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Explanation:
Double difference series as
10-2=8, 25-10=15, 54-25=29, 104-54=50
Again difference – 15-8=7, 29-15=14, 50-29=21 so next difference is 7*4 = 28
So 50+28+104

Incorrect

Explanation:
Double difference series as
10-2=8, 25-10=15, 54-25=29, 104-54=50
Again difference – 15-8=7, 29-15=14, 50-29=21 so next difference is 7*4 = 28
So 50+28+104

A box contains 2 blue, 3 green and 5 red balls. If three balls are drawn at random, what is the probability that all balls are different in color?

Correct

Explanation:
Total balls = 10
So prob. = ^{2}C_{1}×^{3}C_{1}×^{5}C_{1} / ^{10}C_{3}

Incorrect

Explanation:
Total balls = 10
So prob. = ^{2}C_{1}×^{3}C_{1}×^{5}C_{1} / ^{10}C_{3}

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Rice is bought at the rate of Rs 20 per kg. Out of the total quantity of 150 kg, 60 kg of rice is sold at 10% profit. At what rate per kg the remaining need rice need to be sold so that there is a profit of 25% on the total price?

Correct

Explanation:
Let remaining (150-60) = 90 kg at x%. So by allegation method,
(60 kg)…………………(90 kg)
10% …………………….x%
…………….25%
(x-25)…………..………15
So (x-25)/15 = 60 kg/90 kg
Solve, x = 35%
90 kg costs = 90*10 = Rs 900
So at 35% profit SP of 90 kg is (135/100) * 900 = Rs 1215
So SP of 1 kg = 1215/90 = 13.5
OR
(60/150)*10 + (90/150)*x = 25
Solve, x = 35%, since its positive so profit
90 kg costs = 90*10 = Rs 900
So at 35% profit SP of 90 kg is (135/100) * 900 = Rs 1215
So SP of 1 kg = 1215/90 = 13.5

Incorrect

Explanation:
Let remaining (150-60) = 90 kg at x%. So by allegation method,
(60 kg)…………………(90 kg)
10% …………………….x%
…………….25%
(x-25)…………..………15
So (x-25)/15 = 60 kg/90 kg
Solve, x = 35%
90 kg costs = 90*10 = Rs 900
So at 35% profit SP of 90 kg is (135/100) * 900 = Rs 1215
So SP of 1 kg = 1215/90 = 13.5
OR
(60/150)*10 + (90/150)*x = 25
Solve, x = 35%, since its positive so profit
90 kg costs = 90*10 = Rs 900
So at 35% profit SP of 90 kg is (135/100) * 900 = Rs 1215
So SP of 1 kg = 1215/90 = 13.5

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Amit is two and a half times more efficient than Baban. If Amit alone can complete a work in 20 days, then what portion of total work can both of them together complete in 7 days?

Correct

Explanation:
Let efficiency of B is x, then of A is 1 + 2 (1/2) = 7x/2
Ratio of efficiencies of A : B = 7x/2 : x= 7 : 2
So ratio of number of days of A : B = 2 : 7
A completes work in 10 days, so 2x = 20, x = 10
So B alone completes work in 7x days = 70 days
(A+B)’s 1 day’s work = 1/20 + 1/70 = 9/140
So in 7 days, pat of work completed is 7 * (9/140)

Incorrect

Explanation:
Let efficiency of B is x, then of A is 1 + 2 (1/2) = 7x/2
Ratio of efficiencies of A : B = 7x/2 : x= 7 : 2
So ratio of number of days of A : B = 2 : 7
A completes work in 10 days, so 2x = 20, x = 10
So B alone completes work in 7x days = 70 days
(A+B)’s 1 day’s work = 1/20 + 1/70 = 9/140
So in 7 days, pat of work completed is 7 * (9/140)

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

In rowing to a certain distance it takes three fourth time in moving downstream than upstream. What is the ratio of the rate of boat in still water to the rate of current?

Correct

Explanation:
Shortcut:
Let downstream time = 3t/4 and then upstream time = t
B = [t_{u} + t_{d}] / [t_{u} – t_{d}] * R
B = [t+ 3t/4] / [t- 3t/4] * R
B/R = 7/1

Incorrect

Explanation:
Shortcut:
Let downstream time = 3t/4 and then upstream time = t
B = [t_{u} + t_{d}] / [t_{u} – t_{d}] * R
B = [t+ 3t/4] / [t- 3t/4] * R
B/R = 7/1

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

The present population of a town is 1,38,240. If from last three years the population first increased at the rate of 20% for first years and then decreased by 20% in the third year, what was the population of town three years ago?

Correct

Explanation:
x * 120/100 * 120/100 * 80/100 = 138240
Solve, find x

Incorrect

Explanation:
x * 120/100 * 120/100 * 80/100 = 138240
Solve, find x