Numbers

consist of Integers, Whole Numbers and Natural Numbers. Integers are the set of all

Non–Fractional Numbers lying between –∞ and ∞ i.e., –∞... –4, –3, –2, –1, 0, 1, 2, ∞.

Natural Numbers are the set of all Non–Fractional Numbers from 1 to infinity i.e. 1, 2,

3, … ∞. Whole Numbers are the set of all Natural Numbers and Zero 0, 1, 2, … ∞.

The Imaginary Numbers are generally written in the form of “a + bi”, where a and b are

Real Numbers and „i‟ is the Imaginary Number and its value is –1. Operation on

Complex Numbers is similar to that on Real Numbers. Let a + bi and c + di be two

Complex Numbers.

Then the various arithmetic operations are:

Addition : (a + bi) + (c + di) = [(a + c) + (b + d) i]

Example : (3 + 4i) + (4 – 3i) = [(4 + 3) + (4 – 3) i] = 7 + i

Subtraction : (a + bi) – (c + di) = [(a – c) + (b – d) i]

Example : (3 + 4i) – (4 – 3i) = [(3 – 4) + (4 – 3) i] = –1 + 7i

Multiplication : (a + bi) * (c + di) = ac + adi + cbi+ bdi2

= (ac – bd) + (ad + bc) i

Example : (3 + 4i) (4 – 3i) = (12 – (–12)) + (–9+16) i = 24 + 7i

The Numbers that have two factors “Only and Exactly” – the number itself and 1 are all

Prime Numbers. Natural Numbers greater than 1, which are not Prime, are known as

Composite Numbers.

A Common Factor of two or more numbers is a number which divides each of them

exactly. For example, 4 is a Common Factor of 8 and 12. H.C.F. (Highest Common

Factor) is the greatest number that divides each one of them exactly. For example, 6 is

the Highest Common Factor of 12, 18 and 24. It is also called Greatest Common Divisor

(G.C.D.)

The Least Common Multiple (L.C.M.) of two or more given numbers is the least or

lowest number which is exactly divisible by each of them. For example, consider the two

numbers 12 and 18. Multiples of 12 are 12, 24, 36, 48, 72 … and Multiples of 18 are 18,

36, 54, 72, …. Common Multiples are 36 and 72. Therefore, Least Common Multiple of

12 and 18 is 36.

Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between

any two successive numbers is always constant. If the Ratio of any two successive terms

is equal throughout the sequence, then the sequence is said to be in a Geometric

Progression (G.P.)



Tips:

1. To square any number ending with 5, let the number be „A5‟.

(A5)2 = A (A + 1)25.

Ex: 252

= (2 *(2 + 1)) 25 = 625. (A = 2, join 25 instead of multiplying).

2. A number is divisible by 3, if the Sum of the digits in the number is divisible by3.

Ex: 3792 is divisible by 3 since 3+7+9+2 = 21, which is divisible by 3.

3. A number is divisible by 4, if the number formed by the last two digits is divisible by 4

or both the last two digits are zero.

Ex: The number 2616 is divisible by 4 since “16”, the number formed by the

last two digits are divisible by 4.

4. A number is divisible by 6, if the number is even and Sum of its digits is divisible by 3.

Ex: The number 4518 is divisible by 6 since it is even and Sum of its digits

4+5+1+8 = 18, is divisible by 3.

5. A number is divisible by 8, if the number formed by the last 3 digits is divisible by 8.

Ex: The number 41784 is divisible by 8 as the number formed by the last 3

digits is Divisible by 8.

6. A number is divisible by 9 if the Sum of its digits is divisible by 9.

Ex: The number 819 is divisible by 9 as the sum of the digits, 8+1+9 = 18,

is divisible by 9

7. A number is divisible by 12 if it is divisible by both 3 and 4.

Ex: The number 24 is divisible by 3 and 4 is also divisible by 12

8. In Comparison of Fractions, if the Denominators of the Fractions are same, the largest

fraction is the one whose Numerator is the largest.

Ex: 7/8 > 5/8 > 3/8.

9. If the Numerators of the Fractions are same, the largest fraction is the one whose

Denominator is the smallest.

Ex: 5/2 > 5/7 > 5/9.

10. If a Composite number C, which can be expressed as C = am * bn * cp….where a, b,

c…are all Prime Factors and m, n, p are Positive Integers, the number of Factors is equal

to (m + 1) (n + 1) (p + 1) …



Ex: The total numbers of Factors of 576 can be found by writing

576 = (24 * 24) = (23

* 3) * (2

3 * 3) = (2

6 * 3

2)

Therefore, the number of Factors = (6 + 1)* (2 + 1) = 21.

11. To find the L.C.M. and H.C.F. of Fractions:

L.C.M. of Fractions = (L.C.M. of Numerators) / (H.C.F. of Denominators)

H.C.F. of Fractions = (H.C.F. of Numerators) / (L.C.M. of Denominators)

1.1.2. Formulae:

1. Dividend = (Quotient * Divisor) + Remainder

2. In A.P, to find the nth term, Tn = a + (n – 1) d, where a is the first term, and d is the

difference.

3. To find the Sum of n numbers, Sn = n / 2 [2a + (n – 1) d] or Sn = n* (a + l)/2

4. In G.P, to find the nth

term Tn = a rn–1

.

5. Sum Sn = a(r

n – 1) if r >1

–––––––– r – 1

6. L.C.M * H.C.F. = Product of the numbers

L.C.M. = Product of the numbers

H.C.F.

Problem 1:

How many three digit numbers are divisible by 6 in all?

Solution:

Formula:

Tn = a + (n – 1) d

Three digit number divisible by 6 range from 102, 108, 114…… 996

Tn = 996, a = 102, d = 6

996 = 102 + (n – 1) 6 Hence n = 150

Problem 2:

A number when divided by 342 gives a Remainder 47. When the same number is divided

by 18 what would be the Remainder?

Solution:

Dividend = (Divisor * Quotient) + Remainder

= 342Q + 47

= 19 * 18Q + 18 * 2 + 11

= 18(19Q + 2) + 11

Remainder is 11.



Problem 3:

Find the Greatest 5-digit number which is exactly divisible by 463.

Solution:

The Greatest 5-digit number is 99999. Dividing this number by 463, the Remainder is 454.

So, the required number = 99999 – 454

= 99545. 1.1.3. Problems to Solve:

1. A certain number when divided by 899 gives a Remainder 63. What is the

Remainder when the same number is divided by 29?

2. How many numbers between 11 and 90 are divisible by 7?

3. The Sum of the Squares of two numbers is 3341 and the difference of their Squares is

891. Find the number.

4. In a two-digit number, the Sum of the digits is 15. If 9 is added to the number, the

digits are reversed. Find the numbers.

5. What is the Maximum number of Slices you can obtain by cutting a cake with only 4

cuts?

6. Two numbers are such that the Ratio between them is 4:7. If each is increased by 4,

the Ratio becomes 3:5. Find the Largest number.

7. The Denominator of a Fraction is 3 more than the Numerator. If 4 is added to both

the Numerator and the Denominator, the Fraction becomes 4/5. Find the Fraction.

8. The Sum of three numbers is 136. If the Ratio between first and second is 2:3 and

that between second and third is 5:3, find the second number.

9. The Product of two numbers is 120 and the Sum of their Squares is 289. Find the

Sum of the numbers.

10. The Sum of the Squares of three Consecutive Odd numbers is 251. Find the numbers.

1.2. Ratio and Proportion 1.2.1. Concept:

Ratio is a comparison of two quantities. It is a relationship between two quantities with

respect to magnitude. If a and b are two numbers, the Ratio of a to b is a/b and is denoted

by a : b. The two quantities that are being compared are called terms. The first term is

called Antecedent and the second term is called Consequent.

1.2.2. Types of Ratios:

Duplicate Ratio: The Ratio of the Squares of two numbers is called the Duplicate Ratio

of those two numbers. Ex: 32

/ 42.



Triplicate Ratio: The Ratio of the Cubes of two numbers is called the Triplicate Ratio

of that numbers. Ex: 33 / 4

3

Sub–Duplicate Ratio: The Ratio of the Square roots of two numbers is called the Sub–Duplicate Ratio of two numbers. Ex: 3/4 is the Sub–Duplicate Ratio of 9 / 16.

Inverse Ratio: If the antecedent and consequent of a Ratio interchange their places, the

new Ratio is called the Inverse Ratio of the first. Ex: 3 / 5 is the Inverse Ratio of 5 / 3.

Compound Ratio: The Ratio of the Product of the antecedents to that of the consequents

of two or more given Ratios is called the Compound Ratio. Ex: If 3 / 5, 4 / 5 and 5 / 7 are

the given Ratios, then (3*4*5) / (5*5*7) i.e. 12 / 35 is the compound Ratio.

The Equality of two Ratios is called Proportion. Ex: 3 / 5 = 4 / 10, we write

3 : 5 : : 4 : 10.

If a / b = c / d, then each term of the Ratio a / b and c / d is called a Proportional. Here, a,

d are known as Extremes and b, c are known as Means.

If four quantities are in Proportion, then Product of Means = Product of Extremes

If a : x :: x : b, x is called the Mean or Second Proportional of a, b.

If a : b :: b : x, x is called the Third Proportional of a, b.

If a : b :: c : x, x is called the Fourth Proportional of a, b, c.

Tips:

1. If two numbers are in the Ratio of a : b and the Sum of these numbers is x, then these

numbers will be ax / (a + b) and bx / (a + b) respectively.

2. If in a Mixture of „x‟ litres, two liquids A and B are in the Ratio of a : b, then the

Quantities of Liquids A and B in the Mixture will be ax / (a + b) litres and bx / (a + b)

respectively.

3. If the Ratio of two numbers is a : b, then the numbers that should be added to each of the

numbers in order to make this Ratio c : d is given by (ad – bc) / (c – d).

4. If the Ratio of two numbers is a : b, then the number that should be subtracted from each

of the numbers in order to make this Ratio c : d is given by (bc – ad) / (c – d).

5. If the Incomes of two Persons are in the Ratio of a : b and their Expenditures are in the

Ratio of c : d. If the Saving of each Person be Rs. S, then their Incomes are given by,

Rs. (aS (d – c)) / (ad – bc) and Rs. (bS (d – c)) / (ad – bc)

6. If a : b = n1 : d1 and b : c = n2 : d2, then a : b : c = (n1*n2) : (d1*n2) : (d1 : d2)



Problem 1:

Two numbers are in the Ratio of 4 : 5 and the Sum of these numbers is 27.Find the two

numbers.

Solution:

Here, a = 4, b = 5, and x = 27.

The first number = (a * x) / (a + b)

= (4 * 27) / (4+ 5) = 12

The second number will be = 27 – 12

= 15.

Problem 2:

Two numbers are in the Ratio 9 : 7.If 14 is subtracted from each, the Ratio becomes 7 : 5.

Find the numbers.

Solution:

Let the numbers be 9x and 7x.

9x – 14 = 7

7x – 14 5

5(9x – 14) = 7(7x – 14)

=> 45x – 70 = 49x – 98

=> 98 – 70 = 49x – 45x

=> x = 7 Hence the numbers are 9x = 63 and 7x = 49.

Problem 3:

Find the number that must be subtracted from the Ratio of 5 : 6 to make it equal to 2 : 3.

Solution:

We have a : b = 5 : 6 and c : d = 2 : 3.

The required number = (bc – ad) / (c – d) = [(6*2) – (5*3)] / (2 – 3) = 3.

1.2.3. Problems to Solve:

1. Rs. 49 was divided among 150 children. Each Girl got 50 paise and each Boy 25 paise.

How many Boys were there?

2. Five Bananas and Four Apples cost as much as Three Bananas and Seven Apples. Find

the Ratio of the cost of One Banana to that of One Apple.

3. Annual Income of A and B is in the Ratio of 5 : 4 and their Annual Expenses bear a

Ratio of 4 : 3. If each of them saves Rs. 500 at the end of the Year, then find their

Annual Income.

4. If 1 / 3A = 1 / 4B = 1 / 5C, then A: B: C =?

5. A, B, C and D share a Property worth Rs. 77, 500. If A : B = 3 : 2, B: C = 5 : 4

and C : D = 3 : 7, Find Share of B.



6. A Man spends Rs 500 in buying 12 items of Tables and Chairs. The cost of one table is

Rs. 50 and that of one chair is Rs. 40. What is the Ratio of the numbers of Chairs and

Tables purchased?

7. An Alloy contains Copper and Zinc in the Ratio 7 : 3. If the alloy contains 10.5 kg Zinc,

find the Quantity of Copper in the Alloy.

8. If Income of A, B and C are in the Ratio of 9 : 3 : 7 and Income of C exceeds the Income

of B by Rs. 1200 then find the Income of A.

9. Find the Age of Amit and Puneet which are in the Ratio 2 : 3. After 12 years, their Ages

will be in the Ratio 11 : 15. What is the age of Puneet?

10. Two numbers A and B are such that the Sum of 5% of A and 4% of B is 2/3rd

of the Sum

of 6% of A and 8% of B. Find the Ratio of A : B.

1.3. Alligation or Mixture 1. 3.1. Concept:

Alligation means “Linking”. It is a Rule to find the Ratio in which two or more

Ingredients at their Respective Prices should be mixed to give a Mixture at a given Price.

By using this Rule, we can also find the Mean or Average Price of a Mixture when the

Prices of two or more Ingredients which may be mixed together and the Proportion in

which they are mixed are given.

Alligation Rule: Suppose, Rs. D per Unit be the Price of first ingredient (superior quality) mixed with

another ingredient (cheaper quality) of Price Rs. C per unit to form a Mixture whose

Mean Price is Rs. M per unit, then the two ingredients must be mixed in the

Ratio:

Quantity of Cheaper = C.P. (Cost Price) of Costlier – Mean

Quantity of Costlier Mean – C.P. of Cheaper

The two Ingredients are to be mixed in the Inverse Ratio of the Differences of their Prices

and the Mean Price.

The above Rule may be represented schematically as

Cost of Cheaper Cost of Costlier

C D

M

(D–M) (M–C)

Ratio = (D – M): (M – C)



Problem 1:

Sugar at Rs. 15 per kg is mixed with Sugar at Rs. 20 per kg in the Ratio 2 : 3. Find the

Price per kg of the Mixture.

Solution:

c (Rs.15) d (Rs.20)

m (Rs.x)

d – m = 20 – x m – c = x – 15

From the given data, 20 – x = 2 .

x – 15 3

x = 18

Price of the Mixture = Rs. 18 per kg.

Problem 2:

A merchant has 100 kg of salt, part of which he sells at 7% Profit and the rest at 17%

Profit. He Gains 10% on the whole. Find the Ratio.

Solution: 7 17

10

(17 – 10) (10 – 7)

Ratio = (17 – 10): (10 – 7)

Therefore, the Ratio is 7: 3.


1. In what ratio is ghee worth of Rs. 30 per Kg and ghee worth of Rs. 45 per Kg be mixed

so as to get the mixture worth of Rs. 40 per Kg?

2. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar

costing Rs. 7 per kg so that there may be a Gain of 10% by selling the Mixture at Rs. 9.24

per kg?

3. In what Ratio two varieties of Tea, one costing Rs. 27 per Kg and the other costing Rs. 32

per Kg should be blended to produce a blended variety of Tea worth Rs. 30 per Kg?

How much should be the quantity of Second Variety of Tea, if the First Variety is 60 Kg?



4. How much Water is to be added to 14 litres of Milk worth Rs. 5.40 a litre so that the

value of the Mixture may be Rs. 4.20 a litre?

5. In what Ratio must Water be added to Spirit to Gain 10% by selling it at the cost Price?

6. Two vessels A and B contain Milk and Water mixed in the Ratio 4 : 3 and 2 : 3. In what

Ratio must these Mixtures be mixed to form a new Mixture containing half Milk and half

Water?

7. In an Examination out of 480 students 85% of the Girls and 70% of the Boys passed.

How many Boys appeared in the Examination if the total Pass Percentage was 75%?

8. Some amount out of Rs. 7000 was lent at 6% p.a. and the remaining at 4% p.a. If the

Total Simple Interest from both the Fractions in 5 Years was Rs. 1600, find the Sum lent

at 6% p.a.

9. Mira‟s Expenditure and Savings are in the Ratio 3 : 2. Her Income increases by 10%.

Her Expenditure also increases by 12%. By what % does her saving increase?

10. The Average Weekly Salary of all Employees (Supervisors and Laborers) is Rs. 100.

The Average Weekly Salary of all the supervisors is Rs.600 while the Average weekly

Salary of all the laborers is Rs. 75. Find the number of supervisors in the factory if there

are 840 laborers in it

1.4. Chain Rule 1.4.1. Concept:

There are two Types of Variations

1) Direct Variation

2) Indirect Variation

Direct Variation: Apples Cost (Rs.)

20 140

12 x

If number of Apples is less, cost is also less. If the Variation of one thing, while

increasing or decreasing, affects the other thing in the same way, then it is Direct

Variation.

20 / 12 = 140 / x

x = 140*12

20

= 84

Indirect Variation: Men Days

14 8

24 x



If number of Men increase, the Days required for the Completion of the work will be less.

When one value increases, then the other value decreases. It is called Indirect Variation.

24 = 8

14 x

24* x = 8* 14

x = 8 * 14

24

x = 14

3

Problem 1:

If 11.25 m of a Uniform Iron Rod weighs 42.75 kg, what will be the weight of 6 m of the

same rod?

Solution:

Length (m) weight (kg)

11.25 42.75

6 x

Since it is a Direct Proportion,

x _ = 6

42.75 11.25

x = 6 * 42. 75

11.25

x = 22.8 kg

Problem 2:

39 Persons can repair a road in 12 Days, working 5 Hours a day. In how many Days will

30 Persons working 6 Hours a day, complete the same work?

Solution:

Men Days Hours

39 12 5

30 x 6

More Hours less Days (Inverse Proportion)

Less Men more Days (Inverse Proportion)

x = 5 * 39

12 6 30

x = (5 *39 * 12) / (6 * 30)

x = 13 Days.




1. If 5 toys cost Rs. 234, how much will 55 Toys cost?

2. If 45 Men can do a Piece of Work in 20 hrs, in how many Hours can 15 Men do it?

3. If the Wages of 9 Men for 14 Days is Rs. 2100, then find the wages of 18 Men for 7

Days.

4. If 20 Men can build a wall of 56 meters long in 6 Days, what length of a similar wall can

be built by 35 Men in 3 Days?

5. If 12 Men working 9 Hours a day can reap a field in 16 Days, in how many Days will 18

Men reap the field working 8 Hours a day?

6. 20 Men complete one–third of work in 20 Days. How many more Men should be

employed to finish the rest of the work in 25 more Days?

7. In a dairy farm, 40 cows eat 40 bags of husk in 40 Days. In how many Days will one

cow eat one bag of husk?

8. A contract is to be completed in 46 Days and 117 Men were asked to work 8 Hours a day.

After 33 Days, 4/7 of the work is completed. How many additional Men should be

employed so that the work could be completed in time, each man now working 9 Hours a

day?

