SBI PO Quantitative Aptitude Quiz: 27th April

Dear Aspirants,

Quantitative Aptitude Quiz For SBI PO/Clerk Prelims

Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

Q1. A mixture contains milk and water in the ratio 5 : 1. On adding 5 liters of water, the ratio of milk and water becomes 5 : 2. The quantity of milk in the original mixture is?

16 litres

25 litres

22.75 liters

32.5 liters

28 litres

Solution:

Q2. A vessel contains 1 litre mixture of kerosine and disel and the other vessel contains 2 litre mixture of the same. The ratio of kerosine and disel in first and second vessels is 5 : 4 and 7 : 2 respectively. The mixture of second vessel is poured into first vessel. Find the ratio of kerosine to disel in the first vessel.

8 : 11

9 : 8

19 : 8

11 : 19

11 : 8

Solution:

Q3. A chemist has 10 litre of a solution that is 10% sulphuric acid by volume. He wants to dilute the solution to 4% strength by adding water how many litre of water must be added?

15 litre

20 litre

18 litre

25 litre

17 litre

Solution:

Q4. Rishabh bought 25 kg of wheat at the rate of Rs. 6 per kg and 35 kg of wheat at the rate of Rs. 7 per kg. He mixed both type of wheat and sold the mixture at the rate of Rs. 6.75 per kg. What was his gain or loss in the transaction?

Rs. 16, gain

Rs. 16, Loss

Rs. 20, gain

Rs. 10, gain

Rs. 10, loss

Solution:

gain in the transaction
= 6.75 × 60 – (25 × 6 + 35 × 7)
= Rs. 10, gain

Q5. An alloy contains silver and lead in the ratio 5 : 8 and another alloy contains silver and lead in the ratio 5 : 3. If equal amount of both the alloys are melted together, then the ratio of silver to lead in the resulting alloy is

25 : 24

3 : 8

103 : 105

105 : 103

8 : 3

Solution:

Directions (6-10): In each of the following questions two equations are given. You have to solve the equations and Give answer –

Q6. I. 3x² + 16x + 21 = 0
II. 6y² + 17y + 12 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 3x² + 16x + 21 = 0
⇒ 3x² + 9x + 7x + 21 = 0
⇒ (x + 3) (3x + 7) = 0
⇒ x = –3, –7/3
II. 6y² + 17y + 12 = 0
⇒ 6y² + 9y + 8y + 12 = 0
⇒ 3y (2y + 3) + 4 (2y + 3) = 0
⇒ y = – 3/2, –4/3
y > x

Q7. I. 16x² + 20x + 6 = 0
II. 10y² + 38y + 24 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 16x² + 20x + 6 = 0
⇒ 8x² + 10x + 3 = 0
⇒ 8x² + 4x + 6x + 3 = 0
⇒ (2x + 1) (4x + 3) = 0
⇒ x = –1/2, –3/4
II. 10y² + 38y + 24 = 0
⇒ 5y² + 19y + 12 = 0
⇒ 5y² + 15y + 4y + 12 = 0
⇒ (y + 3) (5y + 4) = 0
y = –3, –4/5
x > y

Q8. I. 8x² + 6x = 5
II. 12y² – 22y + 8 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 8x² + 6x – 5 = 0
⇒ 8x² + 10x – 4x – 5 = 0
⇒ (4x + 5) (2x – 1) = 0
⇒ x = ½, –5/4
II. 12y² – 22y + 8 = 0
⇒ 6y² – 11y + 4 = 0
⇒ 6y² – 3y – 8y + 4 = 0
⇒ (2y – 1) (3y – 4) = 0
⇒ y = 1/2, 4/3
y ≥ x

Q9. I. 17x² + 48x = 9
II. 13y² = 32y – 12

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 17x² + 48x – 9 = 0
⇒ 17x² + 51x – 3x – 9 = 0
⇒ (x + 3) (17x – 3) = 0
⇒ x = 3/17, – 3
II. 13y² – 32y + 12 = 0
⇒ 13y² – 26y – 6y + 12 = 0
⇒ (y – 2) (13y – 6) = 0
⇒ y = 2, 6/13
y > x

Q10. I. 8x² + 26x + 15 = 0
II. 4y² + 24y + 35 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 8x² + 26x + 15 = 0
⇒ 8x² + 20x + 6x + 15 = 0
⇒ 4x (2x + 5) + 3(2x + 5) = 0
⇒ (2x + 5) (4x + 3) = 0
⇒ x = – 5/2, –3/4
II. 4y² + 24y + 35 = 0
⇒ 4y² + 10y + 14y + 35 = 0
⇒ 2y (2y + 5) + 7 (2y + 5) = 0
⇒ (2y + 5) (2y + 7) = 0
⇒ y = –5/2, –7/2
x ≥ y

Directions (11 – 15): Read the following graph carefully and answer the questions given below

Percentage distribution of total number of students who completed their graduation from different Universities A, B, C, D, E and F.