# Quant Test for SBI PO 2018 Prelim Exam Set – 24

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.

Click here to know the details of the Examination

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 + 13x + 14 = 0
II. 4y2 + 9y + 2 = 0

x>y
y>=x
x=y or relation cannot be established.
y>x
x>=y
Option B
I. 3x2 + 13x + 14 = 0
3x2 + 13x + 14 = 0
3x2 + 6x + 7x + 14 = 0
x = -7/3, -2
II. 4y2 + 9y + 2 = 0
4y2 + 9y + 2 = 0
4y2 + 8y + y + 2 = 0
y = -2, -1/4
x ≤ y

2. I. 16x2 + 20x + 6 = 0
II. 10y2 + 38y + 24 = 0

y>x
x>=y
x>y
y>=x
x=y or relation cannot be established.
Option C
I. 16x2 + 20x + 6 = 0
Divide both equations by 2
8x2 + 10x + 3 = 0
8x2 + 4x + 6x + 3 = 0
x = -1/2, -3/4
II. 10y2 + 38y + 24 = 0
5y2 + 19y + 12 = 0
5y2 + 15y + 4y + 12 = 0
y = -4, -4/5
x > y

3. I. 3x2 + 4x – 39 = 0
II. 3y2 – 5y – 78 = 0

x=y or relation cannot be established.
y>=x
x>=y
y>x
x>y
Option A
I. 3x2 + 4x – 39 = 0
3x2 + 4x – 39 = 0
3x2 – 9x + 13x – 39 = 0
x = -13/3, 3
II. 3y2 – 5y – 78 = 0
3y2 – 5y – 78 = 0
3y2 – 18y + 13y – 78 = 0
y = -13/3, 6
x = y or relation cannot be established

4. I. 4x2 + 19x + 21 = 0
II. 2y2 – 25y – 27 = 0

y>=x
x=y or relation cannot be established.
y>x
x>y
x>=y
Option C
I. 4x2 + 19x + 21 = 0
4x2 + 19x + 21 = 0
4x2 + 12x + 7x + 21 = 0
x = -3, – 1.75
II. 2y2 – 25y – 27 = 0
2y2 – 25y – 27 = 0
2y2 – 27y + 2y – 27 = 0
y = 13.5, -1
x < y

5. I. x2 – 6x – 91 = 0
II. y2 – 32y + 247 = 0

x=y or relation cannot be established.
x>y
x>=y
y>x
y>=x
Option E
I. x2 – 6x – 91 = 0
x2 – 13x + 7x – 91 = 0
x = 13, -7
II. y2 – 32y + 247 = 0
y2 – 19y -13y + 247 = 0
y = 19, 13
x ≤ y

6. Directions(6-10): The bar graph shown below depicts the number of appeared candidates and passed candidates (in hundreds) in a test from seven different institutions. 7. What is the average number of candidates passed from all the institutions together ?
500
900
600
700
400
Option D
Required average = 4900/7 = 700

8. From which institution is the difference between the appeared candidates and passed candidates the maximum ?
D
E
A
F
C
Option A
Difference between the appeared and passed candidates from institution A = 1300 – 1200 = 100
from B = 1400 – 1000 = 400
from C = 700 – 300 = 400
from D = 1200 – 400 = 800
from E = 1500 – 1200 = 300
from F = 600 – 400 = 200
from G = 1100 – 500 = 600
Hence, from the above calculation , it is institution D .

9. What is the respective ratio of the number of candidates who have failed from institution B to the number of candidates who have appeared from institution F ?
4:7
5:1
1:9
2:3
3:4
Option D
The number of candidates who have failed from institution B = 400
The number of candidates who have appeared from institution F = 600
Required ratio = 2:3

10. The number of candidates passed from institutions C and E together is approximate what percent of the total number of candidates appeared from institutions A and E together ?
60%
54%
75%
55%
57%
Option E
Required % = 1600/2800 * 100 = 57% approx.

11. What is the difference between the number of candidates appeared from institutes B,C,D and F together and candidates passed from institutions A,E and G together?
900
1000
1200
1500
800
Option B
Total number of students appearing from B,C,D and F together = 39 hundred
candidates passed from institutes A,E and G together = 29 hundred
Required difference = 1000