9. If 6 Men working 8 Hours a day earn Rs.840 per week, then how much will 9 Men

working 6 Hours a day earn per week?

10. If 3 Men or 6 Boys can do a Piece of Work in 10 Days working 7 Hours a day, how many

Days will it take to complete a Piece of Work twice as large with 6 Men and 2 Boys

working together for 8 Hours a day?

1.5. Partnership

1.5.1. Concept: In Partnership two or more Persons carry on a Business and Share the Profits of the Business

at an agreed Proportion. Persons who have entered into Partnership with one another are

individually called Partners and collectively called a Firm and the name under which their

Business is carried on is called the Firm Name. The Partnership may be Simple or

Compound.

Simple Partnership is one in which the capital of each Partner is in the Business for same

time. Compound Partnership is one in which the capitals of Partners are invested for

different periods. In this, a Partner can be a working Partner or Sleeping Partner. Sleeping

Partner is one who invests the capital in the Business but does not actively participate in the

conduction of Business. Working Partner besides investing capital takes part in running the

Business. For his work, he is either paid some Salary or given a certain percent of Profit in

addition.



1.5.2. Formulae:

1. If A invested Rs. x and B invested Rs. y then Ratio A : B = X : Y

2. If A invested Rs. x and after 3 months B invested Rs. y then the Share is in the Ratio

A : B = (x * 12) : (y * 9).

2. If Capitals of two Partners are Rs. C1 and Rs. C2, and the total Profit is Rs. P, then

Shares of the Partners in the Profits are Rs. ((C1 * P) / (C1 + C2)) and Rs. ((C2 * P) / (C1

+ C2)).

Tips:

1. If the Capitals of two Partners is Rs.C1 and Rs.C2 for the periods t1 and t2

respectively, then Profit of A / Profit of B = (C1*t1) / (C2*t2)

2. If three Partners invested their capitals in a Business. If the timing of their investment

is in the Ratio t1: t2: t3 and their Profits are in the Ratio P1 : P2 : P3, then the Ratio of

their capitals invested is (P1 / t1) : (P2 / t2) : (P3 / t3).

Problem 1:

Sona, Sweetha, Sumathy entered into a Partnership investing Rs.35000, Rs. 45000 and

Rs. 55000 respectively. Find the respective Shares of Sona, Sweetha Sumathy if the

annual Profit is Rs. 40500.

Solution:

Ratio = 35000: 45000: 55000 = 7: 9: 11

Sona‟s Share = 40500 * 7 / 27 = Rs. 10500.

Sweetha‟s Share = 40500 * 9 / 27 = Rs. 13500.

Sumathy‟s Share = 40500*11 / 27 = Rs. 16500.

Problem 2:

Ramani started a Business investing Rs. 9000. After Five months, Raj joined with a

capital of Rs. 8000. If at the end of the Year, they earn a Profit of Rs. 6970, then what

will be the Share of Raj in the Profit?

Solution:

Ratio of their Share = 9000*12: 8000*7

= 27 : 14

Raj‟s Share = Rs. (6970*14 / 41)

= Rs.2380.



Problem 3:

A and B start a Business with the amount of Rs.10000 and Rs.15000 respectively. After

3 months, A withdraws Rs. 2000 and after 6 months again he withdraws Rs. 4000. At the

end of the Year they get the Profit of Rs. 7000. What will be their Share according to the

Investment?

Solution:

A‟s investment = (10000 * 3) + (8000 * 6) + (4000 * 3)

= 30000 + 48000 + 12000

= Rs. 90000

B‟s investment = 15000 *12

= Rs. 1, 80,000

A: B = 90000 : 180000

= 1 : 2

A‟s Share = 7000 * 1 / 3 = Rs. 2333

B‟s Share = 7000 * 2 / 3 = Rs. 4667


1. Find the amount of Profit each Partner A, B and C will get if they invest Rs. 20000,

Rs. 50000 and Rs. 40000 respectively in a Business. The net Profit for the Year was Rs.

12100 which was divided in Proportion to investments made.

2. A and B enter into a Partnership with Rs. 50000 and Rs. 60000 respectively. C joins

them after 6 months contributing Rs. 70000 and B leaves x months before the end of the

Year. If they Share the Profit in the Ratio of 20 : 18 : 21, find the value of x.

3. A and B started a Business investing Rs. 9000 and Rs. 12000. After 6 months, B

withdrew half of his investment. After a Year, if the total Profit is Rs. 4600, what would

be B‟s Share in it?

4. Rs. 120 is divided among X, Y and Z so that X‟s Share is Rs. 20 more than Y‟s Share and

Rs. 20 less than Z‟s Share. What is Y‟s Share?

5. Rahul started a Business with a capital of Rs. 8000. After Six months, Sanjay joined him

with Investment of some Capital. If at the end of the Year each of them gets Equal

Amount as Profit, how much did Sanjay invest in the Business?

6. Two Partners invest Rs. 12500 and Rs. 8500 respectively in a Business. If one Partner

gets Rs.300 more than the other in the Profit, what is the Total Profit?

7. Gupta and Bansal enter into a Partnership with their capitals in the Ratio 5 : 6. At the end

of 8 months, Gupta withdraws his capital. If they receive the Profits in the Ratio of 5: 9,

find how long Bansal‟s Capital was used.



8. A and B jointly Invest Rs. 2100 and Rs. 3100 respectively in a Firm. A is an Active

Partner and hence he gets 25% of Profit separately. If their Business yields them total of

Rs. 1040 as Profit, what will be the Gain of each of them?

9. A began Business with Rs. 6000 and was joined afterwards by B with Rs. 8000. At the

end of the Year B got one–third of the total Profit. When did B join?

10. X and Y started a Business. X invested Rs. 3000 for 4 months and Y invested Rs. 2000

for 6 months. How much should X be paid out of a total Profit of Rs. 500?

1.6. Area and Volume

1.6.1. Concept: The Area of any figure is the amount of surface enclosed within its bounding lines. Area

is always expressed in square units. The Perimeter of a Geometrical Figure is the Total

Length of the sides enclosing the Figure.

A Closed Figure bounded by three sides is called a Triangle. A Triangle having one of

its angles equal to 90 degrees is called a Right–Angled Triangle. A Triangle in which

all sides are equal is called an Equilateral Triangle. A Triangle whose two sides are

equal is an Isosceles Triangle.

A Closed Figure bounded by four sides is called a Quadrilateral. A Quadrilateral in

which opposite sides are equal and parallel is called a Parallelogram. Square is a

Quadrilateral with all sides equal and all the Four Angles equal to 90o. A Rectangle is a

Quadrilateral with opposite sides equal and all the four angles equal to 90 o

. A Rhombus

is a Quadrilateral in which all sides are equal. A Trapezium is a Quadrilateral in which

any two opposite sides are parallel. Distance between parallel sides is called its Height.

Circle is the path traveled by a point which moves in such a way that its distance from a

fixed point remains constant. The fixed point is known as the Centre and the fixed

distance is called the Radius.

A Solid is a figure bounded by one or more surfaces. It has Three Dimensions namely

Length, Breadth or Width and Height. The Volume of any solid figure is the amount of

space enclosed within its bounding spaces. It is measured in cubic units. Cuboid is a

Solid Figure which has Six Rectangular Faces. It is also called as Rectangular

Parallelopiped.

A Right Circular Cylinder is a solid with circular ends of equal radius and the line

joining their centers perpendicular to them. This is called the Axis of the Cylinder. The

Length of the Axis is called the Height of the Cylinder. This can be best illustrated by

rolling a rectangular paper either lengthwise or breadth-wise in a round way, i.e. a

cylinder is generated by rotating a rectangle by fixing one of its sides. A Right Circular

Cone is a solid obtained by rotating a Right–Angled Triangle around its height.



Pythagoras Theorem:

In a Right Angled Triangle, the Square of the Hypotenuse is equal to the Sum of the

Squares of the other two sides.

E.g.: In a Right–angled Triangle ABC Right angled at B, AC2 = AB

2 + AC

2

1.6.2. Formulae: 1. Area of Square = (Side) * (Side) = (Side)

2

Perimeter of Square = 4*(Side)

2. Area of Rectangle = Length * Breadth

Perimeter of Rectangle = 2 * (Length + Breadth)

3. Area of Triangle = ½ * Base * Height

4. Area of Triangle = (s*(s – a)* (s – b)* (s – c))

where S = a + b + c , a, b and c are sides of the Triangle.

2

5. Area of equilateral Triangle = 3 a2, where „a‟ is side of the Triangle.

4

6. Altitude of Equilateral Triangle = 3 a, where „a‟ is side of the Triangle.

2

7. Area of a circle = r2 ,

where „r‟ is radius of the circle.

8. Circumference of the Circle = 2 r

9. Area of Rhombus = ½* d1 * d2

where d1 and d2 are diagonals.

10. Area of a Parallelogram = b * h

11. Area of a Trapezium = [½ * (a + b) * h]

Where a and b are parallel sides and h is the height.

12. Volume of the Cube = (edge) 3

= a3.

13. Total surface area of Cube = 6 * a2

14. Diagonal of the Cube = a * 3

15. Volume of cylinder = Area of the base * height

16. Area of the curved surface = Circumference of the base * height

17. Area of the total surface = Area of the curved surface + Area of two

circular ends.

18. Volume of Cone = 1/3 * ( r2h)

Where h is height and r is Radius of Base.

19. Area of the largest Circle that can be inscribed in a Square of side „a‟ is a2.

4

20. Area of a Square inscribed in a Circle of Radius r is 2r2 and the side of the Square

is 2 r.

21. The Area of Largest Triangle inscribed in a Semi Circle of Radius r is r2

.



Tips: 1. If the Length and the Breadth of a Rectangle are increased by x% and y%

respectively, then the Area of Rectangle will increase by (x + y + xy / 100) %.

2. If the Length of a rectangle is increased by x%, then its breadth will have to be

decreased by (100x / (100 + x)) % in order to maintain the same area of rectangle.

3. If each of the defining dimensions or sides of any two–dimensional figure is changed

by x%, its area changes by x*(2 + x/100) %.

4. If all the sides of a Quadrilateral are increased (or decreased) by x%, its diagonals also

increase (or decrease) by x%.

5. If the Ratio of the areas of two Squares be a : b, then the Ratio of their sides, Ratio of

their Perimeters and the Ratio of their diagonals, each will be in the Ratio a : b.

6. If the Length and Breadth of a room are l and b respectively, and a Carpet of Width

„w‟ is used to cover the floor, then the required Length of the Carpet is (l * b)/ w.

7. If the Length and Breadth of a room are l and b respectively, then the least number of

Square tiles required to cover the floor = (l*b) / (H.C.F (l, b)).

8. A Rectangular garden l m long and b m broad is surrounded by a path w m wide. The

Area of the Path is given by 2w*(l + b + 2w) sq.m.

9. If all three measuring dimensions of a sphere, cuboid, cube, cylinder or cone are

increased or decreased by x%, y% and z% respectively, then the volume of the figure

will increase or decrease by

(x + y + z + ((xy + yz + xyz) / 100) + xyz/100)%

10. If a Sphere of Radius R is melted to form smaller Spheres each of Radius r, then the

number of small Spheres = Big Sphere‟s Volume / Small Sphere‟s Volume (R / r)3.

Problem 1:

What is the Cost of planting the field in the form of the Triangle whose Base is 2.8 m and

Height 3.2 m at the rate of Rs.100 / m2?

Solution:

Area of Triangular field = ½ * 3.2 * 2.8 m2

= 4.48 m2

Cost = Rs.100 * 4.48

= Rs. 448.

Problem 2:

Area of a Rhombus is 850 cm2. If one of its Diagonal is 34 cm, find the Length of the

other Diagonal.



Solution:

Area of a Rhombus = ½ * d1 * d2

850 = ½ * 34 * d2

= 17 * d2

Therefore,

d2 = 850 / 17

= 50 cm

Problem 3:

The Radius and Height of a Cylinder are increased by 10% and 20% respectively. Find

the Percentage increase in Surface Area.

Solution:

Let the increase in Radius x be 10% and the increase in Height y be 20%.

Percentage increase in Surface Area is

= (x + y + (xy / 100)) %

= (10 + 20 + ((10*20) / 100)) % = 32 %


1. If the Base of an Isosceles Triangle is 10 cm and the Length of equal sides is 13 cm,

find its area.

2. How many tiles of 20 cm by 10 cm will be needed to cover the floor of a room 25 m

long and 16 m wide?

3. The Length and Breadth of a Rectangle are increased by 20% and 5% respectively.

Find the Percentage increase in its area.

4. Find the Area of a Trapezium having Parallel sides 65 m and 44 m and their

separation being 20 m.

5. The Diameter of a Circle is 105 cm less than the Circumference. What is the

Diameter of a Circle?

6. The Area of the Parallelogram is 72 sq.cm and its Height is 12 cm. Find the Base of

the Parallelogram.

7. The Diameter of a Wheel is 2 cm. If it rolls forward covering 10 Revolutions, find the

Distance traveled by it.

8. Find the Area of an Equilateral Triangle each of whose sides measure 6 cm.

9. A Rectangular Carpet has an Area of 120 sq.m. and a Perimeter of 46 m. Find the

Length of its Diagonal.



10. A Copper Sphere of Diameter 18 cm is drawn into a Wire of Diameter 4 mm. Find

the Length of the Wire. 1.7. Probability

1.7.1. Concept: Probability is a concept which measures Uncertainties. Probability is the chance of

occurring of a certain Event. It is expressed quantitatively. Probability has its origin in the

problems dealing with games of chance such as gambling, coin tossing, dice throwing

and playing cards. The possible outcomes of a trial are called Events. Events are

generally denoted by capital letters. Events are said to be “Equally likely” if there is no

reason to expect anyone in preference to other.

In Simple Events, we consider the Probability of happening or non happening of single

events. In the case of Compound Events we consider the occurrence of two or more

Events. If the occurrence of an event precludes or rules out the happening of all the other

Events in the same experiment, it is called Mutually Exclusive Event. Two Events A

and B are said to be Independent if the occurrence of one does not affect the Probability

of the occurrence of the other.

The number of combinations of n objects taken r at a time (r <= n) is denoted by C (n, r)

or nCr and is denoted as

nCr = n! / (r! * (n – r)!)

If A and B are two Events associated with sample space S, then

P (A B) is the Probability that either A or B or both the Events occur.

P (A B) is the Probability that A and B occur simultaneously.

1.7.2 Formulae:

1. P (E) = n(E) E – Event

n(S) S – Sample space

2. P (A B) = P (A) + P (B) – P (A B), where P (A) is the Probability of Event A and

P (B) is the Probability of Event B.

3. If the Events are Mutually Exclusive, then P (A B) = P (A) + P (B), because

P(A B) = 0 for Mutually Exclusive Events.

Tips: 1. P (E) + P (E

1) = 1

2. Independent Events P (A B) = P (A) *P (B)

3. Probability of Occurrence of an Event is always positive.



4. Probability of Occurrence of an Event is from zero to one.

5. Probability of Occurrence of an impossible Event is 0.

Problem 1:

If two Dice are thrown simultaneously, what is the Probability that the first Dice shows

up 5 and the second Dice does not show up 5?

Solution:

We want only 5 on first Dice and any number other than 5 on the second Dice, the

number of favorable cases are 5 viz.,

(5, 1) (5, 2) (5, 3) (5, 4) (5, 6)

Hence the Probability = 5 / 36.

Problem 2:

A bag contains 2 Green, 6 Blue, and 3 Black balls. If a Ball is drawn at random, what is

the Probability that it is not a Black ball?

Solution:

One ball can be drawn out of the 11 balls from the bag in 11C1 ways.

A ball other than a black ball (2 + 6 = 8) can be drawn in 8C1 ways.

Hence, the Probability that the ball drawn is not black is

8C1 / 11C1 = 8 / 11.

Problem 3:

From a well shuffled pack of 52 Cards, a Card is drawn at random. Find the Probability

that it is either a Heart or a Queen.

Solution:

P(A) = Getting a Heart Card; P(B) = Getting a Queen Card.

P(A) = 13 / 52;

P(B) = 4 / 52;

P (A B) = 1 / 52.

Required Probability, P(A B)

= P(A) + P(B) – P (A B)

= (13 / 52) + (4 / 52) – 1 / 52

= 4 / 13.


1. Two Dice are thrown. Find the Probability of getting an Odd number on one Dice and a

multiple of three on the other



2. A man and his wife appear for an interview for two vacancies for the same post. The

Probability of husband‟s selection is 1/7 and the Probability of wife‟s selection is 1/ 5.

What is the Probability that only one of them is selected?

3. A Card is drawn from pack of 100 Cards numbered 1 to 100. Find the Probability of

drawing a number which is a Square.

4. Two Dice are thrown. Find the Probability that the total score is a Prime Number.

5. A Card is drawn from a pack of 52 cards. Find the Probability of getting a King of Club

or a Queen of Heart.

6. A Coin is tossed Three times. Find the chance that Head and Tail show alternatively.

7. One bag contains 4 White and 2 Black balls. Another bag contains 3 White and 5 Black

balls. One ball is drawn from each bag; Find the Probability that both are White.

8. Five Letters are placed at random in Five Addressed Envelopes. Find the Probability that

all the Letters are not dispatched in the Right Envelopes.

9. The Probability that a man will be alive for 25 Years and hence is 0.3 and the Probability

that his wife will be alive for 25 Years and hence is 0.4. Find the Probability that 25

Years hence both will be alive.

10. Find the Probability of getting a total of 7 or 11 in a simultaneous throw of two Dice.

1.8. Time and Work 1.8.1. Concept:

The Time required for the Completion and the number of Men engaged for a Project is

Inversely Proportional to each other, i.e. the more number of Men involved, the lesser is

the time required to finish a job. If „A‟ can do a Piece of Work in n Days, then at a

Uniform Rate of Working, „A‟ will finish 1 / nth

work in 1 day.

A, B and C can do a Piece of Work in T1, T2 and T3 Days, respectively. If they have

worked for D1, D2 and D3 Days respectively, then

Amount of Work done by A = D1 / T1

Amount of Work done by B = D2 / T2

Amount of Work done by C = D3 / T3

Also, Amount of Work done = D1 / T1 + D2 / T2 + D3 / T3

(A, B and C together)

= 1 (if the work is complete)



Tips: 1. If A can do a Piece of Work in „n‟ Days and B can do the same Piece of Work in

„m‟ Days then Work will be done in [m*n / (m + n)]Days, if both Work together.

2. If A, B and C while working alone, can complete a Work in X, Y and Z Days

respectively, then they will together complete the Work in X*Y*Z / (XY + YZ +

ZX).

3. Two Persons A and B, working together, can complete a Piece of Work in X Days.

If A, Working alone, can complete the work in Y Days, then B, working alone will

complete the work in X*Y / (Y – X) Days.

