# SBI Clerk Prelims 2021 Quantitative Aptitude Questions (Day-73)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Prelims 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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Directions (01-05): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give an answer as,

1) I)
2x2 – 30x + 112 = 0

II) y2– 18y + 80 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

2) I)
x2 – 9x – 360 = 0

II) y2– 16y – 192 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

3) I)
x2 – 58x + 840 = 0

II) y2– 48y + 560 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

4) I)
3x2 – 18x + 15 = 0

II) y2– 13y + 42 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

5) I)
x2 + 39x + 224 = 0

II) y2+ 35y + 196 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Bar graph

Directions (06-10): Study the following information carefully and answer the questions given below.

The given bar graph shows the number of students in three different departments in five different colleges. 6) The total number of ECE students in all the colleges together is approximately what percent of the number of CSE students in all the colleges together?

A.87%

B.89%

C.85%

D.83%

E.91%

7) Total number of EEE students in all the colleges together is approximately what percent of the total number of students in E in all the departments together?

A.173%

B.176%

C.179%

D.182%

E.185%

8) In which college, the total number of students in all departments together is maximum?

A.C

B.D

C.E

D.A

E.B

9) What is the difference between the number of CSE students in B and the number of ECE students in C?

A.60

B.80

C.100

D.120

E.None of these

10) What is the ratio of the number of EEE students in D to the number of ECE students in A?

A.17:7

B.27:17

C.37:27

D.47:37

E.None of these

2x2 – 30x + 112 = 0

2x2 – 16x – 14x + 112 = 0

2x(x – 8) – 14(x – 8) = 0

(2x – 14)(x – 8) = 0

x = 8, 7

y2 – 18y + 80 = 0

y2 – 10y – 8y + 80 = 0

y(y – 10) – 8(y – 10) = 0

(y – 8)(y – 10) = 0

y = 8, 10

x ≤ y

x2 – 9x – 360 = 0

x2 – 24x + 15x – 360 = 0

x(x – 24) + 15(x – 24) = 0

(x + 15)(x – 24) = 0

x = – 15, 24

y2 – 16y – 192 = 0

y2 – 24y + 8y – 192 = 0

y(y – 24) + 8(y – 24) = 0

(y – 24)(y + 8) = 0

y = 24, -8

Relationship between x and y cannot be established.

x2 – 58x + 840 = 0

x2 – 30x – 28x + 840 = 0

x(x – 30) – 28(x – 30) = 0

(x – 28)(x – 30) = 0

x = 28, 30

y2 – 48y + 560 = 0

y2 – 20y – 28y + 560 = 0

y(y – 20) – 28(y – 20) = 0

(y – 28)(y – 20) = 0

y = 28, 20

x ≥ y

3x2 – 18x + 15 = 0

3x2 – 15x – 3x + 15 = 0

3x(x – 5) – 3(x – 5) = 0

(3x – 3)(x – 5) = 0

x = 1, 5

y2 – 13y + 42 = 0

y2 – 6y – 7y + 42 = 0

y(y – 6) – 7(y – 6) = 0

(y – 7)(y – 6) = 0

y = 7, 6

x < y

x2 + 39x + 224 = 0

x2 + 32x + 7x + 224 = 0

x(x + 32) + 7(x + 32) = 0

(x + 7)(x + 32) = 0

x = – 7, -32

y2 + 35y + 196 = 0

y2 + 7y + 28y + 196 = 0

y(y + 7) + 28(y + 7) = 0

(y + 28)(y + 7) = 0

y = -28, -7

Relationship between x and y cannot be established.

Required Percentage = [(135 + 210 + 240 + 280 + 165)/(175 + 320 + 195 + 225 + 265)] * 100

= 87%

Required percentage = [(225 + 285 + 315 + 185 + 330)/(265 + 165 + 330)] * 100

= 176%

A = 175 + 135 + 225 = 535

B = 320 + 210 + 285 = 815

C = 195 + 240 + 315 = 750

D = 225 + 280 + 185 = 690

E = 265 + 165 + 330 = 760

Difference = 320 – 240 = 80 