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** Directions (1-5): Study the pie chart below which gives the distribution of the number of students in different sections of class 10 in a school. Answer the questions that follow. **

**Which of the following combination of sections contribute to 50% of the total strength of class 10?**

A) A, B, F

B) B, C, F

C) A, B, C

D) C, D, E

E) None of these**C) A, B, C**

Explanation:

A, B, F = 36^{o}+72^{o}+18^{o}= 126^{o}

B, C, F = 72^{o}+72^{o}+18^{o}= 162^{o}

A, B, C = 36^{o}+72^{o}+72^{o}= 180^{o}

C, D, E = 72^{o}+18^{o}+45^{o}= 135^{o}

180^{o}is the half of complete 360^{o}, so sections A, B, and C contribute to 50% of total strength.**If there are 144 students in section A, then what is the number of students in section G?**

A) 396

B) 108

C) 325

D) 366

E) None of these**A) 396**

Explanation:

Let total number of students in class 10 is x

So (36/360) * x = 144

This gives x = 1440

So number of students in section G = (99/360) * 1440 = 396**If the number of students in section B is 50% more than that in section A, then number of students in section A is what percent less than that in section B?**

A) 16 2/3%

B) 25%

C) 33 1/3%

D) 40%

E) 50%**C) 33 1/3%**

Explanation:

There is no need to see the pie chart to answer this question

Required % = 50/(100+50) * 100**Find the number of students in sections A, B and E together.**

A) 644

B) 524

C) 500

D) Cannot be determined

E) None of these**D) Cannot be determined**

Explanation:

Number of students in section A = 36/360 * (total number of students in class 10)

Since we are not given the total number of students in class 10, we cannot determine the number of students in each section.**If the total number of students of class 10 increase by 50% and students in section A increase by 20%, then what will be the angle for section A in the new pie-chart?**

A) 35.8^{o}

B) 25.5^{o}

C) 36.4^{o}

D) 28.8^{o}

E) None of these**D) 28.8**^{o}

Explanation:

Let total students in class 10 is ‘s’. After increase by 50%, new number of students = 150/100 * s = 3s/2

Angle for section A is 36^{o}, so students in section A = 36/360 * s = s/10. They increase by 20%, so new number of students in section A = 120/100 * s/10 = 6s/50

So new angle for A = [(6s/50)/(3s/2)] * 360 = 28.8^{o}**A jar contains 4 green balls, 4 white balls and 4 yellow balls. If 3 balls are drawn at random, find the probability that the balls are not of the same color.**

A) 44/57

B) 52/55

C) 3/55

D) 10/31

E) None of these**B) 52/55**

Explanation:

The probability can be found by subtracting prob. that all balls are same in color from 1.

Total ways of drawing 3 balls out of total 12 balls =^{12}C_{3}= 220

Case 1: all balls green

Ways are^{4}C_{3}= 4

Case 2: all balls white

Ways are^{4}C_{3}= 4

Case 3: all balls yellow

Ways are^{4}C_{3}= 4

So prob. that all balls are of same color = (4+4+4)/220 = 12/220 = 3/55

Prob. that they are not of same color = 1 – 3/55 = 52/55**A, B and C start a business in which A’s contribution is 3/2 times that of B’s and also A’s contribution is thrice that of C’s. Their total investment in the business is Rs 15,000. Find the share of B in the profit of Rs 3000 at the end of a year.**

A) Rs 2,000

B) Rs 1,500

C) Rs 1,050

D) Rs 1,000

E) Rs 500**D) Rs 1,000**

Explanation:

Let A’s contribution = Rs x, then B’s contribution = 2/3 * x and C’s contribution = 1/3 * x

Then profits in the ratio –

A : B : C = x : 2x/3 : x/3 = 3 : 2 : 1

So share of B = [2/(3+2+1)] * 3000 = 1000**A, B and C together can complete a piece of work in 4 days. All started the work and after working for 2 days, B left. A and C completed the remaining work in 3 days. Find the number of days in which B can alone complete the work.**

A) 10 days

B) 12 days

C) 15 days

D) 18 days

E) 21 days**B) 12 days**

Explanation:

(A+B+C)’s 1 day’s work = 1/4

They work for 2 days. So work done after 2 days = 1/4 * 2 = 1/2

Now B left, so remaining work (1 – 1/2) = 1/2 is to be done by A and C in 3 days.

1/2 work done in 3 days so complete (1) work done by A and C in 6 days.

Now (A+B+C)’s 1 day’s work = 1/4 and (A+C)’s 1 day’s work = 1/6

So B’s 1 day’s work = 1/4 – 1/6 = (3-2)/12 = 1/12**The amount that A and B earns is in the ratio 4 : 5. If income of A is increased by 25% and that of B is decreased by 20%, the ratio of their earnings becomes 5 : 4. Find the total of their earnings.**

A) Rs 75,000

B) Rs 55,000

C) Rs 50,000

D) Data inadequate

E) None of these**D) Data inadequate**

Explanation:

Let A’s income = 4x, B’s income = 5x

After change – A’s income = 125/100 * 4x = 5x and B’s income = 80/100 * 5x = 4x

So new ratio = 5x : 4x = 5 : 4 which is also given, but it does not give value for x, so data inadequate**There is a loss of 25% on an article when it is sold at 2/5 of its Marked Price. How much percent more is Marked Price than Cost Price?**

A) 80.5%

B) 82%

C) 87.5%

D) 89.25%

E) None of these**C) 87.5%**

Explanation:

Let MP = x

Then SP = 2/5 * x = 2x/5

Loss is 25%, so CP = [100/(100-25%)] * 2x/5 = 8x/15

So required % = [(x – 8x/15)/(8x/15)] * 100

= [(7x/15) / (8x/15)] * 100 = (7/8) * 100 = 87.5%

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