# SBI Clerk Mains Quantitative Aptitude Practice Set 1

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# SBI Clerk Mains Quantitative Aptitude Practice Set 1

### Introduction

Career in Banking is one of the most lucrative and most sought after careers. In India, Bank Recruitment Exams are primarily conducted for recruitment of Probationary Officers, Clerks & Specialist Officers. India currently[2019] has 93 commercial and 27 public sector banks out of which 19 are nationalized and 6 are SBI and its associate banks and rest two are IDBI Bank and Bharatiya Mahila Bank, which are categorized as other public sector banks. Recruitment for Bank Probationary Officers, Management Trainees, Clerks and for various other posts generally follow a 3 step recruitment process: Preliminary Exam + Mains Exam + Interview & Group Discussion. The article SBI Clerk Mains Quantitative Aptitude Practice Set 1 presents a practice set for the most sought after SBI PO recruitment. Until the year 2013, All Public Sector Banks used to conduct their own entrance test, GDs and Personal Interview for recruiting candidates. However, after 2014, IBPS started conducting recruitment Tests for 12 PSU Banks. IBPS holds a separate entrance test for recruitment.

Mains exams are very important to clear every government sector or bank related recruitment process in India. Only those candidates who are selected in the Mains round are allowed to move further up in the recruitment process. The marks obtained in the Mains exams are considered for the final merit list. Mains exams usually consist of 4 sections, with 155 questions with a time duration of 3 hours. Mains exams most certainly have negative marking.

### Quiz

Directions(1-5) : A school has four sections A, B, C, D of Class IX students.

Result No. of Students
Section A Section B Section C Section D
Students failed in both Exams 28 23 17 27
Students failed in half-yearly but passed in Annual Exams 14 12 8 13
Students passed in half-yearly but failed in Annual Exams 6 17 9 15
Students passed in both Exams 64 55 46 76

1. If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context?

A. Half yearly examinations were more difficult.
B. Annual examinations were more difficult.
C. Both the examinations had almost the same difficulty level.
D. The two examinations cannot be compared for difficulty level.

Explanation:
Number of students who passed half-yearly exams in the school

= (Number of students passed in half-yearly but failed in annual exams)

+ (Number of students passed in both exams)

= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams)

+ (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

2. How many students are there in Class IX in the school?

A. 336
B. 189
C. 335
D. 430

Explanation:
Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:

= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430.

3. Which section has the maximum pass percentage in at least one of the two examinations?

A. A Section
B. B Section
C. C Section
D. D Section

Explanation:
Pass percentages in at least one of the two examinations for different sections are:
For Section A[ $$\frac{(14 + 6 + 64)}{(28 + 14 + 6 + 64) × 100}$$ ]% = [ $$\frac{(84)}{(112) × 100}$$ ] = 75%
For Section B[ $$\frac{12 + 17 + 55)}{(23 + 12 + 17 + 55) × 100}$$ ]% = [ $$\frac{(84)}{(107) × 100}$$ ] = 78.5%
For Section C[ $$\frac{(14 + 6 + 64)}{(17 + 8 + 9 + 46) × 100}$$ ]% = [ $$\frac{(63)}{(80) × 100}$$ ] = 78.5%
For Section D[ $$\frac{(13 + 15 + 76)}{(27 + 13 + 15 + 76) × 100}$$ ]% = [ $$\frac{(104)}{(31) × 100}$$ ] = 79.39%

4. Which section has the maximum success rate in annual examination?

A. A Section
B. B Section
C. C Section
D. D Section

Explanation:
Total number of students passed in annual exams in a section

= [ (No. of students failed in half-yearly but passed in annual exams)

+ (No. of students passed in both exams)
Success rate in annual exams in Section A
[ $$\frac{No. of students of Section A passed in annual exams}{Total number of students in Section A × 100 }$$ ]%
[ $$\frac{(14 + 64)}{(28 + 14 + 6 + 64) × 100% }$$ ]%
[ $$\frac{(78)}{(112) × 100 }$$ ]%
= 69.64%.