4. If A works Thrice that of B then Ratio of Work done by A and B is 3 : 1.

5. If A can complete a / b part of Work in X Days, then c / d part of the work will be

done in (b*c*X) / (a*d) Days.

6. If „a‟ Men and „b‟ Women can do a Piece of Work in „n‟ Days, then „c‟ Men and

„d‟ Women can do the work in (n*a*b) / [(b*c) + (a*d)].

Problem 1:

Worker „A‟ takes 15 hrs to complete a job. Worker B takes 30 hrs to do the same job.

How long should it take both A and B, working together but independently, to do the

same job?

Solution:

A‟s 1 Hour‟s work = 1 / 15

B‟s 1 Hour‟s work = 1 / 30

(A + B)‟s 1 Hour‟s work = (1 / 15) + (1 / 30)

= (2 + 1) / 30

= 1 / 10

Both A and B will finish the Work in 10 Days.

Problem 2:

A can do a Piece of Work in 6 Days, and B can do it in 12 Days. What time will they

require to do it working together?

Solution:

Part of the work done by A in one day = 1 / 6

Part of the work done by B in one day = 1 / 12

Part of the work done by A and B in one day = 1 / 6 + 1 / 12

= 1 / 4

Time required by A and B together to finish the work = 4 Days.



Problem 3:

12 Men or 15 Women can do a work in 14 Days. In how many Days, 7 Men and 5

Women would complete the work?

Solution:

Here, A = 12,

B = 15,

N = 14

C = 7 and

D = 5.

Required Number of Days = n*a*b / [(b*c) + (a*d)]

= 14*12*15 / ((15*7) + (12*5))

= 168 / 11 Days.


1. A can do a Piece of Work in 30 Days while B can do it in 40 Days. If A and B Work

Together, in how many Days can they complete the work?

2. 10 Men can complete a Piece of Work in 15 Days and 15 Women can complete the

same work in 12 Days. If all the 10 Men and 15 Women work together, in how many

Days will the work get completed?

3. A can finish a work in 18 Days and B can do the same work in 15 Days. B worked

for 10 Days and left the job. In how many Days, A alone can finish the remaining

work?

4. X can do a Piece of Work in 40 Days. He works at it for 8 Days and then Y finishes

it in 16 Days. How long will they together take to complete the work?

5. 4 Men and 6 Women can complete a work in 8 Days, while 3 Men and 7 Women can

complete it in 10 Days. In how many Days will 10 Women complete it?

6. A and B can finish a job in 12 Days while A, B and C can finish it in 8 Days. How

long will it take for C alone to complete the work?

7. 5 Men can complete a work in 2 Days, 4 Women can complete the same work in 3

Days and 5 children can do it in 3 Days. If 1 man, 1 Woman and 1 child work

together, in how many Days will they complete the work?

8. A can do a Piece of Work in 8 Days. A undertook to do it for Rs. 320. With the help

of B he finishes the work in 6 Days. What will be the Share of B?

9. A can do a piece of work in 9 Days. If B is 50% more efficient than A, then in how

many Days can B do the same work?

10. Two Men undertake to do a Piece of Work for Rs. 200. One alone can do it in 6 Days

and the other in 8 Days. With the help of a Boy they finish it in 3 Days. How much is

the Share of the Boy?



1.9. Pipes and Cisterns 1.9.1. Concept:

Pipes are connected to a Tank or Cistern and are used to fill or empty the Tank. Inlet is a

Pipe connected to a Tank or Cistern for filling it. Outlet is a Pipe connected to a Tank or

Cistern for emptying it. The concept is similar to that of Time and work.

If an inlet can completely fill the Tank in X Hours, the part of the Tank filled in 1 Hour is

1 / X.

If an outlet can empty the full Tank in Y Hours, the part of the Tank emptied in 1Hour is

1 / Y.

If both inlet and outlet are open, net part of the Tank filled in 1Hour = 1 / X – 1 / Y.

Tips: 1. If an inlet can completely fill the empty Tank in X Hours, the part of the Tank filled

in 1 Hour is 1 / X.

2. If an outlet can empty the full Tank in Y Hours, the part of the Tank emptied in 1

Hour is 1 / Y.

3. If inlet and outlet are open, net part of the Tank filled in 1 Hour = 1 / X – 1 / Y.

4. Two Pipes A and B can fill a Cistern in X and Y Hours respectively while working

alone. If both the Pipes are opened together, then the time taken to fill the Cistern

is (X*Y) / (X + Y) Hours.

5. Two Pipes A and B can fill a Cistern in X Hours and Y Hours, respectively. There

is also an outlet C. If all the three Pipes are opened together, the Tank is full in Z

Hours. The time taken by C to empty the full Tank is

(X*Y*Z) / (XZ + YZ – XY) Hours.

6. A Cistern has a leak which can empty it in X Hours. A Pipe which allows Y litres

of Water per Hour into the Cistern is turned on and now the Cistern is emptied in Z

Hours. The capacity of the Cistern can be found as (X*Y*Z) / (Z – X) litres.

Problem 1:

A tap can fill a Tank in 6 Hours. After half the Tank is filled three more similar taps are

opened. What is the total time taken to fill the Tank completely?

Solution:

Time taken to fill the half Tank = 3 Hours.

Part filled in 1 Hour = 1 / 6

Part filled by the four taps in 1 Hour = (4* 1 / 6) =2 / 3

Remaining part (1 – 1 / 2) = 1 / 2

2 / 3: 1 / 2 = 1: x or x = (1 / 2* 1* 3 / 2) = 3 / 4 hrs

i.e. All the four taps need 45 minutes more to fill the remaining Tank.

So, totally 3 Hours 45 minutes is the time taken to fill the Tank completely



Problem 2:

12 buckets of Water fill a Tank when the capacity of each Tank is 13.5 litres. How many

buckets will be needed to fill the same Tank, if the capacity of each bucket is 9 litres?

Solution:

Capacity of Tank = (12 x 13.5) litres

= 162 litres

Capacity of each bucket = 9 litres

Number of buckets needed = (162 / 9) = 18

Problem 3:

A Cistern is filled in 9 Hours and it takes 10 Hours when there is a leak in its bottom. If

the Cistern is full, in what time shall the leak empty it?

Solution:

Work done in 1 Hour by the leak and the filling Pipe = 1 / 9

Work done by the leak in 1 Hour = 1 / 9 – 1/10

= 1 / 90

Hence the leak can empty it in 90 Hours.


1. Three Pipes A, B and C can fill a Cistern in 10, 12 and 15 Hours respectively while

working alone. If all the three Pipes are opened together, find the time taken to fill

the Cistern.

2. A Tap can fill a Tank in 25 minutes and another can empty it in 50 minutes. Find

whether the Tank will be filled up or emptied first and in how many minutes?

3. A Tap can fill a bath in 20 minutes and another tap can fill it in 30 minutes. Amit

opens both the taps simultaneously. When the bath should have been full, he finds

that the waste Pipe was open. He then closes the waste Pipe and in another 4

minutes the bath is full. What time will the waste Pipe empty it?

4. A Cistern can be filled by two Pipes in 20 and 30 minutes respectively. Both Pipes

are opened. When must the first Pipe be turned off so that the Cistern may be filled

in 10 more minutes?

5. Three Pipes A, B and C can fill a Cistern in 6 hrs. After working together for 2 hrs,

C is closed and A and B fill the Cistern in 8 hrs. Find the time in which the Cistern

can be filled by Pipe C.

6. Two Pipes A and B together can fill a Cistern in 24 min and 32 min respectively.

Both Pipes being open, Find when the first Pipe must be turned off so that the

Cistern may be just filled in 16 minutes.



7. A leak in the bottom of a Tank can empty the full Tank in 8 hrs. An inlet Pipe fills

Water at the rate of 6 litres a minute. When the Tank is full, the inlet is opened and

due to leak the Tank is empty in 12 Hours. How many litres does the Cistern hold?

8. Two Pipes A and B separately fill a Tank in 6 hrs and 8 hrs respectively. Both the

Pipes are opened together, but 1½ hrs after the start the Pipe A is turned off. How

much time will it take to fill the Tank?

9. If two Pipes function simultaneously, the reservoir will be filled in 12 Hours. One

Pipe fills the reservoir 10 Hours faster than the other. How many Hours does the

faster Pipe take to fill the reservoir?

10. A reservoir is provided by two Pipes A and B. A can fill the reservoir 5 Hours faster

than B. If both together fill the reservoir in 6 Hours, in how many Hours the

reservoir will be filled by A alone?

1.10. Interest 1.10.1. Concept:

When a Person A borrows some money from another Person B, then A has to pay certain

amount to B for the use of this money. This amount paid by A is called Interest. The

total amount of money borrowed by A from B is called the Principal. The money paid

back to B, which comprises of the Principal and the Interest is called the Amount.

The Interest is usually charged according to a specified term, which is expressed as some

percent of the principal and is called the Rate of interest for fixed period of time.

Interest can be of two types:

1. Simple Interest (S.I) 2. Compound Interest (C.I)

When the interest is payable on the principal only, it is called Simple Interest. For

example, Simple Interest on Rs. 100 at 5% per annum will be Rs.5 each Year, that is, at

the end of one Year total amount will be Rs. 105. At the end of second Year it will be

Rs. 110 and so on.

Compound Interest is the method, where the interest for each period is added to the

Principal before interest is calculated for the next period. With this method, the Principal

grows as the interest is added to it.

1.10.2. Formulae:

Simple Interest: Amount = P + I

I = (P*N*R) / 100

Where P – Principal, N – Period – Rate of interest, I – interest

It can be written as R = (I x 100) / (P*N)



Compound Interest:

Amount A = P* (1 + R / 100) N

CI = [Amount – Principal].

Example:

Difference between C.I and S.I for 2 Years

C.I – S.I = P*[R / 100] 2

Example:

Difference between C.I and S.I for 3 Years

C.I – S.I = P*[(R / 100)3 + 3 (R / 100)

2]

Tips (Simple Interest):

1. If a certain Sum in T Years at R% per annum amounts to Rs. A, the Sum will be

P = ((100*A) / (100 + (R*T)))

2. The annual payment that will discharge a debt of Rs. A due in T Years at R% per annum

is Annual payment = Rs.(100*A) / ((100*T) + (R*T*(T – 1)) / 2)

3. If a certain Sum of money becomes n times itself in T Years at Simple Interest, then the

Rate of Interest per annum is R = 100*(n – 1) / T %.

4. If a certain Sum of money becomes n times itself in T Years at a Simple Interest, then the

time T1 in which it will become m times itself is given by T

1 = (m – 1)* T / (n – 1)

5. If a debt of Rs. Z is paid in n number of installments and if the value of each installment

is Rs. A, then the borrowed amount is given by,

Z = (n*a) + [(R*a) / (100*b)] * [n*(n–1) / 2]

6. If a certain Sum of money P lent out at SI amounts to A1 in T1 Years and A2 in T2 Years,

then P = (A1T2 – A2T1) / (T2 – T1).

7. If an amount P1 lent at Simple Interest rate of R1% per annum and another amount P2 at

Simple Interest rate of R2% per annum, Rate of Interest for the whole Sum is

R = (P1R1 + P2R2) / (P1 + P2).

Tips (Compound Interest):

8. When the Rates of Interest are different for different Years, say R1, R2, R3 per cent for

first, second and third Year respectively, then

Amount = P*[1+ (R1/100)]* [1+ (R2/100)]* [1+ (R3/100)]

Problem 1:

Find the S.I on a Principal of Rs.1000 at an Interest Rate of 4% per annum for a period of

4 Years.



Solution:

The formula for S.I = PNR / 100

Where P is principal = Rs.1000

R is rate of interest = 4%

N is time period = 4 Years

S.I = 1000*4*4 / 100 = Rs.160

Problem 2:

Find the Simple Interest on Rs. 5200 for 2 Years at 6% per annum.

Solution:

Here, P = Rs.5200, T = 2 Years and R = 6%

Therefore, Simple Interest = P*R*T/100

= 5200*6*2/100

= Rs. 624.

Problem 3:

Mahesh borrowed Rs. 3000 from his friend Suresh at 15 per cent per annum for 3 Years.

Find the Interest and Money returned by Mahesh to Suresh.

Solution:

Here, P = Rs.3000, R = 15% per annum, T = 3 Years

Interest, I = P*R*T / 100

= 3000*15*3 / 100

= Rs.1350.

Amount, A = P + I.

= Rs.3000 + Rs.1350

= Rs.4350.

Problems to Solve:

2. A man borrowed Rs.8000 at the rate of 12% S.I. and lent the whole Sum to another

Person at the rate of 15%. What will be the Gain after 7 Years?

3. In what time, can we get Rs.1200 as S.I. on Rs.5000 at the rate of 8%?

4. Find the rate percent, if the Compound Interest on Rs.15625 for 3 Years is Rs.1951?

5. Mr. Krishna received a loan at 13% per annum S.I. After 4 Years he returned the

principal and interest. If he returned Rs. 9120, what will be the Principal Amount?

6. Brinda borrowed Rs.1000 to build a hut. She pays 5% Simple Interest. She lets the

hut to Ramu and receives the rent of Rs.12.5 per month from Ramu. In how many

Years Brinda ought to clear off the debt?



7. The difference between the C.I. and the S.I. accrued on an amount of Rs.18000 in 2

Years was Rs. 405. Find the Rate of Interest.

8. If the S.I. on a Sum of money at 5% p.a. for 3 Years is Rs.1500 Find the C.I. on the

same Sum for the same period and at the same Rate?

9. A man borrowed Rs.8000 at 12% per annum on S.I. He lent the whole Sum at 12%

p.a on C.I. What will be his Gain after 2 Years?

10. Rs. 3757 is to be divided between A and B such that A‟s Share at the end of 7 Years

may be equal to B‟s Share at the end of 9 Years. If rate percent be 10% p.a.

Compound Interest. Find B‟s Share

11. The Population of a certain village increases by 5% annually. Its present population

is 8000. What will be the population after 3 Years?

1.11. Average 1.11.1. Concept:

The Average or Mean or Arithmetic Mean of a number of Quantities of the same kind is

equal to their Sum divided by the number of those Quantities. For example, the Average

of 3, 9, 11, 15, 18, 19 and 23 is

[3 + 9 + 11 + 15 + 18 + 19 + 23] / 7 = 98 / 7 = 14.

1.11.2. Formulae: 1) Average = Sum of Quantities

Number of Quantities

2) Sum of quantities = Average * Number of Quantities.

3) Number of quantities = Sum of Quantities

Average

Problem 1:

The Average of 9 Results is 30. If the Average of First Five Results is 48 and that of the

last five is 53, Find the Fifth Result.

Solution:

Fifth Result = (48 * 5 + 53 * 5 – 30 * 9)

= (240 + 265 – 270)

= 235

Problem 2:

A man‟s Average daily Expenditure is Rs.10 during May, Rs.14 during June and Rs.15

during July. Find the Average daily Expenditure for the Three months.



Solution:

As there are 31 Days in May, 30 Days in June and 31 in July.

The total Expenditure = (10 * 31+14 * 30+15 * 31) rupees

= (310 + 420 + 465)

= Rs.1195

The number of Days = 31 + 30 + 31 = 92

The Average daily Expenditure = Rs.1195 / 92 = Rs.13 approx.


1. The Mean of 100 Observations was calculated as 40. It was found later on that one of

the Observations was misread as 83 instead of 53. Find the correct Mean.

2. The Average of 11 Observations is 60. If the Average of First Five Observations is

58 and that of the last Five is 56, then Find the Sixth Observation.

3. The Average Salary of a Staff of 18 Officers and 32 Clerks is Rs. 800. If the Average

Salary of the Officers is Rs.1200, find the Average Salary of the Clerks.

4. A Cricketer has a certain Average for 9 Innings. In the Tenth Inning he scores 100

runs, thereby increasing his Average by 8 runs. Find his new Average.

5. Having scored 98 runs in the 19th

Innings, a Cricketer increases his Average score by

4. What will be his Average score after the 19th

Innings?

6. The Average of Ten numbers is 7. If each number is multiplied by 12 find the

Average of new set of numbers.

7. The Average Temperature from Monday to Thursday is 48o and from Tuesday to

Friday is 52o. If the Temperature on Monday is 42

o, what was it on Friday?

8. The Average of marks obtained by 120 candidates was 35. If the Average of the

passed candidates was 39 and that of the failed candidates was 15, find the number of

those Candidates who passed the Examination

9. On an 800-mile trip, a car traveled half the distance at 80 miles per Hour and the

other half at 100 miles per Hour. What was the Average Speed of the car?

10. The Average Height of 30 Boys out of a class of 50 is 160 cm. If the Average Height

of the remaining Boys is 165 cm find the Average Height of the whole class (in cm).



1.12. Permutations and Combinations 1.12.1. Concept:

If an operation can be performed in „m‟ different ways; following which a second

operation can be performed in „n‟ different ways, then the two operations in succession

can be performed in „n*m‟ ways.

If an operation can be performed in „m‟ different ways and another operation, which is

independent of the first operation, can be performed in „n‟ different ways, then either of

the two operations can be performed in (m + n) ways.

Each of the different arrangements which can be made by taking some or all of given

number of things or objects at a time is called Permutation. Let r and n be positive

integers such that 1 < r < n. Then, the number of Permutations of n different things,

taken r at a time, is denoted by the symbol nPr.

Each of the different groups or selections which can be made by taking some or all of a

number of things (irrespective of other) is called a Combination. The number of

combinations of n different things taken r at a time is denoted by nCr.

Factorial: The continued Product of first n natural numbers is called n!

n! = 1*2*3*4*…… (n–1) * n

0! = 1

1.12.2. Formulae:

1. nPr = n! / (n–r)!

2. nCr = n! / [r! (n–r)!]

3. nCo = nCn = 1 , nC1 = n

Tips:

If „n‟ is Even then the greatest value of

nCr is

nCn/2.

If „n‟ is Odd, then the Greatest value of nCr is

nCn+1/2 or

nCn–1/2.

Problem 1:

Find the number of Permutations and combinations which can be made by taking 6 items

at a time from Eight given Distinct items without Repetition.

Solution:

The number of Permutations is 8P6 = n! / (n – r)!

= 8! / 2!

= 20160



The number of Combinations is 8C6 = n! / r! (n – r)!

= 8! / 6! 2!

= 28

Problem 2:

How many different Permutations can be made out of the letters of the word,

„ASSISTANTS‟ have taken all together?

Solution:

In this word, there are 10 letters composed of 4S‟s, 2A‟s, 2T‟s, 1N, and 1I.

Hence, the number of Permutations

= 10! / 4! 2! 2!

= 37800

Problems to Solve: 1 There are 15 buses running between Delhi and Mumbai. In how many ways can a man go

to Mumbai and return by a different bus?

2. If there are 12 Persons in a party, and if each two of them shake hands with each other,

how many handshakes happen in the party?

3. There are 8 students appearing in an examination of which 3 have to appear in a

Mathematics paper and the remaining 5 in different subjects. In how many ways they are

made to sit in a row if the candidates in Mathematics cannot sit next to each other?

