# Quant Test for SBI PO 2018 Prelim Exam Set – 51

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. 20x2 – 31x + 12 = 0
II. 6y2 – 7y + 2 = 0

x>y
x=y or relation cannot be established.
x>=y
y>=x
y>x
Option A
I. 20x2 – 31x + 12 = 0
20x2 – 31x + 12 = 0
20x2 – 16x – 15x + 12 = 0
x = 3/4, 4/5
II. 6y2 – 7y + 2 = 0
6y2 – 7y + 2 = 0
6y2– 3y – 4y + 2 = 0
y = 1/2, 2/3
x>y

2. I. 2x2 + 5x – 12 = 0
II. 2y2 – 19y + 35 = 0

x=y or relation cannot be established.
y>x
x>=y
x>y
y>=x
Option B
I. 2x2 + 5x – 12 = 0
2x2 + 5x – 12 = 0
2x2 + 8x – 3x – 12 = 0
x = -4 , 3/2
II. 2y2 – 19y + 35 = 0
2y2 – 19y + 35 = 0
2y2 – 14y – 5y + 35 = 0
y = 5/2, 7
x < y

3. I. 5x2 – 13x + 6 = 0
II. 3y2 – 22y – 35 = 0

y>=x
x=y or relation cannot be established.
x>=y
x>y
y>x
Option B
I. 5x2 – 13x + 6 = 0
5x2 – 13x + 6 = 0
5x2 – 10x – 3x + 6 = 0
x = 3/5, 2
II. 3y2 – 22y – 35 = 0
3y2 – 22y – 35 = 0
3y2 – 15y – 7y – 35 = 0
y = -7/3, 5
Relation cannot be established.

4. I. 50x2 – 95x + 42 = 0
II. 50y2 – 65y + 21 = 0

y>x
x=y or relation cannot be established.
x>=y
x>y
y>=x
Option C
I. 50x2 – 95x + 42 = 0
50x2 – 95x + 42 = 0
50x2 – 60x – 35x + 42 = 0
x = 7/10, 6/5
II. 50y2 – 65y + 21 = 0
50y2 – 65y + 21 = 0
50y2 – 65y + 21 = 0
y = 3/5, 7/10
x ≥ y

5. I. 3x2 – 4x – 15 = 0
II. 5y2 – 9y – 18 = 0

x>=y
x=y or relation cannot be established.
y>=x
y>x
x>y
Option B
I. 3x2 – 4x – 15 = 0
3x2 – 4x – 15 = 0
3x2 – 9x + 5x – 15 = 0
x = -5/3, 3
II. 5y2 – 9y – 18 = 0
5y2 – 9y – 18 = 0
5y2 – 15y + 6y – 18 = 0
y = -6/5, 3
x = y or relation cannot be established

6. Directions(6-10): Following pie-charts show the percentage distribution of job vacancies
in IT industries in the year 2000 and 2010. In the year 2000, the total number of vacancies was 5.4 lakh, and in the year 2010, it was 8.6 lakh.  7. What is the total number of vacancies available in Chennai in 2000 and in Mumbai in the year 2010?
3.50 lakh
1.02 lakh
3.01 lakh
2.238 lakh
4.15 lakh
Option D
Sum = 5.4* 8/100 + 8.6* 21/100
= 0.432 + 1.806
= 2.238 lakh

8. What is the difference between the number of vacancies available in Bangalore in the year 2010 and 2000?
3.250 lakh
5.080 lakh
4.303 lakh
2.112 lakh
1.426 lakh
Option E
Difference = 8.6*26/100 – 5.4*15/100
= 2.236 – 0.81 = 1.426 lakh

9. What is the average number of vacancies available in Hyderabad in the year 2000 and 2010?
70000
52420
60000
35420
45500
Option A
Hyderabad(2000) = 5.4*10/100 = 0.54 lakh
Hyderabad(2010) = 8.6*10/100 = 0.688
Average = 0.54 + 0.86/2 = 0.7 lakh = 70,000

10. What is the percentage rise in vacancies available in Hyderabad from year 2000 to 2010? (Give approximate value only).
50%
45%
59%
61%
48%
Option C
Hyderabad(20OO) = 5.4*10/100 = 0.54 lakh
Hyderabad(2010) = 8.6*10/100 = 0.688 lakh
% rise =(0.86-0.54)/0.54*100 ==59%

11. If the number of vacancies in Pune is 48000 in the year 2010 and the percentage distribution is the same as given in the chart, what is number of vacancies available in NCR in 2010?
130000
150000
120000
164571
110000
Option D
Total number of vacancies in 2010= 48000*100/7
=685714.3
Vacancies in NCR = 24% of 685714.3 = 164571