# SBI Clerk Pre Quantitative Aptitude (Day-42)

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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

Directions (1-5) :

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.

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

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

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.

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) :

Diff = Sum*(r/100)2

160 = Sum*(8/100)2

160*100*100/64 = Sum

Sum = Rs. 25000

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

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

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

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

A/C = 3/4

B + 5/D + 5 = 2/3

2D + 10 = 3B + 15

2D – 3B = 5 