Similarly, success rate in annual exams in:
Section B [ $$\frac{(13 + 76)}{(27 + 13 + 15 + 76) × 100% }$$ ]%
Similarly Section D[ $$\frac{(89)}{(131) × 100 }$$ ]% = 67.94%
Clearly, the success rate in annual examination is maximum for Section A

5. Which section has the minimum failure rate in half-yearly examination?

A. A section
B. B section
C. C section
D. D section

Explanation: Option D

Explanation:
Total number of failures in half-yearly exams in a section

= [ (Number of students failed in both exams)

+ (Number of students failed in half-yearly but passed in Annual exams)

] in that section
Failure rate in half-yearly exams in Section A
[ $$\frac{Number of students of Section A failed in half-yearly}{Total number of students in Section A × 100 }$$ ]%
[ $$\frac{(28 + 14)}{(28 + 14 + 6 + 64) × 100% }$$ ]%
[ $$\frac{(42)}{(112) × 100 }$$ ]%
= 37.5%.
Similarly, Section D [ $$\frac{Number of students of Section A failed in half-yearly}{Total number of students in Section A × 100 }$$ ]%
[ $$\frac{(27 + 13)}{(27 + 13 + 15 + 76) × 100% }$$ ]%
[ $$\frac{(40)}{(131) × 100 }$$ ]%
30.53%
Clearly, the failure rate is minimum for Section D.

Directions(6-10) : Find the missing number

6. 5, 10, 13, 26, 29, 58, 61, (….)

A. 122
B. 64
C. 125
D. 128

Explanation: Option A

Explanation:
Numbers are alternatively multiplied by 2 and increased by 3.

So, the missing number = 61 x 2 = 122.

7. 15, 31, 63, 127, 255, (….)

A. 513
B. 511
C. 517
D. 523
D. 128

Explanation: Option B

Explanation:

Each number is double the preceding one plus 1.

So, the next number is (255 x 2) + 1 = 511.

8. 1, 8, 27, 64, 125, 216, (….)

A. 354
B. 343
C. 392
D. 245

Explanation: Option B

Explanation:

Numbers are $${1}^{3}$$, $${2}^{3}$$,$${3}^{3}$$, $${4}^{3}$$, $${5}^{3}$$, $${6}^{3}$$.

So, the missing number is 73 = 343.

9. 3, 7, 6, 5, 9, 3, 12, 1, 15, (….)

A. 18
B. 13
C. -1
D. 3

Explanation: Option C

Explanation:

There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted.

The next number is 1 – 2 = -1.

10. 11, 13, 17, 19, 23, 29, 31, 37, 41, (….)

A. 43
B. 47
C. 53
D. 51

Explanation: Option A

Explanation:

Numbers are all primes. The next prime is 43.

11. Find the roots of the quadratic equation: $${x}^{2}$$+ 2x – 15 = 0?

A. -5, 3
B. 3, 5
C. -3, 5
D. -3, -5
E. 5, 2

Explanation: Option A

Explanation:
$${x}^{2}$$+ 5x -3x- 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = -5.

12. Find the roots of quadratic equation: 2$${x}^{2}$$ + 5x + 2 = 0?

A. -2, -1/2
B. 4, -1
C. 4, 1
D. -2, 5/2
E. None of these

Explanation: Option A

Explanation:
2$${x}^{2}$$ + 4x + x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 => x = -2, -1/2

13. Find the roots of quadratic equation: $${x}^{2}$$ + x – 42 = 0?

A. -6, 7
B. -8, 7
C. 14, -3
D. -7, 6
E. 3, -14

Explanation: Option D

Explanation:
$${x}^{2}$$ + 7x – 6x + 42 = 0
x(x + 7) – 6(x + 7) = 0
(x + 7)(x – 6) = 0 => x = -7, 6

14. Find the roots of quadratic equation: 3$${x}^{2}$$ – 7x – 6 = 0?

A. -6, 3
B. 3, -2/3
C. -5, 2
D. -9, 2
E. None of these

Explanation: Option B

Explanation:
3$${x}^{2}$$ – 9x + 2x – 6 = 0
3x(x – 3) + 2(x – 3) = 0
(x – 3)(3x + 2) = 0 => x = 3, -2/3

15. The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?

A. 9, 10
B. 10, 11
C. 11, 12
D. 12, 13
E. None of these

Explanation: Option A

Explanation:
Let the two consecutive positive integers be x and x + 1
$${x}^{2}$$ + $${x + 1}^{2}$$ – x(x + 1) = 91
$${x}^{2}$$ + x – 90 = 0
(x + 10)(x – 9) = 0 => x = -10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.