study material general apt vol-02

7/28/2019 Study Material General Apt Vol-02

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Joint CSIR-UGC NET for JRF and Eligibility for Lectureship

Section-A

General Aptitude

Volume-2

Contents

1. Mathematical and Numerical Ability 2

2. Logical reasoning (Mathematical) 55

3. Data Interpretation 73

4. Puzzles 97


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*MUDRA* Study Material For NET & SLET Exams of UGC-CSIR

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Section-A, General Aptitude Vol-2

1. Mathematical and Numerical Ability

Numerical ability pertains to the understanding and application of numerical relations

and dealing with numbers as symbols. A certain level of cognitive capacity is essential fro

dealing with numbers as symbols. This cognitive helps us to function at a definite level of

mental abstraction. This is the reason why the psychologists call numerical ability as one

aspect of Abstract Intelligence. The questions for checking a persons level of numerical

ability mostly deal with numerical comprehension, numerical retention, numerical reasoning,

and numerical analysis.

Question in this section will require a higher level of abstraction from the candidate

is as compare to the earlier two sections. One has to understand the logic behind the

relation between the given numbers. One has to understand the logic behind the relation

between numbers to get the correct solution.

NUMBER SERIES

In these tests, the series consist of numbers (digits). Number series are generally

formed by:

A. Addition of figures given in the row.

B. Subtraction of figures from one another.C. Division of figures by one another.

D. Multiplication of figures

E. Logical transposition of figures.

F. Increase/decrease of numbers in a specific pattern.

A careful examination of the example given here will familiarize you with the type of

questions that you are likely to face.

Example: Each of the following series follows a regular pattern. Write down the

number which will complete the sequence and replace the question mark(?)

1. 5 10 15 25 40 ?

2. 3 12 48 192 ?

3. 48 24 72 36 108 ?

4. 1 3 7 15 31 ?

5. 285 253 221 189 ?


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6. 1 2 8 9 15 16 ?

7. 9 4 8 5 7 6 ?

8. 905 576 329 247 ?

Answers with explanations

1. 65. Each term is the addition of the two previous numbers.

2. 278. Multiply each term by 4 to get the next number.

3. 54. Divide by 2 and multiply by 3 alternately.

4. 63. Numbers increase in steps of 2, 4, 8 and 16.

5. 157. Numbers decrease by 32 each time.

6. 22. The interval is alternately 1 and 6.

7. 6 and 7. Two alternating series, one increasing and the other decreasing.

8. 82. The interval between each pair of numbers becomes the succeeding term in the

series.

I. Completing the given series by finding the missing terms

Example

Directions: Find the missing term in each of the following series:

1. 1, 2, 15, ?, 45, 66, 91

a) 25 b) 26 c) 27 d) 28

Solutions: Clearly, the given sequence follows the pattern: +5, +9, +13, +17, +21, +25,

Thus, 1+5=6, 6+9=15,.

So, missing term = 15+ 13=28.

Hence, the answer is (d)

2. 2, 5, 9, 19, 37, ?

a) 73 b) 75 c) 76 d) 78

Solutions: Clearly, we have : 2 2 + 1= 5, 5 2 + 1= 9, 9 2 + 1= 19, 19 2 + 1= 37,.

So, missing term = 37 2 + 1= 75

Hence, the answer is (b)


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3. 4, 8, 28, 80, 244, ?

a) 278 (b) 428 (c) 628 (d) 728

Solutions: The terms of the given series are: 31+1, 3

2-1, 3

3+1, 3

4-1, 3

5+1,.

So, missing term = 361 = 7291 = 728

Hence, the answer is (d).

4. 10000, 11000, 9900, 10890, 9801,?

a) 10241 (b) 10423 (c) 10781 (d) 10929

Solution: Clearly, alternately we add and subtract 10% of a term to obtain the next term ofthe series.

Thus, 10000 + (10% of 10000) = 11000; 11000 - (10% of 11000) = 9900, 9900 + (10% of

9900) = 10890, 10890-(10% of 10890) = 9801

So, missing term = 9801 + (10% of 9801) = 9801 + 980 = 10781.

Hence, the answer is (c).

5. 0, 6, 24, 60, 120, 210, ?

a) 240 b) 290 c) 336 d) 504

Solution: Clearly, the given series is: 13

- 1, 23

- 2, 33

- 3, 43

- 4, 53

- 5, 636

Therefore, missing term = 73 - 7 = 343 - 7 =336.

Hence, the answer is (c).

6. 1, 4, 27, 16, ?, 36, 343

a) 25 b) 87 c) 120 d) 125

Solution: The given series consists of cubes of odd numbers and squares of even numbers,

i.e., 13, 2

3, 3

3, 4

2,

So, missing term = 53

= 125

Hence, the answer is (d).


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7. 4, 6, 12, 14, 28, 30, ?

a) 32 b) 60 c) 62 d) 64

Solution: The given sequence is a combination of two series:

I. 4, 12, 28, ? and II. 6, 14, 30,..

Now, the pattern followed in each of the above two series is: +8, +16, +32, .

So, missing number = (28 + 32) = 60.

Hence, the number is (b).

8. 1, 3, 3, 6, 7, 9, ?, 12, 21

a) 10 b) 11 c) 12 d) 13

Solution: The given sequence is a combination of two series:

I. 1, 3, 7, ?, 21 and II. 6, 14, 30,.

The pattern followed in I is +2, +4,.and the pattern followed in II is +3.

So, missing number = 7 + 6 = 13.

9. Which fraction comes next in the sequence2

1,4

3,8

5,16

7,?

a)32

9b)

17

10c)

34

11d)

35

12

Clearly, the numerators of the fractions in the given sequence form the series 1, 3, 5, 7, in

which each term is obtained by adding 2 to the previous term.

The denominators of the fractions fro the series 2, 4, 8, 16, i.e. 21, 2

2, 2

3, 2

4.

So, the numerator of the next fraction will be (7+2) i.e. 9 and the denominator will be 25

i.e.

32.

Thus, the next term is32

9. Hence, the answer is (a).

study material general apt vol-02

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