Quantitative Aptitude - SPLessons

IBPS Clerk Mains Quantitative Aptitude ...

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IBPS Clerk Mains Quantitative Aptitude Practice

shape Introduction

Quantitative Aptitude is an important section in the employment related competitive exams in India. In particular, exams like IBPS, SBI and other bank related employment exams have Quantitative Aptitude questions along with Reasoning. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.This article presents IBPS Clerk Mains Quantitative Aptitude Practice Sets for acing the IBPS Clerk Mains Examination.

shape Aptitude

Directions (1 – 10): What value should come in place of question mark (?) in the following equations?

1. (\( \sqrt{1444 ÷ 5} \)) × 3.25 = ?


    A. 24.7
    B. 25.4
    C. 26.6
    D. 27.2
    E. 28.5


Answer: Option A

Solution: x = \( \frac{38} {5}\) × 3.25

= 24.7

2. 17.8 + \( \frac{3} {7}\) of 89.6 = ?


    A. 52.8
    B. 54.4
    C. 56.2
    D. 58.6
    E. 60.4


Answer: Option C

Solution: 17.8 + 38.4

= 56.2

3. 373.816 + 274.102 – 199.573 – 108.108 = ?


    A. 338.147
    B. 339.787
    C. 340.237
    D. 341.347
    E. 342.107


Answer: Option C

Solution: x = 647.918 – 307, 681

= 340.237

4. 14.8 × 7.25 = 20% of = ?


    A. 524.5
    B. 528.5
    C. 532.5
    D. 536.5
    E. 540.5


Answer: Option D

Solution: 107.3 = \( \frac{20} {100}\) × x

x = 536.5

5. \( \sqrt{1156} \) x \( \sqrt{42.25} \) = ?


    A. 212
    B. 215
    C. 218
    D. 220
    E. 221


Answer: Option E

Solution: x = 34 × 6.5

= 221

6. 2\( \frac{1} {6}\) + (3 \( \frac{3}{4} \) – 1\( \frac{1}{4}) \) = ?


    A. 4\( \frac{5} {12}\)
    B. 4\( \frac{1} {4}\)
    C. 4\( \frac{7} {12}\)
    D. 5\( \frac{1} {4}\)
    E. None of these


Answer: Option E

Solution: (2 + 3 – 1) + \( \frac{1}{6} \) + \( \frac{3}{4}) \) – \( \frac{1}{4} \)

= \( \frac{2 + 9 – 3}{12} \)

= 4 + \( \frac{8}{12} \)

= \( \frac{42}{3} \)

7. 36251 + 43261 = ? + 52310


    A. 27202
    B. 28102
    C. 29302
    D. 26602
    E. None of these


Answer: Option A

Solution: 36251 + 43261 = x + 52310

79512 = x + 52310

79512 – 52310 = x

x = 27202

8. 7 \( \frac{3} {6}\) of 534 + 262 = 61800 – ?


    A. 56533
    B. 57533
    C. 58533
    D. 37355
    E. None of these


Answer: Option B

Solution: \( \frac{45} {6}\) × 534 + 262 = 61800 – x

4005 + 262 = 61800 – x

x = 57533

9. 72% of 486 – 64% of 261 = ?


    A. 184.66
    B. 183.66
    C. 188.88
    D. 182.88
    E. 186.24


Answer: Option D

Solution: x = 349.92 – 167.04

= 182.88

10. ? ÷ 62 × 12 = 264


    A. 1364
    B. 1284
    C. 1348
    D. 1388
    E. None of these


Answer: Option A

Solution: \( \frac{x} {62}\) × 12

= 264

x = 1364

Directions (1 – 3): Study the table carefully to answer the questions that follow:


Universities Percentage of Students Percentage of Boys Percentage of Girls
A 12 55 45
B 15 60 40
C 8 30 70
D 28 75 25
E 17 20 80
F 20 64 36


1. What is the ratio of total number of boys in University B and D together to the total number of girls in the same universities together?


