SBI Clerk Pre Quantitative Aptitude (Day-42)

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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Quadratic Equation

Directions (1 – 5): 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,

a) If x> y

b) If x ≥ y

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

d) If x< y

e) If x ≤ y

1)

I) x2 + 12x + 20 = 0

II) y2 + 13y + 22 = 0

 

2)

I) 5x2 + 15x + 10 = 0

II) y2 – y – 2 = 0

 

3)

I) x2 – 27x + 182 = 0

II) y2 – 23y + 132 = 0

 

4)

I) 2x2 – 18x + 36 = 0

II) y2 – 12y + 32 = 0

 

5)

I) x2 + 23x + 120 = 0

II) y2 + 13y + 42 = 0

Application Sums

6) If the difference between the simple interest and Compound interest on a certain sum for 2 years at 8 % is Rs. 160. Find the sum?

a) Rs. 25000

b) Rs. 28000

c) Rs. 32000

d) Rs. 36000

e) None of these

7) 6 years ago, the ratio of ages of A and B is 2: 3. After 6 years, the ratio of ages of A and B is 3: 4. Find the sum of the present ages of A and B?

a) 64 years

b) 72 years

c) 68 years

d) 66 years

e) None of these

8) Pipe A opened to fill the tank and after 4 hours is closed and then Pipe B opened and fills the remaining tank in 10 hours. If Pipe B alone fill 25% of the tank in 5 hours, in how many hours pipe A and B together fill the tank completely?

a) 8(5/7) hours

b) 4(5/7) hours

c) 5(5/7) hours

d) 6(5/7) hours

e) None of these

9) A bag contains x+4 pink, 6 green and 8 brown colour balls; if two balls are taken random and the probability of getting both are green colour balls is 5/92, then find the difference between the no. of pink colour balls and the no. of brown colour balls

a) 1

b) 2

c) 3

d) 4

e) 0

10) Average ages of 5 persons A, B, C, D and E is 38 years. Ratio of the ages of A to C is 3:4 and after 5 years the ratio of the ages of B to D becomes 2: 3. If the difference between the ages of A and B is 2 years, then what is the present age of E?

a) 10 years

b) 39 years

c) 40 years

d) 48 years

e) Cannot be determine

Answers :

Directions (1-5) :

1) Answer: C

I) x2 + 12x + 20 = 0

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

x= -2, -10

II) y2 + 13y + 22 = 0

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

y = -2, -11

Relationship between x and y cannot be established.

2) Answer: E

I) 5x2 + 15x + 10 = 0

5x2 + 10x + 5x + 10 = 0

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

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

x = -1, -2

II) y2 – y – 2 = 0

(y + 1) (y – 2) = 0

y = -1, 2

Hence x ≤ y

3) Answer: A

I) x2 – 27x + 182 = 0

(x – 14) (x – 13) = 0

x = 14, 13

II) y2 – 23y + 132 = 0

(y – 11) (y – 12) = 0

y = 11, 12

Hence x > y

4) Answer: C

I) 2x2 – 18x + 36 = 0

2x2 – 12x – 6x + 36 = 0

2x(x – 6) – 6(x – 6) = 0

(2x – 6)(x – 6) = 0

x = 3, 6

II) y2 – 12y + 32 = 0

(y – 4) (y – 8) = 0

y = 4, 8

Hence Relationship between x and y cannot be established.

5) Answer: D

I) x2 + 23x + 120 = 0

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

x = -15, -8

II) y2 + 13y + 42 = 0

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

y = -6, -7

Hence x < y

Directions (6-10) :

6) Answer: A

Diff = Sum*(r/100)2

160 = Sum*(8/100)2

160*100*100/64 = Sum

Sum = Rs. 25000

7) Answer: B

6 years ago, the ratio of ages of A and B = 2: 3 (2x, 3x)

Present ages of A and B = 2x + 6, 3x + 6

After 6 years, the ratio of ages of A and B = 3: 4

According to the question,

(2x + 12)/(3x + 12) = (3/4)

8x + 48 = 9x + 36

x = 12

Sum of the present ages of A and B = 2x + 3x + 12 = 72 years

8) Answer: C

B = 5 * 4/1 = 20 hours

4/x + 10/20 = 1

4/x = ½

1/x = 1/8

A + B = 1/20 + 1/8

= 7/40

Time = 40/7 hours = 5(5/7) hours

9) Answer: B

Given,

6c2/(x+18)c2=5/92

X2+35x-246=0

Simplify the above equation we get x=6

Required difference = 10-8=2 balls

10) Answer: E

A + B + C + D + E = 190 years

A/C = 3/4

B + 5/D + 5 = 2/3

2D + 10 = 3B + 15

2D – 3B = 5

We cannot find the answer.

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