# SBI Clerk Prelims 2021 Quantitative Aptitude Questions (Day-70)

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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Simplification

Directions (01-05): What value should come in the place of (?) in the following questions?

1) (0.8)43 ÷ (0.64)18 * (0.512) = (0.8)?

A.8

B.10

C.12

D.14

E.17

2) 222 – (216)(1/3) + (8000)(1/3) = ?

A.492

B.496

C.498

D.500

E.518

3) 262 – 36 = (?)2 + 3 * √1369

A.21

B.22

C.23

D.25

E.29

4) 30 * 28 + 20 * 60 – ? = 23 * 38

A.1156

B.1146

C.1166

D.1136

E.1126

5) 12.5% of 240 + 37.5% of 320 + 62.5% of 400 = ?

A.400

B.300

C.200

D.480

E.360

Directions (06-10): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,

6) I) x2 + 12x + 20 = 0

II) y2+ 13y + 22 = 0

a) x> y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x< y

e) x ≤ y

7) I) 5x2 + 15x + 10 = 0

II) y2– y – 2 = 0

a) x> y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

8) I)x2 – 27x + 182 = 0

II) y2– 23y + 132 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

9) I) 2x2 – 18x + 36 = 0

II) y2–12y + 32 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

10) I)x2 + 23x + 120 = 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

(0.8)43 ÷ (0.64)18 * (0.512) = (0.8)?

(0.8)43/(0.8)36 * (0.8)3 = (0.8)?

(0.8)43 – 36 + 3 = (0.8)?

? = 10

222 – (216)(1/3) + (8000)(1/3) = ?

484 – 6 + 20 = ?

498 = ?

262 – 36 = (?)2 + 3 * √1369

676 – 36 = ?2 + 3 * 37

?2 = 529

? = 23

30 * 28 + 20 * 60 – ? = 23 * 38

840 + 1200 – ? = 874

? = 1166

12.5% of 240 + 37.5% of 320 + 62.5% of 400 = ?

1/8 * 240 + 3/8 * 320 + 5/8 * 400 = ?

30 + 120 + 250 = ?

400 = ?

x2 + 12x + 20 = 0

x2 + 10x + 2x + 20 = 0

x(x + 10) + 2(x + 10) = 0

(x + 2)(x + 10) = 0

x= -2, -10

y2 + 13y + 22 = 0

y2 + 11y + 2y + 22 = 0

y(y + 11) + 2(y + 11) = 0

(y + 2)(y + 11) = 0

y = -2, -11

Relationship between x and y cannot be established.

5x2 + 15x + 10 = 0

5x2 + 10x + 5x + 10 = 0

5x(x + 2) + 5(x + 2) = 0

(5x + 5)(x + 2) = 0

x = -1, -2

y2 – y – 2 = 0

y2 – 2y + y – 2 = 0

y(y – 2) + 1(y – 2) = 0

(y + 1)(y – 2) = 0

y = -1, 2

Hence x ≤ y

x2 – 27x + 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) – 14(x – 13) = 0

(x – 14)(x – 13) = 0

x = 14, 13

y2 – 23y + 132 = 0

y2 – 12y – 11y + 132 = 0

y(y – 12) – 11(y – 12) = 0

(y – 11)(y – 12) = 0

y = 11, 12

Hence x > y

2x2 – 18x + 36 = 0

2x2 – 12x – 6x + 36 = 0

2x(x – 6) – 6(x – 6) = 0

(2x – 6)(x – 6) = 0

x = 3, 6

y2 – 12y + 32 = 0

y2 – 8y – 4y + 32 = 0

y(y – 8) – 4(y – 8) = 0

(y – 4)(y – 8) = 0

y = 4, 8

Hence Relationship between x and y cannot be established.

x2 + 23x + 120 = 0

x2 + 8x + 15x + 120 = 0

x(x + 8) + 15(x + 8) = 0

(x + 15)(x + 8) = 0

x = -15, -8

y2 + 13y + 42 = 0

y2 + 7y + 6y + 42 = 0

y(y + 7) + 6(y + 7) = 0

(y + 6)(y + 7) = 0

y = -6, -7

Hence x < y

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