    A. 25 : 19
    B. 35 : 21
    C. 20 : 7
    D. 30 : 13
    E. 5 : 3


Answer: Option D

Solution: Total boys in B = \( \frac{15} {100}\) × 32500 × \( \frac{60} {100}\) = 2925

Total boys in D = \( \frac{28} {100}\) × 32500 × \( \frac{75} {100}\) = 6825

Total girls in B = \( \frac{15} {100}\) × 32500 × \( \frac{40} {100}\)= 1950

Total girls in D = \( \frac{28} {100}\) × 32500 × \( \frac{25} {100}\)= 2275

Reqd. Ratio = (\( \frac{2925 + 6825} {1950 + 2275})\)

Reqd. Ratio = \( \frac{9750} {4225}\)

Reqd. Ratio = \( \frac{30} {13}\)

2. The total number of students in the University A is what per cent of the total number of students in University F?


    A. 55%
    B. 67%
    C. 60%
    D. 48%
    E. 58%


Answer: Option C

Solution: \( \frac{12} {100}\) × 32500

= \( \frac{x} {100}\) × \( \frac{20} {100}\) × 32500

x = 60

3. What is the total number of boys from university A,C and E together?


    A. 6030
    B. 5030
    C. 7030
    D. 4030
    E. 3030


Answer: Option D

Directions (4 – 7): The following graph shows the number of students participated in annual talent show of Kurukshetra university from different colleges. Study the graph carefully to answer the given questions.


4. Find the ratio of the total number of students participated in dancing from college P, T & V together to the total students participated in singing from college P, Q & R together.?


    A. 28 : 27
    B. 25 : 26
    C. 25 : 29
    D. 27 : 29
    E. None of these


Answer: Option B

Solution: Ratio = \( \frac{340 + 190 + 220} {240 + 320 + 220}\)

= 25 : 26

5. From which of these colleges the total number of students participated is 2nd minimum?


    A. College P
    B. College Q
    C. College R
    D. College S
    E. None of these


Answer: Option D

Solution: Collage P = 840
Q = 900
R = 780
S = 740
T = 790
U = 730
V = 870
Q > V > P > T > R > S >U

6. If 40% of the total students of college S who participated in acting, are doing solo acting and other are in group acting. 66\( \frac{2} {3}\)% of the students participated in group acting are a part of comedy dramas. Then find the total number of students who are acting in comedy dramas.?


    A. 104
    B. 120
    C. 110
    D. 108
    E. None of these


Answer: Option A

Solution: Total in S = 260

Number of students acting In comedy drama

= \( \frac{60} {100}\) × \( \frac{2} {3}\) × 260

= 104

7. The difference between the no. of students participating in dancing from college P and R and that in singing from T and U?


    A. 165
    B. 170
    C. 82
    D. 94
    E. None of these


Answer: Option B

Solution: Total in dance from P & R

= 340 + 260 = 600

= Number of students singing

From T & U is = 280 + 150 = 430

Difference = 600 – 430

= 170

8. The length and breadth of a plot are 35 m and 16 m respectively. If the rate of fencing is Rs. 7 per meter, what is cost of its fencing?


    A. Rs. 3920
    B. Rs. 602
    C. Rs. 714
    D. Rs. 357
    E. None of these


Answer: Option C

Solution: Perimeter of plot = 2 (c + b) = 2

(35 + 16) = 102 m

Total cost of fencing = 7 × 102 = Rs 714

Directions (9 – 10): In the following number series one number is missing. Find out the missing number.

9. 2, 21, 138, 705,?


    A. 2386
    B. 3540
    C. 2820
    D. 3236
    E. None of these


Answer: Option E

Solution: 2 × 7 + 1 × 7 = 21

21 × 6 + 2 × 6 = 138

138 × 5 + 3 × 5 = 705

705 × 4 + 4 × 4 = 2836

17, 52, 158, 477, ?, 4310


    A. 1433
    B. 1432
    C. 1435
    D. 1434
    E. None of these


Answer: Option C

Solution: 17 × 3 + 1 = 52

52 × 3 + 2 = 158

158 × 3 + 3 = 477

477 × 3 + 4 = 1435

1435 × 3 + 5 = 4310