4. In how many different ways can the letters of the word „ALLAHABAD‟ be permuted?

5. In how many different ways can the letters of the word „TRAINER‟ be arranged so that

the vowels always come together?

6. How many words, each of 3 vowels and 2 consonants, can be formed from the letters of

the word INVOLUTE?

7. How may different words can be formed with the letters of the word „BHARAT‟. In How

many of these B and H are never together?

8. A select group of 4 is to be performed from 8 men 6 women in such a way that the group

must have atleast one woman. In how many different ways can it be done?

9. There are three different rings to be worn in four fingers with at the most one in each

finger. In how many ways can this be done?

10. If there are 6 periods in each working day of a school, in how many ways can one arrange

5 subjects such that each subject is allowed at least one period?



1.13. Percentage 1.13.1. Concept:

The term Percent Means per Hundred or for every Hundred. It is the Abbreviation of the

Latin phrase „per centum‟. The term percent is sometimes abbreviated as p.c. The

symbol % is often used for the term percent.

A Fraction whose Denominator is 100 is called a Percentage and the Numerator of the

Fraction is called rate percent,

Example: 5% = 5 / 100

5 parts out of every hundred parts.

Tips:

1. To convert a Fraction into a percent, multiply by 100 and put % sign.

2. If A is x% more than B, then B is less than A by [X / (100 + x) * 100] %

3. If A is x% less than B, then B is more than A by [X / (100 – x) *100] %

Problem 1:

Find the Percentage Change in Area if the Length is increased by 20% and Breadth is

decreased by 10% of a Rectangle.

Solution:

Increase in Length = 20%

Decrease in Breath = –10%

Percentage = X + Y + (XY / 100) %

= 20 – 10 – (20*10 / 100) = 8%

Problem 2:

The Price of a Hindi book in 1987 was Rs. 100 but due to devaluation of the rupee it has

risen to Rs. 250. What is the Percentage increase in its Price?

Solution:

Percentage Increase = [(Final Price – Initial Price) / Initial Price]*100%

= [(250 – 100) / 100]*100%

= 150 %


1. In an Examination 75 % of the candidates passed in English, 65% in Mathematics

and 27% failed in both subjects. Find the Pass Percentage.

2. A rise of 20% in the Price of rice compels a Person to buy 4 kg less for Rs. 80.

Find the increased Price per kg.



3. A reduction of 20 percent in the Price of mangoes enables a man to buy 25 mangoes

more for Rs. 40. Find the reduced Price of the basket containing 200 mangoes.

4. A reduction of 12.5% in the Price of a dining table brought down its Price to

Rs. 4375. Find the original Price (in Rs) of the table.

5. A house owner was having his house painted. He was advised that he would

require 25 kg of paint. Allowing for 15% wastage and assuming that the paint is

available in 2 kg cans, what would be the cost of paint purchased, if one can costs

Rs. 16?

6. Swati spends 40% of her Salary on food, 25% on house rent, 15% on entertainment

and 5% on conveyance. If her saving at the end of a month is Rs.1200, find her

Salary per month.

7. The Boys and Girls in a college are in the Ratio 3:2. If 20% of the Boys and 25%

of the Girls are adults, find the percentage of students who are not adults.

8. At an election a candidate who gets 35% of the votes is defeated by a majority of

150 votes. Find the total number of votes recorded.

9. The petrol Prices are reduced by 10%. Find by how much a user must increase the

consumption of petrol so as not to decrease his expenditure on petrol.

10. A candidate must get 33% marks to pass. He gets 220 marks and fails by 11 marks.

What is the maximum number of marks?

1.14. Boats and Streams 1.14.1. Concept:

If a boat moves against the stream that is in the Opposite Direction of the stream, it is

called Upstream. If a boat moves along the stream that is in the direction of the stream,

it is called Downstream. If Water of the river is moving, it is called a Stream. If the

speed of the Water in the river is zero, it is called Still Water.

1.14.2. Formulae: If the speed of the boat be x kmph and the speed of the stream be y kmph then

(a) Speed of the boat down stream (a) = ( X + Y) kmph

(b) Speed of the boat upstream (b) = ( X – Y) kmph

(c) Speed of the boat in still Water = (Downstream speed + upstream speed)

2

= ½ (a + b).

(d) Speed of the stream = ½ (a – b).



Problem 1:

A man can row 2 / 7th of a kilometer upstream in 25 minutes and return in 10minutes.

Find the Speed of the man in still Water.

Solution:

Upstream speed = [(2 / 7) / (25 / 60)]

= 24 / 35 kmph

Down stream speed = [(2 / 7) / (10 / 60)] =12 / 7 kmph

Speed in still Water

= [(24 / 35) + (12 / 7)] / 2

= [(84 / 35) x (1 / 2)]

= 1.2 kmph

Problem 2:

A man can row upstream at 8 kmph and downstream at 10 kmph. Find the man‟s speed

in still Water and the speed of current.

Solution:

Speed in still Water = ½ (10 + 8) kmph

= 9 kmph

Speed of current = ½ (10 – 8) kmph

= 1 kmph.


1. If a man can swim downstream at 8 kmph and upstream at 4 kmph, find the Speed in

still Water.

2. A man can row upstream at 12 kmph and downstream at 17 kmph; find the Speed of

the Stream.

3. A man can row a boat at 6 kmph in still Water. If the Speed of the Stream is 4 kmph,

find the time taken to row a distance of 90 km down the Stream.

4. A man rows 13 km upstream in 5 Hours and 28 km downstream in 5 Hours. Find the

Velocity of the Stream.

5. The Speed of a boat in still Water is 9 kmph and the Speed of the Stream is 1.5 kmph.

A man rows to a place at a distance of 10.5 km and comes back to the starting point.

Find the total time taken by him.

6. A man rows to a place 48 km far and comes back within 14 hours. He finds that he

can row 4 km with the stream in the same time as 3 km against the stream. Find speed

of the Stream.

7. A man rows upstream 12 km and downstream 28 km taking 5 hours each time. Find

the velocity of Water.



8. A man who can swim 48 m/min in still Water swims 200 m against the current and

200 m with the current. If the difference between these two times is 10 min, find the

speed of the current in m per min.

9. In a stream running at 2 km/hr, a motor boat goes 10 km upstream and back again to

the starting point in 55 min. Find the speed of the motorboat in still Water.

10. A man can row 30 km upstream and 44 km downstream in 10 hr. Also, he can row

40 km upstream and 55 km downstream in 13 Hours. Find the rate of the current and

speed of the man in still Water.

1.15. Time and Distance 1.15.1. Concept:

The terms „Time‟ and „Distance‟ are related to the speed of a moving object. The Speed

of an object is defined as the distance covered by it in a unit time of interval. It is

obtained by dividing the distance covered by the object, by the time it takes to cover that

distance.

1.15.2. Formulae:

1. Distance = Speed * Time

2. Speed = Distance

Time

3. Time = Distance

Speed Tips:

1. For converting kmph into m/sec multiply by 5/18

2. For converting m/sec to kmph multiply by 18/5.

3. If two Persons A and B start at the same time from two points P and Q towards

each other and after crossing they take t1 and t2 Hours in reaching Q and P

respectively then

A‟s speed = t2

B‟s speed t1

4. If a Person A covers a distance d1 km at s1 kmph and then d2 km at s2 kmph, then

the Average speed during the whole journey is given by,

Average speed = [s1*s2*(d1 + d2)] / [(s1*d2) + (s2*d1)]

5. If A goes from X to Y at s1 kmph. and comes back from Y to X at s2 kmph, then

the Average speed during the whole journey is given by,

Average Speed = 2s1*s2 / (s1 + s2)



6. If a Person goes certain distance (A to B) at a speed of s1 kmph and returns back

at a speed of s2 kmph. If he takes T Hours in all, then the distance between A and

B is = T*[s1*s2] / (s1 + s2)

Problem 1:

In how many seconds does a 280 m long train moving at 208 kmph cross a platform of

length, 250m?

Solution:

Distance covered = 280 + 250

= 530m

Speed = 208*5 / 18 = 57.78 m/sec

Time = Distance / Speed

= 530 / 57.78 = 9.17 Sec.

Problem 2:

The distance between two cities A and B is 330 km. A train starts from A at 8 A.M. and

travels towards B at 60 km / hr. Another train starts from B at 9 A.M. and travels

towards A at 75 kmph. At what time do they meet?

Solution:

The Trains may meet x Hours after 8 a.m.

Let the Trains be P and Q respectively

Distance by P in „x‟ hrs + Distance by Q in (x –1) hrs = 330

60x + 75 (x –1) = 330

x = 3

They meet at 11 A.M.

Problem 3:

Find the Length of the Bridge which a train 120 m long traveling at 54 kmph can cross in

30 seconds.

Solution:

Speed of the Train = 54 kmph

54 * 5 / 18 = 15 m/sec

Distance covered in 30 seconds = 15 *30

= 450m

Length of the Bridge = Distance covered – Length of the Train

= 450 – 120

= 330m


1. Vikas can cover a certain Distance in 1 hr 24 min by covering two third of the

distance at 4 kmph and the rest at 5 kmph. What is the total distance?



2. If a Boy walks from his house to school at the rate of 4 kmph, he reaches the school

10 min earlier than the schedule time. However, if he walks at the rate of 3 kmph, he

reaches 10 min late. What is the distance of the school from his house?

3. I had to be at a certain place at a certain time and I found that I shall be 40 min too

late, if I walk 3 km an hr and 30 min too soon, if I walk 4 km an hr. How far I have to

walk?

4. If a man runs at 3 m per sec, How many kilometers does he run in 1 Hour 40 min?

5. A man travels uphill with an Average speed of 24 kmph and comes down with an

Average speed of 36 kmph. The distance traveled in both the cases being the same,

what is the Average speed during the journey?

6. A train running at the rate of 40 km an Hour meets a Person, walking along the line in

the opposite direction at the rate of 4 km an Hour and passes him in 5 sec. Find the

length of the train

7. A bullock cart has to cover a distance of 80 km in 10 Hours. If it covers half the

distance in 3/5th

of the time, What should be its speed to cover the remaining distance

in the time left?

8. Excluding stoppages, the speed of the train is 45 kmph and including stoppages, it is

36 kmph. For how many minutes does the train stop per Hour?

9. A car starts with the initial speed of 40 kmph, with its speed increasing every Hour by

5 kmph. How many Hours will it take to cover a distance of 385 km?

10. Two Men start to walk together to a certain distance, one at 4 kmph and another at 3

kmph. The former arrives half an Hour before the latter. Find the distance.

1.16. Problems on Ages 1.16.1. Concept:

To solve these Problems, the knowledge of linear equations is essential. There may be

three situations:–

1. Age some Years ago.

2. Present age.

3. Age some Years hence.

Simple linear equations are to be framed and the solution can be obtained.

Problem 1:

The Ages of two Persons differ by 10 Years. If 5 Years ago, the elder one was 2 times as

old as the younger one, Find their Present Ages.



Solution:

Let the age of the Younger Person be x Years.

Then, age of the Elder Person = (x + 10) Years

2(x–5) = (x + 10 – 5)

2x – 10 = x + 5

x = 15

Hence, their Present Ages are 15 Years and 25 Years.

Problem 2:

The Present Age of a mother is 4 Years more than Four Times the age of her Daughter.

Four Years hence, mother‟s age will be 6 Years more than twice the age of the Daughter.

Find the Present Age of the mother?

Solution:

Let the Daughter‟s present age be x Years.

Then, Mother‟s Present Age = (4x + 4) Years

Four years hence (4x + 4 + 4) = 2(x + 4) + 6

4x + 8 = 2x + 14

x = 3.

Hence, Mother‟s Present Age = (4x + 4) = (4*3 + 4) = 16 Years.


1. One Year ago, a father was four times as old as his son. In 6 Years time his age will

exceed twice his son‟s age by 9 Years. Find the Ratio of their present ages

2. The Average age of five members of a family is 21 Years. If the age of the

grandfather is included, the Average is increased by 9 Years. Find the age of the

grandfather

3. The difference between the ages of two Persons is 10 Years. 15 Years ago, the elder

one was twice as old as the younger, Find the present age of elder

4. The ages of Ram and Shyam differ by 16 years. Six years ago, Mohan‟s age was

thrice as that of Rams. Find their present age?

5. One Year ago, Promila was four times as old as her daughter. Six Years hence,

Promila‟s age will exceed her daughter‟s age by 9 Years. Find the Ratio of the present

ages of Promila and her daughter

6. The Ratio of Laxmis age to the age of her mother is 3:11. The difference of their ages

is 24 years, find the Ratio of their ages after 3 years?

7. Three years ago, X‟s age was double of Y‟s. Seven years hence the Sum of their

united ages will be 83 years. find the age of X today



8. The Ratio of father‟s ages to son‟s age is 3:1. The Product of their ages is 243. What

will the Ratio of their ages be 3 years hence?

9. Narmatha‟s father is now four times her age. In 5 years, he will be 3 times her age. In

How many years, will he be twice her age?

10. Three Years ago, the Average age of a family of 5 members was 17 Years. A baby

having been born, the Average of the family is the same today. What is the age of the

baby? 1.17. Profit and Loss

1.17.1. Concept: The aim of every Business is to earn Profit. The terms that describe Profit is Cost Price

and Selling Price. The Cost Price of an article is the Price at which an article has been

purchased. It is abbreviated as C.P. The Selling Price of an article of an article is the

Price at which an article has been sold. It is abbreviated as S.P.

If the selling Price of an article is more than the cost Price, there is a Gain or Profit. If

the cost Price of an article is greater than the selling Price, it is defined as Loss.

1.17.2. Formulae: 1. Profit (P) = S.P. – C.P

2. Loss (L) = C.P. – S.P

3. Profit %( P %) = (P /C.P)* 100 %

4. Loss % = (L /C.P)* 100 %

5. S.P. = [100 + P%] * C.P.

100

6. C.P. = [ 100 ]* S.P.

100 + P%

7. S.P. = [(100 – Loss %) /100] * C.P.

8. C.P. = [100 / 100 – Loss%] * S.P.

Problem 1:

A Girl buys an article for Rs. 17.50 and sells it for Rs. 18.60. Find her Gain percent.

Solution:

C.P = Rs.17.50

S.P = Rs.18.60

So, Gain = Rs. (18.60 – 17.50)

= Rs.1.10

Gain% = ((1.10 / 17.50)*100) %

= 6.28%



Problem 2:

An Article is bought for Rs. 80 and sold at a Gain of 20%. What is Selling Price of the

book?

Solution:

Gain = Rs. 5, so Gain % = (5/20) * 100 =25%


1. A Man bought 5 apples for Rs. 3 and sold each for Rs. 2. What did he Gain or Loss?

2. A man sold two pens at Rs.20 each. He sold one at a loss of 10% and the other at a Gain

of 10%. Find his Goss or Gain %.

3. If Cost Price of 8 articles is equal to the selling Price of 10 articles. Find the Gain or

Loss.

4. Vikas bought paper sheets for Rs. 7200 and spent Rs. 200 on transport. Paying Rs. 600,

he made 330 boxes, which he sold at Rs. 28 each. Find his Profit percentage.

5. A sells a bicycle to B at a Profit of 20% and B sells it to C at a Profit of 25%. If C pays

Rs.1500, What did A pay for it?

6. The profit earned after selling an article for Rs.625 is the same as loss incurred after

selling the article for Rs. 435. What is the Loss percent?

7. A machine is sold at a Profit of 10%. Had it been sold for Rs. 80 less, there would have

been a loss of 10%. Find the C.P of the machine.

8. A cloth merchant says that due to slump in the market, he sells the cloth at 10% loss but

he uses a false metre scale and actually Gains 15%. Find the actual length of the scale.

9. If a Person makes a Profit of 10% on 1/4th

of the quantity sold and a loss of 20% on the

rest, then what is his Average percent Profit or Loss?

10. When a producer allows 36% commission on retail Price of his Product, he earns a Profit

of 8.8%. What would be his Profit percent, if commission is reduced to 24%?

1.18. Calendar

1.18.1. Concept: The Process of Finding the day of a given date depends upon the number of Odd Days

present. The Days more than the complete number of weeks in a given period is called

Odd Days. The Year (except century) which is divisible by 4 is called a Leap Year,

where as century is a leap Year by itself when it is divisible by 400. Leap Year has 366

Days, i.e. 52 Weeks and 2 Odd Days. An Ordinary Year has 365 Days, i.e. 52 weeks and

1 Odd day.



Tips: 1. A Century has 76 Ordinary Years and 24 Leap Years

i.e. 76 + 48 Odd Days = 124

= 17 weeks and 5 Odd Days.

2. 200 Years contain 3 Odd Days.

3. 400 Years contain 1 Odd day.

4. Since the order is Cyclic (i.e. every Century has 0, 1, 3 or 5 Odd Days)

5. The First Day of a Century must either be Monday, Tuesday, Thursday or Saturday.

6. Last Day of a Century cannot either be Tuesday, Thursday or Saturday.

7. To find the Day of the Week on a particular date when reference date given:

Find the net number of Odd Days for the period between the reference date and the

given date.

8. To find the Day of the Week on a Particular Date when no Reference Day is given:

Count the net number of Odd Days on the given date and write the day

corresponding to the Odd day.

Odd Days:

Sunday = 0

Monday = 1

Tuesday = 2

Wednesday = 3

Thursday = 4

Friday = 5

Saturday = 6

Month code:

Ordinary Year has the Odd Days as follows

January = 0 Odd Days February = 3

March = 3 April = 6

May = 1 June = 4

June = 6 August = 2

September = 5 October = 0

November = 3 December = 5

Month code for leap Year adds 1 with ordinary Year’s corresponding month code.

Problem 1:

11th

January 1997 was a Sunday. What day of the Week was on 7th

January 2000?



Solution:

Total number of Days between 11th

January 1997 and 7th

January 2000

= (365 – 11 Days) in 1997 + (2*365 Days) in 1998 and 1999 + 7 Days in 2000.

= (50weeks + 4 Odd Days) + 2*(52 weeks and 1 Odd day) + 0 Odd day

= 6 Odd Days

Hence, 7th

January 2000 would be 6 Days ahead of Sunday, i.e. it was on Sunday.

Problem 2:

What Day of the Week was 5th

June 1999?

Solution:

(A+B+C+D)-2, Take the Remainder

7

A = Divide the year by 7 and take the Remainder

1985/7, Remainder is 4

B = Divide the year by 4. Take the Quotient and divide it by 7. Take the Remainder.

1985/4 = 496; 496/7, Remainder is 6

C = Divide the Date by 7 and Take the Remainder.

2/7, Remainder is 2

D = Month Code

June Code is 4

(4+6+2+4)-2 = (14-2)/7 = 5

7

Hence, 5th

June 1999 was Saturday.


1. 2 nd

July 1985 was Wednesday. What Day of the Week was 2nd

July 1984?

2. India got Independence on 15th August 1947. What was the Day of the Week?

3. The First Republic Day of India was celebrated on 26th January 1950 What

was the Day of the Week on that Date?

