RRB Clerk Prelims Quantitative Aptitude Questions 2021 (Day-01)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for RRB Clerk Prelims 2021 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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Trains

1) Train A crosses 300 m long train B running in opposite direction in 36 seconds and train A also crosses a man standing in a platform in 12 seconds. What is the speed of train B?

A.60 kmph

B.70 kmph

C.90 kmph

D.Cannot be determined

E.None of these


SI AND CI

2) If the compound interest accrued of an amount of Rs 4000 in two years is Rs 1290, what is the annual rate of interest?

A.11%

B.9%

C.15%

D.18%

E.6%


Approximation

Direction (3-5):  What approximate value will come in place of question mark (?) in the following questions (You are not expected to calculate the exact value.)

3) 460.12/9.01÷. 194.86/12.93 x 244.74 / 6.99 = ?

A.170

B.140

C.160

D.119

E.90


4) (?)2 + (7.99)3 = (45.03)2 – (6.02)3

A.38

B.40

C.36

D.43

E.33


5) √960 + √1025 – √485 + √1090 – √840 = ?

A.45

B.48

C.53

D.57

E.62


Quadratic equation

Directions (6-10): In each of the questions, two equations I and II are given. You have to solve both the equations and give answer

A)  a > b

B)  a < b

C)  a ≥ b

D)  a ≤ b

E)  a = b or relationship between ‘a’ and ‘b’ cannot be established.


6) I) a+ a – 2 = 0

II) 2b– 15b +25 = 0


7) I) 8a– 22a – 21 = 0

II) b+ 14b – 51 = 0


8) I) a+ 8a + 16 = 0

II) 3b2– 2√(6b) + 2 = 0


9) I) a+ 8a + 15 = 0

II) b2+ 12b + 35 = 0


10) I) a– 9a + 20 = 0

II) b2– 12b + 35 = 0


Answers :

1) Answer: D

Distance = Speed * time

Length of train A = x

Speed of train A = y

Speed of train B = z

x + 300 = (y + z) * 5/18 * 36

x + 300 = 10y + 10z

x = y * 5/18 * 12

3x = 10y


2) Answer: C

Amount = P+I

= 4000+1290 = 5290

A = P(1+R/100)t

=> 5290 = 4000(1+R/100)2

(1+R/100)2 = 5290/4000

= 1.3225

= (1+R/100) = 1.15

=> R = 15%


3) Answer: D

?≈ (460/9) ÷ (195/13) * (245/7)

? = (51 ÷ 15)x 35

?= 119


4) Answer: C

Solution: (?)2 + 512 = 2025 – 216

=> (?)2 = 1297

362 = 1296

=> ?≈36


5) Answer: A

? ≈ 31 + 32 – 22 + 33 – 29

?= 45


6) Answer: B

a2 + a – 2 = 0

Or, a2 + 2a – a – 2 = 0

Or, a(a + 2) – 1(a +2 ) =0

Or, (a – 1) (a + 2) = 0

:. a =1, -2

2b2– 15b +25 = 0

Or, 2b– 10b – 5b +25 = 0

Or, 2b (b – 5) – 5(b – 5) = 0

Or, (b – 5) (2b – 5) = 0

:. b = 5, 5/2

Hence a < b


7) Answer: E

8a2 – 22a – 21 = 0

Or, 8a2 – 28a +6a – 21 = 0

Or, 4a (2a – 7) + 3(2a – 7) = 0

Or, (4a + 3) (2a – 7) = 0

:. a = – 3/4 , 7/2

b2+ 14b – 51 = 0

Or, b+ 17b – 3b – 51 = 0

Or, b (b + 17) – 3(b +17) = 0

Or, (b – 3) (b + 17) = 0

Or, b= -17, 3

Hence relation cannot be established between a and b.


8) Answer: B

a2 + 8a + 16 = 0

Or, a2 + 4a +4a + 16 = 0

Or, (a + 4)2 = 0

Or, a + 4 = 0

:. a= -4

3b2– 2 √(6b) + 2 = 0

Or, (√3b)2 – 2√(6b) + (√2)2 = 0

Or, (√3b – √2)= 0

Or, √3b – √2 = 0

:. b= √2/3

Hence b > a


9) Answer: C

a2 + 8a +15 = 0

Or, a2 + 5a + 3a +15 = 0

Or, a(a+5) + 3 (a + 5) = 0

Or, (a +3) (a +5) = 0

:. a = -3, -5

b2+ 12b + 35 = 0

Or, b2 + 5b + 7b +35 = 0

Or, b(b + 5) + 7( b+5) = 0

Or, (b + 7) (b + 5) = 0

b = -5 , -7

Hence a ≥ b


10) Answer: D

a2 – 9a +20 = 0

Or, a2 – 5a – 4a + 20 = 0

Or, a(a – 5) – 4(a – 5) = 0

Or, (a – 4) (a -5) = 0

:. a=4,5

b2– 12b + 35 = 0

Or, b2 – 5b – 7b +35 = 0

Or, b(b – 5) – 7(b – 5) = 0

Or, (b – 7) (b – 5) = 0

b = 5 , 7

Hence a ≤ b

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