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Directions (1-5): What value should come in place of the question mark (?) in the following questions?

(62)^{2} – √11,025 = (?)^{2} + 375

Correct

Incorrect

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): What value should come in place of the question mark (?) in the following questions?

4626.4 + 629.06 = (?)% of 6600 + 3473.46

Correct

Incorrect

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

Directions (1-5): What value should come in place of the question mark (?) in the following questions?

40% of [{(20% of 80) – 8}% of 275] = 11% of ?

Correct

Incorrect

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

The ratio of A’s 4 years ago and B’s age 4 years hence is 6 : 11. After 17 years from now B’s age will be 2 more than twice A’s present age. Find the ratio of A’s present age to B’s age 4 years hence.

Correct

Explanation:
(A-4)/(B+4) = 6/11 so 11A – 6B = 68
Also (B+17) = 2A + 2, so 2A – B = 15
Solve, both equations, A = 22, B= 29
So required ratio is 22/(29+1) = 2/3

Incorrect

Explanation:
(A-4)/(B+4) = 6/11 so 11A – 6B = 68
Also (B+17) = 2A + 2, so 2A – B = 15
Solve, both equations, A = 22, B= 29
So required ratio is 22/(29+1) = 2/3

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

The metal to be used for covering a cylinder having external radius 5 cm, height 21 cm and thickness 1 cm is to be cast from a cylinder. What should be the height of the cylinder of radius 3 cm from which this casting can be done?

Correct

Explanation:
Volume of metal to be casted = (22/7)*[(5)^{2} – (4)^{2}]*21
So (22/7)*3^{2}*h = (22/7)*[(5)^{2} – (4)^{2}]*21
Solve, h = 21

Incorrect

Explanation:
Volume of metal to be casted = (22/7)*[(5)^{2} – (4)^{2}]*21
So (22/7)*3^{2}*h = (22/7)*[(5)^{2} – (4)^{2}]*21
Solve, h = 21

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

A, B and C can alone complete a work in 15, 25 and 30 days respectively. A and B started the work and after some days A is replaced by C. Now the work is completed in a further of 6 4/11 days. How much of the total work did B did?

Correct

Explanation:
Let A replaced by C after x days, so A and B worked for x days, and then B and C for 6 4/11 = 70/11 days. So
(1/15 + 1/25)*x + (1/25 + 1/30)*(70/11) = 1
(8/75)*x + (7/15) = 1
Solve, x = 5 days
So B worked for (5 + 70/11) = 125/11 days
In 125/11 days, B did (125/11) * (1/25) = 5/11 of work

Incorrect

Explanation:
Let A replaced by C after x days, so A and B worked for x days, and then B and C for 6 4/11 = 70/11 days. So
(1/15 + 1/25)*x + (1/25 + 1/30)*(70/11) = 1
(8/75)*x + (7/15) = 1
Solve, x = 5 days
So B worked for (5 + 70/11) = 125/11 days
In 125/11 days, B did (125/11) * (1/25) = 5/11 of work

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

An article is marked at Rs 18,000. A trader bought it at successive discounts of 25% and 10% respectively. He spent Rs 1,350 on its transportation to his shop and then sold the article for Rs 15,000. What is trader’s profit% in the whole transaction?

Correct

Explanation:
He bought the article for [(100-25)/100] [(100-10)/100] * 18000 = 12,150
Spent 1350 on repairs, so total CP = 1350 + 12150 = 13,500
SP = 15,000
So profit% = (1500/13500) * 100

Incorrect

Explanation:
He bought the article for [(100-25)/100] [(100-10)/100] * 18000 = 12,150
Spent 1350 on repairs, so total CP = 1350 + 12150 = 13,500
SP = 15,000
So profit% = (1500/13500) * 100

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A person lent out certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and 2 years is 16 : 49. The rate of interest per annum is

Correct

Explanation:
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 16/49
Solve, r = 6 1/4

Incorrect

Explanation:
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 16/49
Solve, r = 6 1/4