4. Smt. Indira Gandhi died on 31st October 1984.What was the Day of the Week?

5. What was the Day of the Week on 26th June, 2002?



1.19. Clocks 1.19.1. Concept:

The Circumference of a Dial of a Clock is divided into 60 Equal Parts called Minute

Spaces. The Clock has two hands – the Hour Hand and the Minute Hand. The Hour

hand (or short hand) indicates time in Hours and the Minute Hand (or the long hand)

indicates time in Minutes. In an Hour, the Hour hand covers 5 minute spaces while the

minute hand covers 60 minute spaces. Thus in One Hour or 60 minutes, the minute hand

gains 55 minute spaces over the Hour hand. In every Hour the hands are twice at Right

angles. In every Hour both the hands coincide once.

1.19.2. Formulae:

To find the Angle between the hands

= 30 h – 11m or 11m – 30 h, where h denotes Hour and m denotes Minutes.

2 2

Tips:

1. The Minute Hand moves 12 times as fast as the Hour Hand.

2. The Two Hands of the Clock will be together between H and (H + 1) o „clock at (60H /

11) minutes past H o‟clock.

3. The Two Hands of the Clock will be at Right Angles between H and (H + 1) o‟ Clock at

(5H 15)12 / 11 Minutes past H o‟ Clock.

4. If the Minute Hand of a Clock overtakes the Hour Hand at intervals of M minutes of

correct time. The Clock gains or loses in a Day by [(720 / 11) – M]*(60*24 / M) minutes.

Problem 1:

Find the Angle between the Minute Hand and Hour Hand of a Clock at 07:20.

Solution:

Angle between the two hands at 7: 20

= 30h – (11m / 2)

= 30x7 – (11*20 / 2)

= 100 Degrees.

Problem 2:

How many times do the Hands of a Clock Coincide in a Day?

Solution:

The Hands of a Clock Coincide 11 times in every 12 Hours

The Hands Coincide 22 times in a Day.




1. Find the Angle between the Hour Hand and the Minute Hand of a Clock when the

time is 3.25?

2. At what Time between 4 and 5 o‟clock will the Hands of a Watch Point in

Opposite Directions?

3. At what Time between 9 and 10 o‟clock will the hands of a Watch be together?

4. At what time between 5.30 and 6 o‟clock will the hands of a Watch be 90 ?

5. How much does a Watch Lose per day, if its hands coincide every 64 Minutes?

2. Analytical Reasoning

Concept: As it is very much evident from the word „analytical‟ itself that this type of reasoning is

based on the analysis of the statements which are there in the question, the approach for

solving analytical reasoning questions varies from question to question as analytical

reasoning questions are of different types.

There are two ways of solving analytical reasoning questions. The first way is the one

which consists of the general method for solving any type of analytical reasoning

questions. This method consists of a standard approach which must be followed in all the

analytical reasoning questions (based on the need).The other approach is based on the

specific type of the questions.

Tips for Solving Analytical Reasoning questions:

1. Read the statements very thoroughly and repeatedly to clearly understand the

meaning (known or unknown) which they carry.

2. Do not jump from one statement to the other before completely understanding the

statements.

3. Use notations, symbols and abbreviations wherever required, preferably.

4. Mark the keywords which are present in the statements.

5. Organize the information given in the question in the form of suitable Tables, Maps,

or Tree.

6. Do not make unnecessary assumptions.

2.1. Deduction

2.1.1 Concept: Deduction Means reducing two statements into one or three statements into one. It is also

known as Syllogism. Syllogism is a noun which means a form of reasoning in which a

conclusion is drawn from two statements, i.e., Deductive Reasoning. In more clear terms

Syllogism is a mediate deductive inference in which two propositions are given in such

an order that they jointly or collectively imply the third. Thus Syllogism can be defined



as „a form of reasoning in which the conclusion establishes a relation between two terms

on the basis of both terms being related to the same third term as derived in the premises.‟

For example,

1. All human beings are mortal.

2. The child is a human being.

3. The child is mortal.

The conclusion is reached through the medium of a middle term, i.e., „human being‟,

with both subject (children) and the predicate (mortal). Therefore, in a Syllogism two

premises are necessary to arrive at a conclusion.

Types:

The types of deduction are: Deduction in two statements and Deduction in three

statements.

If the statement starts with,

All, No –––– Universal Statement

Some, Many–––– Particular Statement

No, Not–––– Negative Statement

Universal Affirmative: If the statement starts with „ALL‟ or „NO‟ it is Universal

Affirmative.

Universal Negative: If the statement starts with „NO‟ it is Universal Negative. If the

statement starts with „ALL‟ and if the statement has the word „Not‟ it is Universal

Negative.

Particular Affirmative: If the statement starts with „SOME‟ or „MANY‟ it is Particular

Affirmative.

Particular Negative: If the statement starts with „SOME‟ or „MANY‟ and if the

statement has the word „Not‟ it is Particular Negative.

√ –DISTRIBUTED X –UNDISTRIBUTED

Subject Predicate

Universal

Affirmative √ X

Universal

Negative √ √

Particular

Affirmative X X

Particular

Negative X √

Rules for Deductions:

The middle term must be distributed at least once.

Two statements should have only Three Distinct terms.



The Middle term should not come in the conclusion

If both the statements are „Particular‟ there would be No Conclusion

If both the statements are „Negative‟ there would be No Conclusion.

If one Statement is Particular, the Conclusion would be Particular

If one Statement is Negative, the Conclusion would be Negative.

Stat. 1 All No Some All All Some

Stat. 2 All No Many Some No No

Conclusion All No

Conclu.

No

Conclu. Some No Some,

Not

Universal Affirmative – One Conclusion.

Universal Negative – Two Conclusions.

Particular Affirmative – Two Conclusions.

Particular Negative – One Conclusion.

Problem 1:

All pens are pots.

All pots are jacks

Solution:

Statement 1: Universal affirmative

Subject (pens) – Distributed

Predicate (pots) – Undistributed

Statement 2: Universal affirmative

Subject (pots) – Distributed

Predicate (jacks) – Undistributed

Conclusion: “All pens are jacks”

2.1.2 Problems to Solve

1. All bats are keys

All locks are keys

No keys are doors

2. All horses are fruits

Some boxes are trees

Some trees are horses

3. All dogs are Parrots

No train is dog

Some buses are trains



4. Some guns are Pistols

All Pistols are Bombs

All Crackers are Pistols

5. Some fools are professionals

All engineers are fools

6. Some dogs are cats

None of the cats are cows

7. Shyam is bachelor

All bachelors are intelligent

8. All sports are enthusiastic

Cricket is not a sport

9. No magazines are caps

All caps are cameras

10. All poles are Guns

Some Boats are not Poles

2.2 Linear Sequencing 2.2.1. Concept:

Arranging the items in a linear fashion is called Linear Sequencing, i.e. arranging it in

single row manner. There may be 4–5 questions asked from a single linear pattern given.

The direct condition should be found first and then the question can be solved, because

the remaining conditions are based on this direct condition.

Problem 1:

There are five Persons (A, B, C, D, and E) in a room who are seated in a single row in the

following manner: A does not sit immediately next to E. E and D sit immediately next to

each other. C is at the extreme right end of the row. A is immediately to the left of C.

Who must sit in the middle seat of the row?

Solution:

First draw linear arrangement as given below; – – – – –

C is at the extreme right.

i.e. __ __ __ __ C

A is to the immediate left of C.

i.e. __ __ __ A C

E does not sit next to A, So occupy 1 or 2 seat

i.e. __ E __ A C or E __ __ A C

D sits next to E.

E takes 2 seats: i.e. either DEBAC or BEDAC

E takes 1 seat: EDBAC

Hence Middle Seat: B or D




1. Eleven Students A, B, C, D, E, F, G, H, I, J and K are sitting in the first row of the class

facing the teacher. D is to the immediate left of F, who is second to the right of C. A is

second to the right of E, who is at one end. J is the immediate neighbor of A and B is

third to the left of G. H is to the immediate left of D and third to the right of I.

I. Who is sitting in the middle of the row?

II. Which of the following groups of friends are sitting to the right of G?

a. CHDE b. CHDF c. IBJA d. ICHD

2. In a Car Exhibition, seven cars of seven different companies such as Cadillac,

Ambassador, Fiat, Maruthi, Mercedes, Bedford and Fargo were displayed in a row,

facing the east such that:

Cadillac car was to the immediate right of Fargo. Fargo was fourth to the right of Fiat.

Maruthi car was between Ambassador and Bedford. Fiat, which was third to the left of

Ambassador Car, was at one end.

I. Which of the following was the correct position for the Mercedes?

a) Immediate right of Cardilac b) Immediate left of Bedford

c) Between Bedford and Fargo d) Fourth to the right of Maruthi

II. Which cars are on the either side of the Cardilac cars?

a) Ambassador and Maruthi b) Maruthi and Fiat

c) Fiat and Mercedes d) none of these

3. There are seven books one each on Psychology, Hindi, English, Sociology, Economics,

Education and Accountancy, lying on the table one above the other. Sociology is on top

of all the books. Accountancy is immediately below Education which is immediately

below Sociology. Economics is immediately above Psychology but not in the middle.

Hindi is immediately below psychology.

I. Economics is in–between which of the following books?

a) Accountancy and Education b) Psychology and Hindi

c) English and Psychology d) Psychology and Sociology

II. If Sociology and English, Accountancy and Hindi and Education and Psychology

interchange their positions, which book will be between Psychology and Sociology?

a) Accountancy b) Psychology

c) Hindi d) Economics

4. In a pile of reading material, there are novels, story books, dramas and comics.

Every novel has a drama next to it, every story–book has a comic next to it and there

Is no story–book next to a novel. If there be a novel at the top and the number of

Books is 40, what would be the order of the books in the pile?



a) NSCD b) NDSC c) CSDN d) DNCS

5. There are 5 houses – H1, H2, H3, H4, H5 in a row H1 is to the right of H2 & H5 is to

the left of H3 and right of H1, H2 is to the right of H4. Which is to the house in the

middle?

6. The buses stand in a row. One of the 2 buses at the extreme end is red and the other

blue. A yellow bus is standing to the right of green. An orange bus is to the left of

blue bus & Green bus stands between the red and yellow buses.

Orange bus stands at ________ Position from right.

Which bus stands at the centre?

Yellow bus stands to the left of _____?

7. 5 people A, B, C, D, and E read a novel. One who finishes first gives it to C & one

who reads last had taken it from A. E has not read it first or lastly two people read

between B & A.

Whom did B pass the novel to?

Who read the novel last?

8. There are 6 Persons (A, B, C, D, E & F) in a Team who are seated in a linear manner.

A sits in between E and D. D sits in one extreme of the row. There are two places in

between B and E.

How many arrangements are possible?

If C takes the 2nd

spot and E shifts the place to B How many people are sitting in

between E and D?

9. M, N, O, P, Q, R & S are 7 Persons who are sitting in a row. O sits to the immediate

left of S, and R is to the immediate right of P. N has equal number of people on

either side of him. M is to the immediate left of N and there are exactly three Persons

between M and S.

If the seven Persons sit facing forward, viewing from front, then who sits to the

immediate right of Q?

If Q wants to sit to the immediate left of N, then with whom should he exchange

the seat?

How many Persons are there between R and Q?

10. Nine Persons A, B, C, D, E, F, G, H, I are seated in a row. The seats are numbered

from 1–9 from left to right as per the following restrictions. C, F, I sit together. E is to

the immediate right of H. B sits in the fourth seat from the left end and D sits in the

third seat from right end. E is not between B and D.

If G sits to the immediate left of D then who will sit in the centre seat of the row?

If F sits between C and I, then in how many ways altogether can all of them be

seated in the row?

Which of the following is at one end of the row?

11. Five Persons L, M, N, O, and P are sitting in a line. L is to the right of N; M is to the

left of P. N is to the right of O; M is to the right of L.

Who is sitting second from your right side?



Identify the Persons sitting on the extremities.

2.3. Double Lineup 2.3.1. Concept:

The question will consist of at least two different variables. We have to form a table.

There will be more than one question per passage.

Problem 1:

P, Q, R, S, T, U, V, W are 8 employees allocated with eight different lockers numbered

1–8. The lockers are arranged in four rows with two lockers in each row. Lockers 1 and

2 in top row from left to right and lockers 7 and 8 in bottom row from left to right.

Lockers 3 and 4 are in the second row from top arranged from right to left and so do 5

and 6. P has been given locker 1 while v has been allocated locker 8. T has locker just

above that of Q which is just above that of R whereas W locker is in bottom row.

1. Draw the table?

2. Which locker is allocated to employee Q?

3. How many arrangements are possible?

4. If W and P are interchanged what does the table look like?

5. If S is given locker 4 how many solutions are possible?

Solution:

4 Rows and 4 Columns are given.

Fitting the data in the table results in

Left Right

P 1 2 T

S / U 4 3 Q

S / U 6 5 R

W 7 8 V

1) Employee Q is allocated locker 3

2) There are 2 arrangements possible

3) W would take the top row and P would take the bottom row

4) If S is given locker 4 there is only 1 solution possible.

2.3.2. Problems to Solve 1) On the shelf are placed six volumes side by side labeled A, B, C, D, E and F. 3

volumes C, B and E have blue covers while the rest have red covers. D and F are new

volumes while the rest are old volumes. A, C and D are law reports while the rest are

gazetteers

Which is the red colored new law report volume?

Which two old gazetteers have blue covers?



2) P,Q,R,S,T and U are 6 post offices. In each of which one is a post man out of A, B,

C, D, E and F workers. No postman works in two post offices. A works neither in

post office P nor in S, while B works neither in post office T nor in U. C works

neither in post office Q nor in R but D works in T and A works in Q. If B works in

post office P and C does not work in post office U then E works in

(a) S (b) U (c) R (d) R or U

3) 4 people (A, B, C & D) sit and have dinner in a room. The various dishes provided

are (W, X, Y & Z). Each of them has one dish each as follows, If A has Y, B cannot

have W, C has Z. What are the different arrangements possible?

4) There were 6 friends A, B, C, D, E & F. Every one is good in one of the following

Games, Baseball, Volleyball, Carrom, Hockey, TT and Polo. Each owns a different

colored car Yellow, Green, Black, White, Blue and Red. D plays Polo and has yellow

car. C does not play TT & Hockey and owns neither blue nor yellow car. E has white

car and plays Baseball. B does not play TT and has red car. A plays Carrom and he

has black car.

Who plays Volley ball?

Which colored car F owns?

5) 6 friends Ramesh, Dinesh, Lokesh, Nilesh, Shaliesh & Hitesh work in different

companies namely P, Q, R, S, T & U and each one wears company sponsored

different coloured tie such as Blue, Green, Pink, Yellow, Purple & Red though not

necessarily in the same order. The one wearing Blue tie works in Company S and the

one wearing Green tie works in Company P. Hitesh does not work in company R.

Ramesh wears Pink tie and works in Company Q. Nilesh does not work in Company

T and purple colour tie is not sponsored by Company R. Shaliesh works in company

U and neither Nilesh nor Dinesh works in company S. Company T does not Sponsor

Purple or Yellow colored tie and Lokesh works in Company P.

Which Colour Tie is sponsored by company R?

Which of the following Colour Company Person combination is correct?

a) Green – R – Nilesh b) Blue – S – Lokesh

c) Red – T – Dinesh d) Yellow – R – Shaliesh

Which of the following is true?

a) Company U sponsored Green tie b) Shailesh wears Red tie

c) Nilesh works in T d) Red Colour is sponsored by T

6) Among Ram, Lakhan and Kishan are a doctor, a teacher and an engineer. They are

married to Radha, Sita and Gita not in that order. Each of their wives is working. The

day Ram and Lakhan saw the film “Sita aur Gita”, Sita and Gita saw the film Ram

Lakhan. Kishans wife is an artist and regularly holds exhibitions at Venkatappa art

gallery, Bangalore.

Ram and Gita are having a love affair secretly.

Teacher‟s wife teaches in St. Joseph‟s Convent.

Radha is married to the Engineer.



Sita is a good cook and one can find her recipes in every issue of Women‟s Era

magazine.

Who is the Doctor?

a) Ram b) Lakhan c) Kishan d) Can‟t Say

Who is Ram‟s wife?

a) Gita b) Sita c) Radha d) Data insufficient

Who is the Artist?

a) Gita b) Sita c) Radha d) Can‟t Say

Who is Radha‟s Husband?

a) Ram b) Lakhan c) Kishan d) Can‟t Say

Who is married to the teacher?

a) Radha b) Sita c) Gita d) Can‟t Say

Who is the Engineer?

a) Ram b) Lakhan c) Kishan d) Data Insufficient

7. I like the following 5 Places in India A, B, C, D, E because of the following

Characteristics. B and E have Seashores. B, D & A are metropolitan. E is a

peaceful city. A & E are is of historic significance.

Which city has seashore and historic significance?

Which city is both a metropolitan and has seashore?

Which city is not peaceful but has historic significance?

8. Six Products – U, V, W, X, Y and Z – are display windows of a shop. There are six

display windows number 1, 2,3,4,5 and 6. One Product is to be put in one window.

Moreover U cannot be immediately to the left or immediately to the right of V. W

must be immediately to the left of X, Z cannot be in window number 6.

If X is placed in window no.3, W must be placed in which window?

If U is placed in window no.3, immediately to the right of X which Product

must be placed in window no.5?

9. There are four friends Dinesh, Sunil, Rahul and Atul. One of them studies Commerce

and plays Golf and Table tennis, Dinesh and Sunil study Maths. Dinesh plays cricket.

Both the Maths students play Football. Rahul is a student of Physics. The Physics

student plays Football and Badminton. All the friends play two games each and study

one subject each.

Who is the student of commerce?

Who does not play Football?

Who plays cricket?

Who studies Maths and plays Cricket?

10. Five executives of CHOGM hold a conference in Delhi. Mr. A can speak Spanish and

Hindi Mr. B understands Spanish and English Mr.C converses in English and Hindi.

Mr. D speaks French and understands Spanish quite well. Mr. E, a native Indian, can

also speak French.

Which of the following can act as an interpreter when Mr. C and Mr. D wish to

converse ?

a) Only Mr. A b) Only Mr. B c) Only Mr. E

d) Mr. A or Mr. B e) any of the other three executives.



Which of them can not converse without an interpreter?

a) Mr. B and Mr. E b) Mr. A and Mr. B c) Mr. A and Mr. C

d) Mr. B and Mr. D e) Mr. A and Mr. E

Of the languages spoken at this conference, choose the least common language:

a) English and Spanish b) English and French c) Hindi and Spanish

d) English and Tamil e) French and Spanish

2.4. Ordering

2.4.1. Concept: Items are to be arranged based on the conditions given. The questions will be based on

the Person‟s weight, height, age, rank etc. Here the terms like greater than, less than, not

equal to will come into play.

Types:

Ordering:

If More than one Variable is given in the problem, use ordering method according to the

condition.

Elimination:

If the Problem contains only one variable, then use elimination or ordering method,

according to the condition.

