# SBI Clerk Prelims 2021 Quantitative Aptitude Questions (Day-60)

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

Direction (1-7): What value should come in place of the questions mark (?) in the following questions.

1) 1189 – 1079 – 9589 + 3986 + 8924 – 4627 + 6683 – 3195 = ?

A.2122

B.1682

C.2292

D.3452

E.2402

2) 572 ÷ 26 * 12-200 = (2)?

A.5

B.7

C.6

D.8

E.None of these

3) 36% of 245 – 48% of 175 = 10 – ?

A.4.2

B.6.8

C.4.9

D.5.6

E.5.8

4) 3/8 of 4/9 of 1092 + 4/13 of 5/7 of 2002 = ?

A.258

B.702

C.564

D.622

E.440

5) 162 + 342 + 472 + 512= ?

A.5986

B.6222

C.6146

D.6066

E.6342

6) 16% of 450 x ?% of 880 = 3168

A.6

B.2

C.11

D.8

E.5

7) 15.3 *1.7 + 22.6 * 1.2 – 17.8 * 1.5 = ?

A.27.81

B.25.85

C.27.01

D.25.59

E.26.43

Mensuration

8) The radius of the circle is equal to two-fifth of the side of the square. The area of the circle is 616 sq cm. Find the perimeter of a square?

A.152 cm

B.166 cm

C.140 cm

D.178 cm

E.None of these

Percentage

9) In an exam, Rakesh scored 52% marks and failed by 23 marks. In the same exam, Radhika secured 64% marks and scored 34 more than the passing marks. If mohan scored 82% of maximum marks in the exam, then find the marks scored by mohan

A.357.5

B.369

C.373

D.370.5

E.389.5

Mensuration

10) Perimeter of a rectangle is x and circumference of a circle is 8 cm more than the perimeter of the rectangle. Ratio of radius of the circle and length of the rectangle is 1: 2 and ratio of length and breadth of rectangle is 7: 3. Find the length of the rectangle?

A.14 cm

B.21 cm

C.28 cm

D.35 cm

E.7 cm

? = (1189+3986+8924+6683) – (1079+9589+4627+3195)

= 20782 – 18490 = 2292

(2)? = 22 * 12 – 200

= 264 – 200

= 64

= 26

=> ? = 6

10 – ? = 0.36*245 – 0.48*175

= 88.2 – 84 = 4.2

? = 10-4.2

?= 5.8

= 3/8 * 4/9 * 1092 + 4/13 * 5/7 * 2002

= 182 + 440 = 622

?= 256 + 1156 + 2209 + 2601

?= 6222

0.16 * 450 * ? * 880/100 = 3168

=> 633.6 * ? = 3168

=> ? = 5

? = 26.01 + 27.12 – 26.7 = 26.43

Area of circle = πr 2 = 616

(22/7)*r2 = 616

r2 = 616*(7/22) = 196

Side = Radius*(5/2) = 14*(5/2) = 35 cm

Perimeter of the square = 4a

= > 35*4

= > 140 cm

Let the maximum marks be M

Passing marks = 0.52M + 23 = 0.64M – 34

=>0.12M=57

=> M = 475

Required mark = 475*82/100

= 389.5

The ratio of radius of circle and length of rectangle = 1 : 2

The ratio of length and breadth of the rectangle = 7 : 3

The ratio of radius of circle and length and breadth of rectangle = 7 : 14 : 6 (7y, 14y, 6y)

2*(l + b) = x

2*(20y) = x

40y = x — > (1)

2*(22/7)*7y = x + 8

44y = x + 8 — > (2)

By substituting eqn (1) in eqn (2),

44y = 40y + 8

4y = 8

y = 2

The length of rectangle = 14y = 28 cm

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