SBI Clerk Pre Quantitative Aptitude (Day-32)

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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Application Sums

1) The difference between simple interest and compound interest accrued on an amount of Rs. 1800 in 2 years was Rs. 30.42. What is the rate of interest per annum?

a) 11 %

b) 13 %

c) 12 %

d) 14 %

e) None of these

2) If Janani was one-fifth as old as Raja 7 years back and Janani is 19 years old now. How old is Raja after 5 years?

a) 67 years

b) 72 years

c) 52 years

d) 62 years

e) None of these

3) A boy rows 500 m in 450 seconds against the stream and returns in 6 and half minutes. What is his rowing speed in still water?

a) 2.4 km/hr

b) 3.1 km/hr

c) 4.3 km/hr

d) 5.2 km/hr

e) None of these

4) A takes 9 days in completing 25% of work alone. Time taken by A in completing 1/3rd of the work is equal to the time taken by B in completing half of the work. How many days will be taken in completing the work if both A and B started working together?

a) 72/5 days

b) 78/5 days

c) 63/5 days

d) 12 days

e) None of these

5) One ball is picked up randomly from a bag containing 11 red, 8 black and 13 blue balls. What is the probability that it is neither red nor blue?

a) 2/5

b) 1/4

c) 3/5

d) 1/3

e) None of these

Quadratic Equation

Directions (6 – 10): In each of the following questions, two equations are given. You have to solve both the equations to find the relation between x and y and give answer as,

a) If x < y

b) If x > y

c) If x ≤ y

d) If x ≥ y

e) If relationship between x and y cannot be determined

6) I) x2 = 196

II) y2 + 2y – 48 = 0

 

7) I) x³ = 6859

II) y³ = 3375

 

8) I) 2x2 + 19x + 42 = 0

II) 4y2 + 43y + 30 = 0

 

9) I) 72 – 30x = – 2x2

II) y2 – 40/6 = 7/3


10) I) 2x2 – x – 1 = 0

II) 2y2 – 4y + 2 = 0

Answers :

1) Answer: B

Difference = P * r2/1002

30.42 = 1800 * r2/1002

3042/18 = r2

r = 13 %

2) Answer: B

Janani’s age today = 19 years

7 years back, Janani’s age = 12 years

Raja’s age = 12/1 * 5 = 60

Current age of Raja = 60 + 7 = 67

After 5 years, Raja’s age = 67 + 5 = 72 years

3) Answer: C

Speed in upstream = 500/450 m/sec

Speed in downstream = 500/ (6 * 60 + 30) m/sec = 500/390 m/sec

Rate in still water = ½ * (Upstream speed + Downstream speed)

= ½ * (500/450 + 500/390)

= ½ * (50/45 + 50/39) * 18/5 km/hr

= 4.3 km/hr

4) Answer: A

A takes 9 days in completing 25% of work alone

So, for 100% work A alone will take= 9 * 4 = 36 days

Time taken by A in completing 1/3 work = 36 * 1/3= 12 days

12 days taken by B in completing half of work

So, B alone can complete the work= 2 * 12 = 24 days

Required time= 1/36 + 1/24 = 72/5 days

5) Answer: B

Total number of balls = 11 + 8 + 13 = 32

Let, E be the event where the ball can be selected which is neither red nor blue

Therefore, P(E) = 8/32 = 1/4

Directions (6-10) :

6) Answer: E

I) x² = 196

=> x = +14, – 14

II) y2 + 2y – 48 = 0

=> (y – 6) (y + 8) = 0

=> y = 6, -8

Hence, relationship between x and y cannot be determined

7) Answer: B

I) x³ = 6859

=> x = 19

II) y³ = 3375

=> y = 15

Hence, x > y

8) Answer: E

I) 2x2 + 19x + 42 = 0

=> 2x + 12x + 7x + 42 = 0

=> 2x (x + 6) + 7 (x + 6) = 0

= > (2x + 7) (x + 6) = 0

=> x = -7/2, -6

II) 4y2 + 43y + 30 = 0

=> 4y2 + 40y + 3y +30 = 0

=> 4y(y + 10) + 3(y + 10) = 0

=> (4y + 3) (y + 10) = 0

=> y = -3/4, -10

Hence, relationship between x and y cannot be determined

9) Answer: D

I) 72 – 30x = -2x²

=> 2x² – 30x + 72 = 0

=> x² – 15x + 36= 0

= > x = 3, 12

II) y2 – 40/6 = 7/3

=> y2 – 20/3 = 7/3

=> y² = 27/3

=> y = 3, -3

Hence, x ≥ y

10) Answer: C

I) 2x² – x – 1 = 0

=> 2x² – 2x + x – 1 = 0

=> 2x (x – 1) + 1 (x – 1) = 0

= > ( 2x + 1) ( x – 1)= 0

=> x = -1/2, 1

II) 2y2 – 4y + 2 = 0

=> 2y2 – 2y – 2y + 2 = 0

=> 2y (y – 1) – 2 (y – 1) = 0

=> (2y – 2) (y – 1) = 0

=> y = 1, 1

Hence, x ≤ y

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