Problem:

K, L, M, N, O and P are 6 Men. Neither M nor N is the tallest. L is taller than P but

shorter than K. O is as tall as P but taller than M. Which is the tallest?

Solution:

The Persons are K, L, M, N, O and P.

Statement 1: Neither M nor N are tallest

So, Remaining: K L O P

Statement 2: L is taller than P

So, Remaining: K L O

Statement 3: L is shorter than K

So, Remaining: K O

Statement 4: O is as tall as P.

Since P Eliminated, O is not.

Hence: K is the tallest.



2.4.2. Problems to Solve

1. Four People Arti, Banti, Chetan and Dolly have Rs. 100 with them. Arti and Banti

together have as much money as Chetan and Dolly. Arti has more money than Banti.

Chetan has half as much money as Dolly. Arti has Rs. 5 more than Dolly.

How much does Banti have?

a) 12.03 b) 11.6 c) 13.3 d) None of these

Who has the second highest amount?

a) Arti b) Chetan c) Dolly d) Banti

2. A, B, C, D, E, and F are six students in a class. B and C are shorter than F but heavier

than A. D is heavier than B and taller than C. E is shorter than D but taller than F. F

is heavier than D. A is shorter than E but taller than F.

Who among them is the Tallest?

Who is the third tallest from the top when arranged in descending order?

Who among them is the lightest?

3. I have 5 balls B1, B2, B3, B4 & B5 all of different weights. B1 is twice as much as

B2, B2 is 4½ times heavier than B3, B4 is twice of B3, but ½ of B5‟s weight. B5 is

less than B1 but more than B3 in weight. Find which is lightest.

4. A, B, C, D and E are five friends. B is elder to E, but not as tall as C. C is younger to

A, and is taller to D and E. A is taller to D, but younger to E. D is elder to A but is

shorter in the Group.

Which of the following pair is elder to D?

a) BA b) BC c) BE d) EA e) None of these

If another friend F is taller than C how many of them will be between F and E

according to height?

a) None b) one c) two d) three e) None of these

5. J, K, L, M, and N are five Boys in a class. They are ranked in the order of their

heights from tallest to shortest and in the order of cleverness from cleverest to dullest.

K is taller than N, but not as clever as J and L, whereas M is the cleverest of all but

shorter than J. While L is shorter than M but taller than K, L is not as clever as J. No

two Persons got the same ranks in any parameters.

Who is the third tallest?

How many people are shorter than K?

If L is the third cleverest then who is the dullest?

6. H is six months younger to Sita while B is three months younger to A, when Sheila

was born, B was 2 months old. A belongs to the age group of five while H belongs

to the age group of six. Who among them is the oldest?

7. Raksha is younger than Saksha and older than Sita. Anju is younger than Manju and

older than Sanju. Manju is younger than Sita and older than Jugal. Jugal is younger

than Sanju and older than Mughal. Saksha is younger than Beena and older than

Heena. Heena is older than Raksha.



Who is the Youngest?

a) Anju b) Sanju c) Jugal d) Mughal

Who precedes Sanju, Jugal, Mughal in age?

a) Manju b) Sita c) Anju d) Raksha

8. There are five friends – Sachin, Kunal, Mohit, Anuj and Rohan. Sachin is Shorter

than Kunal but taller than Rohan. Mohit is the tallest. Anuj is little shorter than

Kunal and little taller than Sachin?

Who is the Shortest?

Who is the second tallest?

9. A Blacksmith has five iron articles A, B, C, D and E, each having a different weight.

A weighs twice as much as B

B weighs four and a half times as much as C

C weighs half as much as D

D weighs half as much as E

E weighs less than A but more than C

Which is the lightest in weight?

Arrange the articles in descending order of their weights.

10. Among 5 friends, Mohan is older than Raju but not as old as Lalit. Lalit is Older than

Neelesh and Kabir. Neelesh is younger than Raju but not the youngest. Who is the

fourth in the descending order of age?

2.5. Seating Arrangement

2.5.1. Concept: It is the manner in which the people are seated in Place. Based on the question, draw the

shape like Square, Triangle, etc. Mostly it is in the form of a circle. We have to draw

the Shape which is given in the question and arrange them according to the conditions

given.

Problem 1:

P, Q, R, S and T are 5 people who sit at a round table.

P sits two tables to the left of R and Q sits two tables to the right of R.

1. If S is in between Q and R what is the arrangement?

2. How many different arrangements are possible?

3. If S is not sitting between Q and R then who is sitting between Q and S?

Solution:

1. If S is between Q and R then the arrangement is QSRTP.

2.Two different arrangements are possible.

3.Either P is sitting when viewed anticlockwise or T and R when viewed clockwise.




1. There are five different houses A to E in a row. A is to the right of B and E is to left of

C and right of A. B is to the right of D. Which of the houses is in the middle?

a) A b) B c) D e) E

2. P, Q, R, S, T and U are seated in a circle with P opposite to S. T has two seats in

between himself and U. U is to the immediate left of P. What is the formation like?

How many possibilities are there?

3. If 6 people A, B, C, D, E and F sit at a round table with the following conditions

given how does the arrangement look like:

A sits opposite to E.

D sits to the immediate left of A and B sits two places to the right of A.

4. Eight friends A, B, C, D, E, F, G and H are sitting in a circle facing the centre. B is

sitting between G and D. H is third to the left of B and second to the right of A. C is

sitting between A & G and B & E are not sitting opposite to each other. Who is third

to the left of D?

5. A group of eight members sit in a circle. D is between A and F and is opposite to G. E

is to the right of A but on the left of C, whose right hand neighbor is G. B enjoys

having H to his left and F to his right. Find the member who is diagonally opposite to

A.

6. Six friends A, B, C, D, E and F are seated in a closed circle facing the centre. E is to

the left of D. C is between A and B. F is between E and A. Who is to the left of B?

7. Six Persons A, B, C, D, E and F are standing in a circle. B is between F and C; A is

between E and D; F is to the left of D. Who is between A and F?

8. L, M, N, P, Q, R and S are sitting in a circle and playing cards. N who is the neighbor

of P, is not the neighbor of R. Q is second to the left of R. N is second to the left of S ,

who is the neighbor of M. Who is immediate right of S?

9. 6 people A, B, C, D, E and F sit around a table for dinner. Since A does not like C,

he doesn't sit either opposite or beside C. B and F always like to sit opposite each

other. If D is adjacent to F then who is adjacent to C?

(a) E and B (b) D and A (c) D and B (d) none

10. P, Q, R, S, T, V and W are sitting around a circle facing the center. R is third to the

right of V who is second to the right of P. T is the second to the left of Q who is

second to the left of W. V is sitting between S and W. Who is third to the left of T?



2.6. Selections 2.6.1. Concept:

Selecting a part out of whole. Infer from the statements without Assumptions. Select the

item from the whole according to the condition and start to solve from the direct

condition.

Write the condition in a shortcut like,

A and B together Means A = B

If A come and B will not come Means A ≠ B

If B is selected, A and C will not be selecteed which Means B ≠ A, C

Problem:

Three students J, B, S have to select 3 subjects out of 6 subjects P, Y, M, C, H, and L. If

L is selected then P cannot be selected. If S selects C then J does not select L. If B

selects Y then S cannot select C and if J selects P then S selects H and L. If S selects L

and B selects Y then S cannot select?

Solution:

S selects L;

B selects Y;

L is selected P cannot be selected. (given)

P can‟t be selected

If B selects Y then S cannot select C. (given)

So C can‟t be selected

Answer: P and C


1. Four Students are to be selected from T, U, V, W, X, Y and Z. If T is selected then Z

cannot be there. Y cannot come with W. U should be present in the team. Y and X

come together. V cannot be a part of the team.

Find the number of possibilities

If T comes in the team who are its members?

2. In a class of 7 members they are going to select 4 members 2 from Boys and 2 from

Girls. The Boys are A, B, C, D and Girls are X, Y, Z.

Y & C can be together

A& X are always together

C & D cannot be together but C & B are always together.

If A is selected who are the remaining members?

If B is selected who are the remaining members?

Who is the Person who will be always there in the team?

3. A team of 5 is to be selected for a Basketball tournament. The good players among,

Boys are Amit, Bhel, Chaitanya, Dilip & Imran, Among Girls are Priti, Rani, Kiran

and Sonia. The selection criteria are such that Amit and Sonia have to be together.



Priti however cannot be with Rani, similarly we cannot have Rani with Bhel or Dilip

with Kiran, and however Chaithanya and Imran have to be together.

If two of the members have to be Boys, the team who will be the members of the

team?

Now if one of the members is Rani, who would be the others?

Now let two members be Girls and Dilip is one of the members too, then the who

would be the members of the team ?

4. City high school must put together a debating team consisting of four debaters. There

are candidates of equal ability X, Y & Z who are seniors and A, B, C, D who are

juniors. The school requires that they should be two seniors and two seniors in the

team. It is also necessary that all the debaters be able to work with one another.

Debaters Y & A can‟t work together.

Debaters Z & C can‟t work together.

Debaters A & B can‟t work together.

If debater B is selected and debater Y is rejected, who are the members the team

will consist of?

If debater A is in the team, which other debaters must be in the team as well?

If Y & Z are selected, which of the debaters must be in the team with them?

5. In a college a new football team was formed consisting of five members, 3 from Junior

and 2 from Senior. In senior team the members are P, Q, R, S, in junior team the

members are A, B, C and D. The selection of the team is based on the following

conditions:

P does not go along with R

Q and R work together

If B is selected, S is not selected

C and D are always together.

If P is selected the remaining members are:

a) ACDS b) RBAB c) CDAQ d) QRBA

6. From amongst five doctors A, B, C, D and E, four engineers G, H, K and L and six

teachers M, N, O, P, Q and R some teams are to be selected. Of these A, B, G, H, O, P

and Q are females and the rest are males. The formation of teams is subjected to the

following conditions:

Wherever there is a male doctor, there will be no female teacher.

Wherever there is a male engineer, there will be no female doctor.

There should not be more than two male teachers in any team.

If the team consists of two doctors, three female teachers and two engineers,who

are the members of the team ?

a) A B O P Q G H b) C D K L O P Q c) C D O P Q G H d) D E G H O P Q

If the team consists of two doctors, one engineer and four teachers, all the

following teams are Possible except?

a) A B G M N O P

b) A B H M O P Q

c) A B H M R P Q



d) A B K N R P Q

If the team consists of two doctors, two female teachers, and two engineers, all

the following teams are possible except :

a) A B G H O Q

b) A B G H P Q

c) A B K L P Q

d) O P G H A B

7. A team of five is to be selected from amongst five Boys P, Q, R, S and T and four

Girls A, B, C and D. Some criteria for selection are:

P and D have to be together.

A cannot be put with C.

S and B cannot go together.

R and T have to be together.

C cannot be put with Q.

Unless otherwise stated, these criteria are applicable to all the questions below:

If the two of the members have to be Boys, who will be the members of the team?

a) P Q D A B

b) P S D B C

c) Q S D C B

d) R T D A B

If R be one of the members, the other members of the team are?

e) A D P S

f) B D P S

g) B D R T

h) D P R T

8. A Company wanted to start a new project. There was a need of five member team of

3 Managers and 2 Executives. The managers (A, B, C, D, E) and Executives (W, X,

Y, Z) were to be selected with the following conditions:

A and C work together

X and Y work together but X does not work with Z

H does not work with X; C and D work together

If H is selected then A cannot be selected.

If B is selected then who are the remaining Persons?

9. A school decided to go for NSS camp for a week. Boys team consists of A, B, C, D,

E and Girls team consists of M, N, O, P. Six member team of 3 Girls and 3 Boys

were to be selected with the conditions as

A and B cannot be together

M and P cannot be together

P and B go together

D and E go together

M and N go together



10. In a class there are 5 Boys A, B, C, D, E and 5 Girls L, M, N, O, P were to be selected

for a competition. The selected team consists of 3 Boys and 2 Girls with the

conditions as:

From Boys side B and Girls side M should be selected

A does not go with B

A and E go together

If B is not selected then L is selected

M does not go with O

N and P do not go together


3. Logical Reasoning Concept: Logic:

Our mind always seeks a rational explanation of the objects which we come in to contact

with our day-to-day life. The perception and conception of the objects or elements

become knowledge for us if our mind conceives them in a systematic order so that certain

relationships can be established between the objects or elements. Knowledge of a subject

is said to be systematic and scientific when the different parts of knowledge are related

together in a particular way to make a system. This particular way of relationship

between the elements, objects or subjects leads our mind towards the formation of

thoughts which are expressed in language. This process is known as logic. So logic can

be defined as the science of thought.

Reasoning:

When a thought is associated with the three attributes of What, Why and How its validity

can be judged. These three attributes are often termed as „reasoning‟. Reasoning is the

function of the mind passing from known to unknown by establishing a systematic

relationship between the elements.

3.1. Family Tree 3.1.1. Concept:

Deals with Hierarchical structure of a family. It is also called as Blood relation problem.

Problem 1:

There are six members in a family A, B, C, D, E, and F who are related to each other. B

is F‟s daughter–in–law. D is A‟s only grand child. C is D‟s only uncle. A has only 2

children F and C, one male and one female. E is father of C. Then,

1.Who is the grand mother of D?

2.How many males and females are present?

Solution:

Denotes Female Denotes Male

I. B is F‟s daughter-in-law

F Father/Mother-in-law

Daughter-in-law

B



II. D is A‟s only grand daughter

A

?

D

III. C is D‟s only uncle

A

C ?

IV. A has only 2 children F&C, one male and one female

A

D

V. E is the father of C

B

x

x Daughter-in-law

1. A is the grand mother of D.

2. There are 3 males and 3 females in the family.

C

C F

B

A E

C F

D B



3.1.2. Problems to Solve: 1. B is Brother of D. D is sister of E. E is brother of F. How is F related to B?

2. Vivek is the father of M and Q. M is the son and Q is the daughter. L is the spouse

of one of his child. N is of the same sex as Q. N and his/her spouse W, have two

children S and R who are of the same sex as W. R and L are of different sexes.

a) Who is the father of S and R?

b) Whose spouse is L?

c) If all the above characters belong to one family, how many males are there

d) Who is the daughter–in–law of Vivek ?

e) If N is Vivek‟s sister, who is S‟s uncle?

3. A man was going with a Girl. Somebody asked his relationship with the Girl. He

replied, “My paternal uncle is the paternal uncle of her paternal uncles”. What is

the relationship between the man and the Girl?

4. Out of A, B, C, D, E and F, C is sister of F. B is brother of E‟s husband. D has no

brothers but is the father of A (male) and grandfather of F. We have 2 fathers, 3

brothers, one mother and 1 daughter.

a) Find which are the brothers?

b) Who is E‟s husband?

5. Mohan is son of Arun‟s father‟s sister. Prakash is son of Rekha who is mother of

Vikki and grandmother of Arun. Purab is father of Neha and grandfather of Mona.

Rekha is Purab‟s wife.

How is Mohan related to Rekha?

a) Grandson b) Son c) Nephew d) Data Inadequate

How is Vikki „s wife related to Neha?

a) Sister b) Niece c) Sister in law d) None

6. Looking at a portrait of a man, Sanjay said, “His mother is the wife of my father‟s

son. Brothers and sisters I have none.” At whose portrait was Sanjay looking?

7. If B‟s mother was A‟s mother‟s daughter. How was A related to B?

8. Introducing a lady, a man said, „Her mother is the only daughter of my mother–in–law‟. What is the man to the lady?

9. F‟s grandfather is A. A has two sons – E and C. C is married. How is F related to

E?

10. Anil introduces Rohit as, “the son of the only brother of his father‟s wife”. How is

Rohit related to Anil?



3.2. Codes 3.2.1. Concept:

Coding is a method of transmitting a message from one place to another. Decoding is the

ability to decipher a certain code. In these types of questions, certain code values are

assigned to a word or a group of words and the original words should be found out.

Codes may be numerical or alphabets. It is easy to decode if we are able to find the

format of arrangement. The arrangement can be any one of the following:

1. Reversed order

2. Gaps in between

3. +1 and–1 format

4. Tree format

Problem 1:

The term FIRE is coded as DGPC, then how will SNOW be coded as?

Solution:

The term FIRE is coded as DGPC, if we note it we find that the code is two alphabets

preceding the original word.

That is D E F, G H I, P Q R, C D E (Gaps in Between arrangement)

So SNOW can be coded as QLMU.

Problem 2: If NUMERICAL is written as MVLFQJBBK, then how. „ASTROLOGY‟ would be

written in this code?

Solution: In this Code, the letters in Odd place, i.e. in first, third, fifth, seventh and ninth place

letters have been coded to their preceding letters and the remaining ones have next letters

as code.

Hence,

The answer is “ZTSSNMNHX”.


1. If in a certain language TEACHER is coded as VGCEJGT, how would „DULLARD‟

be coded in that code?

2. In a certain code, TRIPPLE is written as SQHOOKD How would DISPOSE be

written in that code?



3. In a certain code COMPUTER is written as RFUVQNPC How would MEDICINE be

is coded as

4. If in a certain language CARROM is written as BZQQNL. How would HOUSE be

written in that code language?

5. In a certain code RIPPLE is written as 613382 and LIFE as 8192. How should

PILLER be written in that code?

6. In a certain code, PALE is written as 2134 and EARTH as 41590. How will PEARL

be written in that code?

7. In a certain code 15789 is written as EGKPT and 2346 is written as ALUR. How is

23549 written in that code?

8. If room is called bed, bed is called window, window is called flower, flower is called

cooler, then on what would a man sleep?

9. If cloud is called white, white is called rain, rain is called green, green is called air, air

is called blue, blue is called Water, Where will the bird fly?

10. If in a certain language, CALCUTTA is coded as GEPGYXXE, Which word would

be coded as FSQFCE?

3.3. Conditionality and Grouping 3.3.1. Concept:

Conditions are given based on which the actions take place. There are four basic

conditions. Based on them the decisions are made. Based on these conditions the groups

are formed.

Condition 1: If A occurs then B also will occur.

Condition 2: If A occurs then B will not occur.

Condition 3: If A has not occurred then B will occur.

Condition 4: If A has not occurred then B also will not occur.

Problem 1:

From a group of 6 Boys A,B,C,D,E,F and 5 Girls L,M,N,O,P a group of 6 is to be

selected based on the following conditions:

O and P have to be together

C cannot go with O



A and D have to be together

D cannot go with L

C and M have to be together

Band N have to be together

B and E cannot be together

If the team consists of four Girls the members of the team are:

i) BELNOP ii) EFLNOP iii) BFLNOP

If the team consist of 5 Boys and 1 Girl that Girl would be:

i) L ii) M iii) N iv) O

Solution:

O and P will be there so C cannot be there as well as M.

The Girls are L, N, O and P

D and A cannot be there.

So choice is (iii) BFLNOP

If B is selected 2 Girls M and N have to be selected. So B is rejected. Therefore M is

selected.

