# Quant Test for SBI PO 2018 Prelim Exam Set – 30

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. 3x2 – 25x + 52 = 0
II. 2y2 – 7y + 3 = 0

x>=y
y>x
x>y
x=y or relation cannot be established.
y>=x
Option C
I. 3x2 – 25x + 52 = 0
II. 2y2 – 7y + 3 = 0
I. 3x2 – 25x + 52 = 0
=> 3x2 – 25x + 52 = 0
=> 3x2 – 12x – 13x + 52 = 0
=> x = 4, 13/3
II. 2y2 – 7y + 3 = 0
=> 2y2 – 7y + 3 = 0
=> 2y2 – 6y – y + 3 = 0
y = 1/2, 3
x > y

2. I. 6x2 + 7x + 2 = 0
II. 15y2 – 38y – 40 = 0

x=y or relation cannot be established.
x>y
y>x
y>=x
x>=y
Option A
I. 6x2 + 7x + 2 = 0
II. 15y2– 38y – 40 = 0
I. 6x2 + 7x + 2 = 0
=>6x2 + 7x + 2 = 0
=> 6x2 + 4x + 3x + 2 = 0
x = -2/3, -1/2
II. 15y2 – 38y – 40 = 0
=> 15y2 – 38y – 40 = 0
=> 15y2 + 12y – 50y – 40 = 0
y = -4/5, 10/3
x = y or relation cannot be established

3. I. x2 – √3969 = √6561
II. y2 – √1296 = √4096

x>y
x=y or relation cannot be established.
x>=y
y>x
y>=x
Option B
I. x2 – √3969 = √6561
II. y2 – √1296 = √4096
I. x2 – √3969 = √6561
=>x2 – √3969 = √6561
=> x2 – 63 = 81
=> x2 = 144
x = -12, 12
II. y2 – √1296 = √4096
=>y2 – √1296 = √4096
=> y2 – 36 = 64
=>y2 = 100
y = -10, 10
x = y or relation cannot be established

4. I. x2 + √5x – 10 = 0
II. 2y2 + 9√5y + 50 = 0

y>=x
y>x
x>=y
x=y or relation cannot be established.
x>y
Option C
I. x2 + √5x – 10 = 0
II. 2y2 + 9√5y + 50 = 0
I. x2 + √5x – 10 = 0
=> x2 + √5x – 10 = 0
=> x2 + 2√5x – √5x – 10 = 0
x = -2√5, √5
II. 2y2 + 9√5y + 50 = 0
=> 2y2 + 9√5y + 50 = 0
=> 2y2 + 4√5y + 5√5y + 50 = 0
y = -2√5, -5√5/2
x ≥ y

5. I. 4x2 – 12x + 5 = 0
II. 6y2 – 13y + 6 = 0

y>x
x>=y
y>=x
x>y
x=y or relation cannot be established.
Option E
I. 4x2 – 12x + 5 = 0
II. 6y2 – 13y + 6 = 0
I. 4x2 – 12x + 5 = 0
=> 4x2 – 12x + 5 = 0
=> 4x2 – 2x – 10x + 5 = 0
x = ½, 5/2
II. 6y2 – 13y + 6 = 0
=> 6y2 – 13y + 6 = 0
=> 6y2 – 4y – 9y + 6 = 0
y = 2/3, 3/2
x = y or relation cannot be established

6. Directions(6-10): Following pie-charts show the distribution of the total number of
students selected in an entrance exam from seven different colleges in 2012 and 2013. (The total
number of students selected from College C7 in 2012 and 2013 are 180 and 270 respectively).
The data given in the pie-chart which are in degrees.
2012 2013 7. What is the per cent rise in the number of students selected from College C4 from 2012 to 2013 ?
60%
52%
77%
54%
68%
Option E
C4(2012) = 1500 × 36°/360°
= 150
C4(2013) = 1800 × 50.4°/360°
= 252
% rise = (252 -150)/150 * 100
= 10200/150
=68%

8. In which of the following colleges is the per cent rise or fall in the number of students selected from 2012 to 2013 the maximum?
C2
C6
C4
C5
C1
Option B
C1 = (414 – 315)/315 *100
= 31.42%
C2 = (210 – 144)/144 *100
= 31.42%
C3 = (240 – 198)/198 *100
= 17.5%
C4 = (252 – 150)/150 *100
= 68%
C5 = (288 – 270)/270 *100
= 6.66%
C6 = (234 – 135)/135 *100
= 73.33%
C7 = (270 – 180)/180 *100
= 50%

9. The total number of students selected from Colleges C5 and C7 together in the year 2012 is approximately what per cent of the number of students selected from College C2 in the year 2013?
303.15%
312.5%
259.8%
300%
299.2%
Option B
C5 + C7 = 1500 * (64.8° + 43.2°)/360°
= (1500 * 108°)/360°
= 450
C2 = 1800 * 28.8°/360°
= 144
Required % = 450/144 × l00 = 312.5%

10. The total number of students selected from all seven colleges together in the year 2013 is approximately what per cent of the total number of students selected from all seven colleges in 2012?
120%
110%
105%
100%
115%
Option A
Total (2012) = 180 × 360°/43.2°
= 1500
Total (2013) = 270 × 360°/54°
= 1800
Required % = 1800/1500 × 100
= 120%

11. What is the difference between the average number of students selected from Colleges C1, C2 and C3 in the year 2012 and the average number of students selected from Colleges C5, C6 and C7 in the year 2013 ?
10
7
5
9
6
Option D
Average of C1, C2 and C3
=(75.6 + 50.4 + 57.6)/(360 * 3) * 1500
= 255
Average of C5, C6 and C7
=(57.6 + 46.8 + 54)/(360 * 3) * 1800
= 264
Difference = 264 – 255 = 9