# Quant Test for SBI PO 2018 Prelim Exam Set – 45

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 2018. Details of the exam are as under: Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam.

Directions(1-5): Find the relation between x and y and choose a correct option.

1. I. 8x2 + 6x + 1 = 0
II. 5y2 + 8y – 4 = 0

y>=x
x>y
x>=y
y>x
x=y or no relation cannot be established.
Option E
I. 8x2 + 6x + 1 = 0
8x2 + 6x + 1 = 0
8x2 + 4x + 2x + 1 = 0
x = -1/4, -1/2
II. 5y2 + 8y – 4 = 0
5y2 + 8y – 4 = 0
5y2 + 10y – 2y – 4 = 0
y = -2, 2/5
x = y or relationship cannot be determined

2. I. 12x2 – 5x – 3 = 0
II. 4y2 – 11y + 6 = 0

y>=x
y>x
x>=y
x=y or no relation cannot be established.
x>y
Option A
I. 12x2 – 5x – 3 = 0
12x2 – 5x – 3 = 0
12x2 + 4x – 9x – 3 = 0
x = -1/3, 3/4
II. 4y2 – 11y + 6 = 0
4y2 – 8y – 3y + 6 = 0
y= 3/4, 2
x ≤ y

3. I. 2x2 – (6 + √3)x + 3√3 = 0
II. 3y2 – (9 + √3)y + 3√3 = 0

x=y or no relation cannot be established.
y>x
x>y
y>=x
x>=y
Option A
I. 2x2 – (6 + √3)x + 3√3 = 0
2x2 – 6x – √3x + 3√3 = 0
2x (x- 3) – √3 (x – 3) = 0
x = 3, √3/2 (0.7)
II. 3y2 – (9 + √3)y + 3√3 = 0
3y2 – 9y – √3y + 3√3 = 0
3y (y – 3) – √3 (y – 3) = 0
y = 3, √3/3 (0.6)
x = y or relationship cannot be determined

4. I. 9/√x + 8/(√x +1) = 5
II. 12/√y – 4/√y = 2

x>=y
y>=x
x>y
x=y or no relation cannot be established.
y>x
Option E
I. 9/√x + 8/(√x +1) = 5
9/√x + 8/(√x +1) = 5
[9(√x +1) + 8√x]/[√x * (√x +1)] = 5
17√x + 9 = 5 (x + √x)
5x – 12√x – 9 = 0
5x – 15√x + 3√x – 9 = 0
5√x (√x – 3) + 3 (√x – 3) = 0
√x cannot be -3/3
So √x = 3
x = 9
II. 12/√y – 4/√y = 2
12/√y – 4/√y = 2
8/√y = 2
√y = 4
y = 16
y > x

5. I. 4x2 – 23x + 30 = 0
II. 4y2 – 3y – 45 = 0

y>=x
x>y
x>=y
y>x
x=y or no relation cannot be established.
Option E
I. 4x2 – 23x + 30 = 0
4x2 – 23x + 30 = 0
x2 – 15x – 8x + 30 = 0
x = 15/4, 2
II. 4y2 – 3y – 45 = 0
4y2 – 3y – 45 = 0
4y2 + 12y – 15y – 45 = 0
y = -3, 15/4
x = y or relation cannot be established

6. Directions(6-10): Study the following pie-chart and answer the following questions.
Total number of Employees = 12600 Percentage-wise distribution of Employees 7. What is the average number of employees in Accounts, Admission and Research department together?
1620
1700
1525
1722
1202
Option D
Number of employees in Accounts Department
=17 * 12600/100 = 1890
Average = 2142 + 1134 + 1890/3
= 5166/3 = 1722

8. What is the difference between the total number of employees in the department of HR and Admission together and the total number of employees in Accounts and Examination department together?
2268
2200
2040
2525
3500
Option A
Difference = (38% of 12600 – 20% of 12600)
= 18% of 12600
= 18 * 12600/100
= 2268

9. The number of employees in Examination department is approximately what percentage of the total number of employees in the department of HR and Academic Affairs together?
40
55
30
60
80
Option B
Number of employees in examination department
= 21 * 12600/100 = 2646
Number of employees in the HR Department = 11 * 12600/100 =1386
Number of employees in Academic Affairs 27 * 12600/100 = 3402
Total number of employees in both the departments Academic Affairs and HR together = 3402 + 1386 = 4788
Required % = 2646/4748*100 = 55.26 == 55

10. If 30 per cent of the number of employees of Research department is females, then what is the number of male employees in the Research department?
1470
1350
1454
1600
1323
Option E
Number of employees in Research Department = 15 * 12600/100 =1890
female employees in Research department = 1890 * 30/100 = 567
Hence, number of male employees in Research department = 1890 – 567
= 1323

11. The number of employees in the department of Academic Affairs is approximately what per cent more than the number of employees in Examination department?
18%
10%
29%
31%
14%
Option C
Number of employees in Academic affairs
= 27 * 12600/100 = 3402
Number of employees in Examination Department = 21 * 12600/100 = 2646
Required % = (3402 – 2646)/2646*100 = 28.57 == 29%