3.3.2. Problems to Solve: 1. In a college a new football team was formed consisting of five members, 3 from

Junior and 2 from Senior. In senior team the members are A, B, C and D. In junior

team the members are P, Q, R, S. The selection of the team is based on the following

conditions:

A does not go along with C

B and C work together

If Q is selected, D is not selected

R and S are always together.

If A is selected the remaining members are:

(a) PRSD

(b) CBPR

(c) RSPB

(d) BCQP

If B and C are selected, how many arrangements are possible?

2. A school decided to go for NSS camp for a week. Boys‟ team consists of P, Q, R, S,

T and Girls‟ team consists of W, X, Y, Z. Six member team of 3 Girls and 3 Boys

was to be selected with the conditions as:

P and Q cannot be together

W and Z cannot be together

Z and Q go together

S and T go together

W and X go together

3. From among 6 Boys A, B, C, D, E, and F and Five Girls P,Q,R,S and T a team of six

is to be selected under the following conditions:

A and D have to be together

C cannot go with S

S and T have to be together

B cannot be teamed with E



D cannot go with P

B and R have to be together

C and Q have to be together

If four members have to be Girls, then find the members of the team?

4. A college selection committee sat to finalize Cricket Team of five members 2 from

team X and 3 from team Z. Team X consist of 4 members A, C, E, G and Z team

consist of 5 members B, D, F, H, J. The conditions given were:

A and E go together but cannot go with B

D and F go together but cannot play with H

H and J go together but cannot play with E

E does not play with D

5. In a class, 5 Boys A, B, C, D, E and 5 Girls L, M, N, O, and P were to be selected for

a competition. The selected team consist of 3 Boys and 2 Girls with the conditions

as, From Boys side B and Girls side M should be selected

A does not go with B

A and E go together

If B is not selected then L is selected

M does not go with O

N and P do not go together


6. Eight students A, B, C, D, E, F, G and H are planning to enjoy car racing. There are

only two cars and following are the conditions

One car can accommodate maximum five and minimum four students.

A will sit in the same car in which D is sitting and H is not in the same car.

B and C can‟t sit in the same car in which D is sitting

F will sit in the car of four people only along with A and E but certainly not

with G.

If H and G are sitting in the same car, who are the other two students sitting in the

same car?

(a) B and C

(b) C and D

(c) B and D

(d) E and B

(e) None of these

If E and A are sitting in the same car, which of the following statements is true?

(a) Five students are sitting in the same car

(b) B is sitting in the same car

(c) F is not sitting in the same car

(d) G is not sitting in the same car

(e) None of these

Which of the following statement is superfluous for the above seating arrangements?

(a) Only (i)

(b) Only (ii)

(c) Only (iii)



(d) Only (IV)

(e) None of these

7. At an Electronic Data Processing Unit, five out of the eight problem sets P, Q, R, S,

T, U, V and W are to be operated daily. On any one day, except for the first day of a

month, only three of the program sets must be the ones that were operated on the

previous day. The program operating must also satisfy the following conditions:

If Program P is to be operated on a day, V cannot be operated on that day.

If Q is to be operated on a day, T must be one of the programs to be operated

after Q.

If R is to be operated on a day, V must be one of the programs to be operated

after R.

The last program to be operated on any day must be either S or U.

Which of the following could be the set of programs to be operated on the first day of

a month?

(a) V, Q, R, T, S

(b) U, Q, S, T, W

(c) T, U, R, V, S

(d) Q, S, R, V, U

(e) P, R, V, S, U

Which of the following is true of any day‟s valid program set operation?

(a) P cannot be operated at third place

(b) Q cannot be operated at third place

(c) R cannot be operated at fourth place

(d) T cannot be operated at third place

(e) U cannot be operated at fourth place

If R is operated at third place in a sequence, which of the following cannot be the

second program in that sequence?

(a) S

(b) T

(c) U

(d) W

If the program sets R and W are to be operated on the first day, which of the

following could be the other programs on that day?

(a) P, T, V

(b) Q, S, V

(c) Q, T, V

(d) T, S, U

(e) T, S, V

8. A team of five is to be selected from amongst five Boys A, B,

C, D and E and four Girls P, Q, R and S Some criteria for selection are:

A and S have to be together.

P cannot be put with R.

D and Q cannot go together.



C and E have to be together.

R cannot be put with B.

Unless otherwise stated, these criteria are applicable to all the questions below:

If the two of the members have to be Boys, the team will consist of?

a) A B S P Q

b) A D S Q R

c) B D S R Q

d) C E S P Q

If R be one of the members, the other members of the team are?

e) P S A D

f) Q S A D

g) Q S C E

h) S A C E

If two of the members are Girls and D is one of the members, the members of the

team other than D are?

i) P Q B C

j) P Q C E

k) P S A B

l) P S C E

If A and C are members, the other members of the team cannot be?

m) B E S

n) D E S

o) E S P

p) P Q E

9. From amongst five doctors A, B, C, D and E, four engineers G, H, K and L and six

teachers M, N, O, P, Q and R, some teams are to be selected. Of these A, B, G, H, O,

P and Q are females and the rest are males. The formation of teams is subject to the

following conditions:

Wherever there is a male doctor, there will be no female teacher.

Wherever there is a male engineer, there will be no female doctor.

There shall not be more than two male teachers in any team.

If the team consists of two doctors, three female teachers and two engineers, the

members of the team are?

(a) A B O P Q G H

(b) C D K L O P Q

(c) C D O P Q G H

(d) D E G H O P Q

If the team consists of two doctors, one engineer and four teachers, all the following

teams are possible except:

(a) A B G M N O P



(b) A B H M O P Q

(c) A B H M R P Q

(d) A B K N R P Q

If the team consists of two doctors, two female teachers, and two engineers, all the

following teams are possible except :

(a) A B G H O Q

(b) A B G H P Q

(c) A B K L P Q

(d) O P G H A B

If the team consists of three doctors, two male engineers and two teachers, who could

be the members of the team?

(a) A B C K L M R

(b) B C D K L N R

(c) C D E K L M N

(d) C D E K L P R

If the team consists of two doctors, two engineers and two teachers, all the following

teams are possible except?

(a) A B G H O P

(b) A B G H M N

(c) C E K L N R

(d) C D K L O P

10. A causes B or C, but not both

F occurs only if B occurs

D occurs if B or C occurs

E occurs only C occurs

J occurs only if E or F occurs

D causes G, H or both

H occurs if E occurs

G occurs if F occurs

If B occurs which must occur?

a) D b) D & G c) G & H d) F & G e) J

If J occurs which must have occurred?

a) E b) either B or C c) both E, F d) Both B and C e) None of the

above

Which may occur as a result of cause not Mentioned.

1) D 2) A 3) E

a) 1 only b) 2 only c) 1 & 2 d) 2 & 3 e) 1, 2, 3



3.4. Series

3.4.1. Concept: Series involves the numbers or letters that are arranged in a particular sequence. The

sequence can be any of the following;

1. Ascending

2. Descending

3. Gap

4. Squares and cubes

5. +/– order series.

Problem 1:

Fill in the series 1, 9, 25, _, 81.

Solution:

As given in the problem it consists of Squares of digits with a digit left in between

It is 12, 3

2, 5

2, _, 9

2

So the solution is 72

= 49.


1. Y, W, U, S, Q, __? __?

2. A, C, F, H, ?, M

3. Z, X, V, T, R, __ ? ___ ?

4. A, D, G, J, M, __?

5. 1, 4, 9, 16, 25, __ ?

6. 1, 6, 13, 22, 33, ___ ?

7. 121, 225, 361, ___ ?

8. 0.5, 1.5, 4.5, 13.5, __?

9. 1, 2, 6, 24, __ ?

10. 5, 9, 17, 29, 45,__ ?

3.5. Direction Sense 3.5.1. Concept:

These problems test our sense of Direction. The best way of solving these questions is to

follow the instructions given in the question carefully and make a diagram accordingly

with the help of which the question can be solved.

Problem 1:

„A‟ starts from his office and walks 3 km towards north. He then turns right and walks 2

km and then turns right and walks 5 km. He then turns right and walks 2 km and then

again turns right and walks 2 km. In which direction is he from the starting point?

a) South b) North–east c) South–East d) He is at the starting point



Solution: He is at the starting point.

Problem 2:

Deepa moved a distance of 75 metres towards the north. She then turned to the left and

walking for about 25 metres, turned left again and walked 80 metres. Finally, she turned

to the right at an angle of 450. In which direction was she moving finally?

a) North-east b) North-west c) South d) South-east e) South-west

Solution:

Deepa started from A, moved 75 m upto B, turned left and walked 25 m upto C. She then

turned left again and moved 80 m upto D. Turning to the right at an angle of 450

, She

was finally moving in the direction DE i.e., South-west.

Hence, the answer is (e).


1. Arman walked towards east for a distance of 5 km, turned towards right and walked for

a distance of 10 km. Then again turned to his right and walked 15 km. Next he turned

to his left and walked 5 km. In which direction is he now with reference to his starting

point?

2. Rajesh is standing to the west of Amir and north of Ruchir and Sathish is to the west of

Ruchir and south of Salman. In which direction is Sathish with reference to Amir?

3. One day Kannan left home and cycled 10 km southwards, turned right and cycled 5 km

and turned right and cycled 10 km and turned left and cycled 10 km. How many

kilometers will he have to cycle to reach his home straight?

5

3

5



4. A man leaves for his office from his house. He walks towards East. After moving a

distance of 20 m, he turns towards South and walks 10 m. Then he walks 35 m towards

the West and further 5 m towards the North. He then turns towards East and walks 15

m. What is the straight distance in meters between his initial and final positions?

5. Ram started walking towards South. He took a right turn after walking 10 metres. He

again took a left turn walking 20 metres. Which direction is he facing now?

6. A walks southwards then right, then left and then right. In which direction is he from

the starting point?

7. I go 5 km east, then turn right and go 8 km. Then I turn left and go 5 km, and then I

turn left and go 8 km. At what distance am I from the starting point now?

8. If A is to the South of B and C is to the east of B, in what direction is A with respect to

C?

9. The Door of my house faces East. I walk from my back door 100 m, then right and

walk 100m and turn left and walk 50 m and reach a point X. In which direction is I

from the starting point?

10. Gaurav walks 20 m towards North. He then turns left and walks 40 m. He again turns

left and walks 20 m. Further, he moves 20 m after turning to the right. How far is he

from his original position?

3.6. Statement Logics 3.6.1. Concept:

The ability of Analysis and interpretation of data logically is tested in these problems.

Types: 1) statement– conclusions

2) statement– assumptions

3) statement– arguments

Conclusions

In each questions there is a statement and two conclusions. Logically apply the

conclusions and give answers.

a) If conclusion 1 follows

b) If conclusion 2 follows

c) If neither 1 nor 2 follows

d) Both 1 and 2 follows

Example:

Statement: Quality has a Price tag. India is allocating lot of funds for education.



Conclusions:

1) Quality of education in India would improve soon

2) Funding alone can enhance Quality of education

Arguments Each question given below has 1 statement and two arguments. We should decide which

of the arguments is strong

Give answer:

a) If only 1 is strong

b) If only 2 is strong

c) If neither 1 or 2 is strong

d) If both 1 and 2 is strong Answer (a)

Example:

Should Private sector be permitted to operate telephone services?

Arguments:

1) Yes, they are operated in advanced western countries

2) No, It is risky to put them in private hands

Assumptions Each question contains 1 statement and 2 assumptions. Check the data and answer.

Give answer:

a) If only 1 implicit

b) If only 2 implicit

c) Both 1 and 2 implicit

d) Neither 1 nor 2 implicit

Answer: (a)

Example:

Statement: The pen is mightier than the sword

Assumptions:

1) The pen is made up of stronger metal than the sword

2) The power of the mind is much stronger than brute physical power

Answer: (1)




Statement Conclusion

a) Only conclusion I follow

b) Only conclusion II follows

c) Either I or II follows

d) If neither I nor II follows

e) If both I and II follows

1. Statement: Morning walks are good for health.

Conclusion: I) all healthy people go for morning walks.

II) Evening walks are harmful.

2. Statement: Company X has marketed the Product. Go ahead; purchase it if Price and

quality are your considerations.

Conclusions: I) The Products must be good in quality.

II) Then Price of the Product must be reasonable.

3. Statement: The Best way to escape from a problem is to solve it

Conclusion: I) Your life will be dull if you don‟t face a problem.

II) To escape from problems you should have solutions

4. Statements: Vegetable Prices are soaring in the Market

Conclusions: I) Vegetables are becoming a rare commodity

II) People cannot eat Vegetable

Statement Argument

a) If only 1 is strong

b) If only 2 is strong

c) If either 1 or 2 is strong

d) If neither 1 nor 2 is strong

e) If both 1 and 2 is strong

5. Statements: Should there be internal assessment in colleges?

Argument: 1. Yes, it will enable the teachers to have a better control over the

Students.

2. No, since it will encourage favoritism among teachers.

6. Statements: Should military service be made compulsory in our country?

Argument: 1) No. It is against the policy of non violence.

2) Yes. Every citizen should protect his country.

7. Statements: Should there be world Government?

Argument: 1) Yes. It will help in eliminating tensions among the nations



2) No. Then only the developed countries will dominate the

Government

Statement Assumption

a) If only 1 implicit

b) If only 2 implicit

c) Either 1 or 2 implicit

d) Neither 1 nor 2 implicit

e) Both 1 and 2 implicit

8. Statements: To keep myself up to date , I always listen to 9.00 pm news on radio–

candidate tells the interview board.

Assumption: 1) The candidate doesn‟t read newspaper.

2) Recent news is broadcasted only on radio

9. Statements: Double your money in five months – An advertisement.

Assumption: 1) The Assurance is not genuine.

2) People want their money to grow.

10. Statements: I cannot contact you on phone.

Assumption: 1) Telephone facility is not available.

2) Now a Days it is difficult to contact on phone

4. Verbal Reasoning Concept:

It is the general notion that English is a foreign language and so it is very difficult. But

this notion is far from truth. A student must understand that English is not a tough

language. A language is a practical medium of expression of our ideas and thoughts.

People who study and use a language are mainly interested in how they can do things

with the language, how they can make meanings, get attention to their problems and

interests, influence their friends and colleagues and create a rich social life for

themselves.

They are only interested in the grammatical structure of the language as a means of

getting things done. Grammar puts together the patterns of the languages.Acquiring a

simple, direct and forceful knowledge in writing and reading calls for constant practice.

The questions in English Language are designed to test candidate‟s understanding of

English and its usage. The aim is to find how well candidates do understand the

grammatical rules and their usages, how well they do comprehend the idea conveyed in

the given passage and how well candidates can express the given idea.

Verbal Ability is to understand the prose and to work with specialized and Technical

Vocabulary. Verbal Ability falls under following categories:



Synonyms

Antonyms

Analogy

Sentence Completions

Idioms

Spotting errors

Reading Comprehension

4.1 Antonyms 4.1.1 .Concept:

The word „Antonym‟ means a word that is opposite in meaning to another word. For

example, The Antonym for the word „Old‟ is „New‟.

Tips: 1. Before you look at the answer choices, try to clearly define the given word. Remember

that you are looking for the answer choice that has a meaning opposite to that of a given

word.

2. Consider slight variations in the meaning of each word.

3. If you have trouble with a definition, try using the word in the sentence. Try to predict

an answer before looking at the answer choices. If an answer choice matches your

predicted answer, it is most likely correct.

4. Use the Latin roots, prefixes, and suffixes to figure out what hard words mean. Look

for cognates from French, Spanish, or Italian if you recognize them.

5. Always consider all the answer choices before you select an answer.

Example:

MAGNIFY

a) Forgive

b) Comprehend

c) Extract

d) Diminish

e) Electrify

Solution:

Magnify (enlarge, expand)

a) Forgive ( excuse, absolve)

b) Comprehend ( understand, know, realize)

c) Extract ( take out, remove, pullout)

d) Diminish (reduce, make smaller)

e) Electrify ( thrill, excite, amaze)

The answer is Diminish.



4.1.2 Questions to Practice: 1. ABATE

a) Aggravate

b) Amplify

c) Enlarge

d) Enhance

2. BUCOLIC

a) Urban

b) City like

c) Slovenly

d) Polished

3. CATACLYSM

a) Steadfastness

b) Privilege

c) Benefit

d) Blessing

4. DEPRECATE

a) Praise

b) Approve

c) Eulogize

d) Depreciate

5. EMOLUMENT

a) Penalty

b) Punishment

c) Monument

d) Retribution

6. FRACTIOUS

a) Complaisant

b) Genial

c) Sagacious

d) Accommodating

7. ILLUSORY

a) True

b) Elusive

c) Real

d) Factual

8. PROVIDENTIAL

a) Ill–starred

b) Unfortunate

c) Unlucky

d) Simulation



9. TURBULENT

a) Quiet

b) Quiescent

c) Calm

d) Truculent

10. VENAL

a) Honourable

b) Likeable

c) Incorruptible

d) Patriotic

4.2 Synonyms: 4.2.1 Concept:

The word „Synonym‟ means a word or phrase with the same or nearly the same

meaning as another in the same language. For example, „Shut‟ and „Close‟ are synonyms.

Example: ASTUTE a) Sheer

b) Noisy

c) Astral

d) Unusual

e) Clever Astute: (smart, intelligent, shrewd)

Sheer: (pure, absolute, complete)

Noisy: (loud, piercing,)

Astral: (planetary, lunar)

Unusual: (strange, Odd, curious)

Clever: (smart, bright, intelligent)

Answer – Clever

4.2.2 Questions to Practice:

1. VARIEGATE

a) Set type

b) Multi–colour

c) Differ

d) Reject

e) Reply in Kind

2. FILCH

a) Pretend

b) Dirty

c) Embarrass

d) Steal

e) Honour



3. INFINITE

a) Verbal

b) Indefinite

c) Endless

d) Strange

e) Vague

4. DEMISE

a) Residence

b) Dismissal

c) Accident

d) Act

e) Death

5. FRUGALITY

a) Extravagance

b) Ripening

c) Thrift

d) Miserliness

e) recurrent

6. UNEQUALED

a) Outstanding

b) Different

c) Praised

d) Resentment

e) Miserliness

7. ADVERSITY

a) Opponent

b) Hardship

c) Opening

d) Public announcement

e) Agency

8. FASTIDIOUS

a) Speedy

b) Precise

c) Squeamish

d) Hungry

e) Slow

9. DISCONCERT

a) Sing in harmony

b) Pretend

c) Cancel Program

d) Confuse

e) Interrupt

10. GARRULOUS



a) Laconic

b) Strangling

c) Ecstatic

d) Frozen

e) Wordy

4.3 Analogy 4.3.1 Concept:

Analogy questions ask you to determine the relationship between the two words in a pair

and then to recognize a similar or parallel relationship between the members of a

different pair of words. The relationship between the words in the original pair will

always be specific and precise.

Tactics: State the relationship between the capitalized words in a clear sentence.

If there is more than one answer, look for a narrower approach

Consider secondary Meaning as well primary Meaning.

Familiarize yourself with common analogy types.

4.3.2 Analogy Types

1. Definition

2. Defining Characteristic

3. Class and members

4. Antonyms

5. Synonyms

6. Part to Whole

7. Function

8. Manner

9. Action & its significance

10. Worker and article created

11. Worker and tool

12. Worker and action

13. Sex

14. Symbol

Example

Book: Library:: Cannon:

a) Artillery

b) Powder

c) Shell

d) War

Solution:

Library is the place to keep books.

Cannon Means gun so Artillery is the place to keep gun, mortar

4.3.3 Questions to Answer:



1. EXPLOSION: DESTRUCTION

(a) Talk : Exaggeration

(b) Girl : Woman

(c) Success : Failure

(d) Engagement : Marriage

2. AGENDA: MEETING

(a) Program : Function

(b) Performance : Ticket

(c) Map : Scale

(d) Foot note : Article

3. CURTAIN: DRAPERY (a) Cockroach : Insect (b) Bed sheet : Bed (c) Pillow : Cushion (d) Mat : Floor 4. BALANCE: WEIGH

(a) Airplane : Height (b) Radar : Detection

(c) Satellite : Revolution (d) Television : Picture

5. CROWN: ROYAL (a) Throne : Regal (b) Wrap : Ermine (c) Pen : Author (d) Crucifix : Religion

6. REVOLVER: HOLSTER (a) Book : Bag (b) Eye : Eyelid (c) Juice : Glass (d) Nostril : Nose 7. SYMPHONY: MUSIC

(a) Mural : Painting (b) Ode : Prose (c) Preface : Book (d) Editorial : Journal

8. TRAITOR: DISLOYALTY

(a) Executioner : Reliability (b) Rebel : Defiance (c) Manager : Administration (d) Hope : Pessimism

9. TEXTILE: MILL

(a)Eggs : Hen ` (b)Coal : Mine

(c)Food : Agriculture (d)Brick : Kiln



10. MODESTY: ARROGANCE

(a) Passion : Emotion (b) Practice : Perfection (c) Cause : Purpose (d) Debility : Strength

4.4. Sentence Completion

4.4.1 Concept:

It tests your

a) Ability to use vocabulary

b) Ability to recognize logical consistency among the elements in a sentence

What makes Sentence Completion difficult is

A. Vocabulary:

Use of words like nonplussed, Harbingers, Eclectic.

Example:

Economy – restrain

Toy – Dally

B. Grammatical complexity :

They include entire range of grammatical possibilities. Use of clauses, prepositional

phrases, gerunds etc. in a convoluted manner

C. Tone:

Writer‟s attitude towards the subject matter i.e. ironic, playful, skeptical, somber etc.

D. Style:

Ideas maybe expressed in different manners, ornately on sparely, poetically or prosaically

formally or informally etc. An author‟s style depends on such details as word choice,

imagery repetition, rhythm, sentence structures and length.

Thought extenders – Continue the idea

Thought contrasters – Reverse the idea

Signal Words:

Additionally Although

Also But

And Despite

As well nevertheless

Besides In spite of

Further more On the contrary



More over Still

Too While

Likewise Which

Tactics: Read the sentence and think of a word that makes sense.

Look at all possible answers.

In double blank, go through the answer, testing the first word in each choice.

Watch for single words that link one part of the sentence to another.

Break down complex sentences into simpler components.

Check whether the Metaphor controls the choice of words.

Example:

Because experience had convinced her that he was both self–seeking and avaricious, she

rejected the likelihood that his donation had been ______.

a) Redundant

b) Frivolous

c) Inexpensive

d) Ephemeral

e) Altruistic

Solution:

The sentence presents simple case of cause and effect.

Opposite of self seeking (selfishness), Avaricious (Greediness) “Altruistic” (unselfish,

selfless)


1. In one shocking instance of _______ research, one of the nation‟s most influential

researchers in the field of genetics reported on experiments that were never carried

out and published deliberately ________ scientific papers on his nonexistent work.

(a) Comprehensive….. abstract

(b) theoretical …… challenging

(c) fraudulent …… deceptive

(d) Derivative …….. authoritative

(e) Erroneous ……. Impartial

2. Measurement is, like any other human Endeavour, a complex activity, subject to

error, not always used ________ and frequently misinterpreted and _________.

(a) Mistakenly ….. decided

(b) erratically …… analyzed

(c) Systematically ……. Organized

(d) Innovatively …….. refined

(e) Properly …….. misunderstood

3. In a revolutionary development in technology, several manufacturers now make

biodegradable forms of plastic; some plastic six–pack rings, for example, gradually

________ when exposed to sunlight.

(a) Harden

(b) Astagnate



(c) Inflate

(d) Propagate

(e) Decompose

4. To alleviate the problem of contaminated chicken, the study panel recommends that

the federal government shift its inspection emphasis from cursory bird–bird visual

checks to a more _______ random sampling for bacterial and chemical

contamination.

(a) Rigorous

(b) Perfunctory

(c) Symbolic

(d) Discreet

(e) Dubious

5. The Chinese, who began systematic astronomical and weather observations shortly

after the ancient Egyptians, were assiduous record–keepers, and because of this, can

claim humanity‟s longest continuous _______ of natural events.

(a) Refinance

(b) Documentation

(c) Maintenance

(d) Domination

6. An institution concerned about its reputation is at the mercy of its members, because

the misdeeds of individuals are often used to _______ the institutions of which they

are a part.

(a) Reform

(b) Discredit

(c) Coerce

(d) Intimidate

7. Job failure Means being fired off a job, being asked to resign, or leaving _______ to

protect yourself because you had very strong evidence that one of the first two was

_______

(a) Voluntarily…impending

(b) Abruptly…significant

(c) Knowingly…operative

(d) Eventually…intentional

8. During the widespread fuel shortage, the Price of gasoline was so _______ that

suppliers were generally, thought to be _______ the consumer.

(a) Reactive…shielding

(b) Stable…blackmailing

(c) Excessive…gouging

(d) Depressed…cheating

9. Normally an individual thunderstorm lasts about 45 minutes, but during certain

conditions the storm may _________, becoming ever more severe, for as long as four

Hours.

(a) Wane

(b) Moderate



(c) Persist

(d) Vacillate

(e) Disperse

10. Perhaps because something in us instinctively distrusts such displays of natural

fluency, some readers approach John Updike‟s fiction with __________.

(a) Indifference

(b) Suspicion

(c) Veneration

(d) Recklessness

(e) Bewilderment

4.5 Spotting Errors 4.5.1 Concept:

Grammatical errors form the most probable errors in error-spotting exercises. Mistakes

in preposition and tenses are more common

Examples:

1) The car is / almost / the same / like mine.

Ans: as mine

2) The weather / feels / as / spring

Ans: like spring


1. Ashoka the Great / was regarded one/ of the greatest emperor the/ world has ever

A B C D

Produced / No error

E

2. None of the two Boys / who were / present there / came to his help. / No error.

A B C D E

3. More we get, more we want/ in this way/ there is no end to human desires/ No error.

A B C D

4. When I arrived / on the station / I saw a man who had robbed me / because he

A B C

thought I carried the King‟s wallet. / No error

D E

5. He had ordered you/ to paint it green/ but you painted/ the blue house. / No error.

A B C D E

6. Little money/ that he was left behind after/ the marriage of his daughter/ was not

A B C

Sufficient for a big family like his./ No error.

D E



7. No less than/ four thousand people/ lost their lives/ in the earthquake/ no error.

A B C D E

8. He wished me/ to dine with him that evening/ but which/ I declined/ No error.

A B C D E

9. I have read several plays of Shakespeare/ who was one of / the greatest dramatists/

A B C

the world has ever produced/ No error.

D E

10. Nobody in their senses/ would have/ uttered/ such silly remarks/ No error.

A B C D E

4.6. Idioms 4.6.1 Concept:

A Group of words whose meaning is different from the meanings of the individual words

are idioms.

For example, „Catch in the act‟ means „Finding someone doing something wrong or bad.‟

Example:

The manager‟s bark is worse than his bite

a) He shouts

b) He punishes

c) His action is worse than his speech

d) His speech is worse than action

Solution:

The manager‟s bark is worse than his bite

d) His speech is worse than action

Questions to Answer:

1. Mary thought that the test was a piece of cake.

(a) Test was about eating a cake.

(b) She ate cake before the test.

(c) The task was relatively easy.

(d) She was thinking of eating a cake.

2. I‟m trying to sell off these books; I don‟t want them on my hands any longer.

(a) The books are heavy.

(b) I will not carry the books.

(c) He has paid for the books.

(d) He wants to get rid of the responsibility.

3. During our winter break, my friends and I hit the slopes.

(a) Hit each other.



(b) Went snow–skiing

(c) Fell in the slopes.

(d) Hit the mountains.

4. He had climbed almost to the top of the rock, but lost his nerve and turned back.

(a) Got scared

(b) Fell down

(c) Falls down from the rock

(d) His nerve was cut.

5. In spite of many different jobs he had to do, Jack always managed to keep his eye

On the ball

(a) He is a cricket player

(b) Being focused.

(c) Hit in the eye with a ball.

(d) He is searching the ball.

6. My father insisted that I put my nose to the grindstone next semester.

(a) Hit my nose if I fail

(b) Work really hard

(c) Grind the stone

(d) Copy during the exam

7. Throughout the Summer, I lived a stone‟s throw from a popular beach.

(a) Throwing stones and playing.

(b) People threw stones at his house.

(c) Stones were piled on the beach

(d) Lived near the beach.

8. I haven‟t studied at all for the exam tomorrow; I‟m really going to be in soup.

(a) Drink soup before exam.

(b) Fall into the soup

(c) Get into trouble.

(d) Prepare soup

9. Sandy is often too bogged down with her studies to spend time with her friends.

(a) Happy

(b) Overwhelmed

(c) Forgetful

(d) Smart

10. The Police found the case a tough nut to crack.

(a) Police could not crack nuts.

(b) The case was difficult to solve.

(c) Police found some nuts in the crime scene.

(d) The case was easy to solve.



4.7. Reading Comprehension 4.7.1 Concept:

Questions on reasoning comprehension resemble questions set on English

Comprehension. The only difference is that questions are designed to evaluate the

Reasoning Power of the candidates. The Questions are generally of the following types:

1. What is the assumption of the Passage?

2. Which of the statements weakens the Argument?

3. Which of the statements strengthens the Argument?

4. Whether a Particular statement can be deduced from the passage or not.

What is an Assumption?

Many statements are made by presuming something to be true. For example if we say,

“The Change in the Public Mood is Noticeable”, it assumes that “There has been a

Change in the Public Mood”. If we don‟t assume it, the statement becomes incorrect.

Which statement Weakens or Strengthens the Argument?

Any Statement that goes Contrary to the assumption or to the Supporting Argument

weakens it. Any argument which gives ideas similar to the Supporting Argument

strengthens it.

What is a Deduction?

If we can infer a statement either by correlating information or by discovering an

implication, it is said to be a deduction.

Example: The emotional appeal of imperialism never completely stilled the British conscience. However,

liberal thinkers throughout the nineteenth century argued that democracy was incompatible with

the maintenance of authoritarian rule over foreign peoples. To think imperially was to think in

terms of restrictive and protective measures; in defiance of the revealed truths of classical

economics. Thus when the British government took over responsibility for India from the East

India Company in 1858, many politicians were conscious of saddling Britain with a heavy

burden. In the first seventy years of the nineteenth century, enlightened British liberals looked

forward to the day when India would stand on its own feet. Even in the heyday of colonialism

British radicals continued to protest that self-proclaimed imperialists however honorable their

motives, would place faith accomplishment before the country and commit blunders of

incalculable consequence. The danger, they felt was all the greater because British foreign policy

still remained a stronghold of the aristocracy, while that related and persuasive lobby, the British

officer class, also had a vested interest in imperial expansion.

It took the humiliation of the Boer war to teach the British government what it would cost to hold

an empire by force. However this fact did not escape Gandhi, the supreme tactician of the Indian



liberation movement. He saw what some perceptive British thinkers had much earlier recognized,

namely, that Britain could not long continue to rule India expect with the co-operation of many

sections of its population. Once that cooperation was withdrawn, the foundation of British

authority in India would crumble. Furthermore, the Indian nationalist leaders were able to exploit

the aversion of the British liberal conscience to methods used by the local colonial rulers in

combating Indian non co-operation.

1. What does the term authoritarian rule mean?

(a) Rule of the dictionary of law

(b) Dictatorial rule of an aristocrat un-accomplished by the rule of law

(c) Arbitrary exercise of power by officials

(d) Rule having stability

(e) None of these

Answer : ( B)

2. What according to the author was the attitude of the British liberals towards the British

imperialist and colonial policy?

(a) One of active co-operation

(b) One of only verbal co-operation

(c) One of total indifference

(d) One of repeated protests

(e) One of disagreement

Answer : ( D)

3. Who was the supreme tactician of the Indian liberation movement?

(a) Mrs. Annie Besant

(b) The enlightened British liberals themselves

(c) Lokmanya Balgangadhar Tilak

(d) Mahatama Gandhi

(e) None of these

Answer : ( D)

4. What according to you would be the most suitable title for this passage?

(a) British imperialism and India

(b) British liberals attitude towards imperialism

(c) Role of mahatma Gandhi in Indian freedom movement

(d) The emotional appeal of British imperialism

(e) British as a colonial power.

Answer : ( D)

4.7.2 Question to Answer:

Read the following passage and pick up correct answers for each of the questions which

follow:



Passage 1:

A standard comprises characteristics attached to an aspect of a process or Product by

which it can be evaluated. Standardization is the development and adoption of standards.

When they are formulated, standards are not usually the Product of a single Person, but

represent the thoughts and ideas of a group, leavened with the knowledge and

information which are currently available .Standards which do not meet certain basic

requirements become a hindrance rather than an aid to progress. Standards must not only

be correct, accurate, and practice in requiring no more and no less than what is needed for

satisfactory results, but they must also be workable in the sense that their usefulness is

not nullified by external conditions. Standards should also be acceptable to the people

who use them. If they are not acceptable, they cannot be considered to be satisfactory,

although they may posses all the other essential characteristics.

1. According to the above paragraph, a processing standard that requires the use of materials

that cannot be procured is most likely to be

(a) Incomplete (b) Unworkable

(c) Inaccurate (d) Unacceptable

2. According to the above paragraph, the construction of standards to which the performance

of job duties should conform is most often.

(a) Work of people responsible for seeing that the duties are properly performed.

(b) Accomplished by the Person who is best informed about the functions

(c) The responsibility of the people who are to apply them.

(d) Attributable to the efforts of various informed Persons.

3. According to the above paragraph, when standards call for finer tolerances than those

essential to the conduct of successful Production operations the effect of the standards on

the improvement of Production operations.

(a) Negative (b) Negligible

(c) Nullified (d) Beneficial

4. Which one of the following is the most suitable title for the above paragraph?

(a) The evaluation of formulated standard

(b) The attributes of satisfactory standards

(c) The adoption of acceptable standards

(d) The use of process or Product standards.

Passage 2:

The chief condition of happiness, barring certain physical prerequisites, is the life of reasons

– the specific growth and power of man. Virtue, or rather excellence ,will depend on clear

judgment , self control, symmetry of desire , artistry of Means; it is not the possession of the

simple man, nor the gift of innocent intent, but the achievement of experience in the fully

developed man. Yet there is a road to it, a guide to excellence, which may save many detours

and delays: it is the middle way the golden Mean. The qualities of character can be arranged

in triads in each of which they first and the last qualities will be extremes and vices, and the

middle quality a virtue or an excellence. So between cowardice and rashness is courage:

between stinginess and extravagance is liberality; between sloth and greed is ambition;



between humility and pride is modesty; between secrecy and loquacity is honesty; between

moroseness and buffoonery is good humour between quarrelsomeness and flattery is

friendship; between Hamlets‟ indecisiveness and Quixote‟s impulsiveness is self control

„Right‟, in ethics or contact is not different from right in mathematics or engineering ; it

Means correct and fit what works best to get the best results.

1. What is the main idea of passing?

(a) The qualities of character are there – extremes and middle

(b) In some respects ethics and mathematics resemble

(c) Happiness can be achieved by following the middle path

(d) None of these

2. What is the implied Meaning of the passage?

(a) Happiness depends upon physical and mental qualities

(b) Self control is necessary

(c) Excellence should be achieved

(d) Rational approach lies in following the middle path

3. The author has not said:

(a) The middle path between humility and pride is modesty.

(b) Middle path avoids delay in achieving excellence; right in ethics

means that works to get the best results.

(c) Courage is the middle path of indecisiveness and impulsiveness

(d) None of the above.

4. Which of the following is not the middle path of different qualities?

(a) Liberality (b) Ambition

(c) Friendship (d) Secrecy

5. Who of the following is not the writer of either Hamlet or Don Quixote?

(a) Ben Johnson (b) Shakespeare

(c) Cervantes (d) None of these

6. Which of the following is the most suitable title for the passage?

(a) Qualities of Character

(b) Chief Condition of happiness

(c) Golden Mean

(d) None of these Passage 3:

To a greater or lesser degree all the civilized communities of the modern world are made up of a small class of rulers, corrupted by too much power, and of a large class of subjects, corrupted by too much passive and irresponsible obedience. Participation in a social order of this kind makes it very difficult for individuals to achieve that non attachment in the midst of activity, which is the distinguishing mark of the ideally excellent human being; and where there is not at least the considerable degree of non attachment in activity, the ideal society of the Profits cannot be realized. A desirable social order is one that delivers us from avoidable evils. A bad social order is one that



leads us into temptation which, if matters were more sensibly arranged, would never rise. Our present Business is to discover what large scale charges are best calculated to deliver us from the evils of too much power and off too much passive and irresponsible obedience. It has been shown that the economic reforms, so dear to advance thinkers are not in themselves Sufficient to produce desirable changes in the character of the society and of the individuals composing it unless carried out by the right sort of Means and in the right sort of governmental administrative and educational context, such reforms are either fruitless or actually fruitful evil. In order to create the proper contexts for economic reform, we must change our machinery of government, our methods of public administration, our system of education and our metaphysical and ethical beliefs

1. The main idea of the passage is (a) Ideal society is one which is free from evils. (b) Man‟s participation is necessary (c) No–attachment is the distinguishing mark. (d) Reforms are necessary for freeing us from power and obedience.

2. What is the inference that one derives?

(a) Changes in the character of society are necessary (b) Ideal society is envisage by Profits (c) It is the mark of ideally excellent human beings (d) None of these. 3. The author does not say (a) By participating in this kind of society one cannot remain non–attached. (b) Subjects indulge in obedience (c) Bad social order leads to temptations (d) None of these. 4. Who do you think the writer is? (a) Capitalist (b) Socialist (c) Anarchist (d) None of these. 5. The passage implies (a) Power and obedience corrupt the society. (b) Power is of no significance. (c) Changes in society are natural. (d) Economics is necessary.

tnp - venkata subbu.b

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