**Hello Aspirants,**

Welcome to Online Quant Section in AffairsCloud.com. We are starting **IBPS Clerk course 2015 **and we are creating sample questions in **Quantitative Aptitude** section, type of which will be asked in IBPS Clerk Main Exam.

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**There are 3 red balls, 4 blue balls, 5 white balls. Three balls are drawn at random. What is the probability that there are at least 2 red balls?**

A) 21/58

B) 7/55

C) 9/52

D) 12/55

E) None of these

**B. 7/55**

Explanation:

Case 1 : 2 red balls:

Prob =^{3}C_{2}×^{9}C_{1}/^{12}C_{3}

Case 2 : 3 red balls:

Prob =^{3}C_{3}/^{12}C_{3}

Add the two cases.

**A man starts from point P at 6:00 AM and reaches point Q at 9:00 AM. Another man starts from point Q at 7:00 AM and reaches point P at 11:00 AM. At what time both men will meet?**

A) 8:00 AM

B) 8:08 AM

C) 8:34 AM

D) 9:18 AM

E) 7:42 AM

**B) 8:08 AM**

Explanation:

Let the distance from A to B is ‘d’ km.

First train reaches destination in 3 hrs, so speed of first train = d/3 km/hr.

Speed of second train = d/4 km/hr

Speed of first train = d/3 km/hr, so after 1 hr, i.e. at 7AM, the first train has covered d/3 km.

The distance left = (d – d/3) km = 2d/3 km. relative speed = d/3 + d/4.

So time they will meet at = 7:00 AM + (2d/3) / (d/3 + d/4)

= 7:00 AM + 1 (1/7)*60 = 8:08 AM

**80 litres of a mixture contains 20% water and the rest milk. What amount of water must be added so that the resulting mixture contains only 50% milk?**

A) 40 litres

B) 50 litres

C) 52 litres

D) 48 litres

E) None of these

**D) 48 litres**

Explanation:

Mixture contains 20% water.

The water which will be added will be 100% water

The new mixture contains 50% milk, this gives 50% water

So by allegation rule

Mixture Water

20 100

. 50

50 30

This gives 50 : 30 = 5 : 3

Let x litres of water to be added to 80 litres mixture.

So 80/x = 5/3

x = 48

**3 liquids A, B, and C are present in the ratio 3 : 4 : 5. After 12 litres of the mixture taken out and 8 litres of A is poured in, the ratio of A to C becomes 7 : 5. Find the original quantity of the mixture?**

A) 48 litres

B) 36 litres

C) 39 litres

D) 60 litres

E) 32 litres

**B) 36 litres**

Explanation:

Let A = 3x, B = 4x, C = 5x

When 12 litres taken out, left out quantities are

A = 3x – 3/(3+4+5) * 12 = 3x – 3

B = 4x – 4/(3+4+5) * 12 = 4x – 4

C = 5x – 5/(3+4+5) * 12 = 5x – 5

Now after 8 litres of A poured in, A becomes (3x – 3) + 8 = 3x + 5

Now given, A : C = 7 : 5. So

3x+5 / 5x-5 = 7/5

Solve, x = 3

Total initial quantity = 3x + 4x + 5x = 12x = 12*3 = 36

**2/5th of work is completed in 10 days by 20 men working 6 hours each day. In how many days the remaining work will be completed by 10 women working 8 hours each day such that 4 women do as much work as is done by 2 men?**

A) 22

B) 42

C) 38

D) 45

E) 50

**D) 45**

Explanation:

4 w = 2m, so 1w = 1/2 m

And then 10 w = 5 m, so we have to find the number of days for 5 men to complete remaining work (1 – 2/5) = 3/5th.

M1*D1*H1*W2 = M2*D2*H2*W1

20*10*6*(3/5) = 5*D2*8*(2/5)

Solve, D2 = 45 days

**2x**^{2}+ 13x + 15 = 0, 2y^{2}– 5y – 12 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined

**D) x ≤ y**

Explanation:

2x^{2}+ 13x + 15 = 0

2x^{2}+ 10x + 3x + 15 = 0

Gives x = -5, -3/2

2y^{2}– 5y – 12 = 0

2y^{2}– 8y + 3y – 12 = 0

Gives y = -3/2, 4

Put on number line

-5 -3/2 4**2x**^{2}+ 5x + 2 = 0, 3y^{2}+ 5y – 2 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined

**E) x = y or relationship cannot be determined**

Explanation:

2x^{2}+ 5x + 2 = 0

2x^{2}+ 4x + x + 2 = 0

Gives x = -2, -1/2

3y^{2}+ 5y – 2 = 0

3y^{2}+ 6y – y – 2 = 0

Gives y = -2, 1/3

Put on number line

-2 -1/2 1/3**5x**^{2}– 7x – 6 = 0, 5y^{2}+ 23y + 12 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined

**C) x ≥ y**

Explanation:

5x^{2}– 7x – 6 = 0

5x^{2}– 10x + 3x – 6 = 0

Gives, x = -3/5, 2

5y^{2}+ 23y + 12 = 0

5y^{2}+ 20y + 3y + 12 = 0

Gives y = -4, -3/5

Put on number line

-4 -3/5 2**3x**^{2}– 4x – 4 = 0, 2y^{2}– 11y + 15 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined

**B) x < y**

Explanation:

3x^{2}– 4x – 4 = 0

3x^{2}– 6x + 2x – 4 = 0

Gives, x = 2, -2/3

2y^{2}– 11y + 15 = 0

2y^{2}– 6y – 5y + 15 = 0

Gives, y = 5/2, 3

Put on number line

-2/3 2 5/2 3**x**^{2}+ x – 12 = 0, 2y^{2}– 9y + 10 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined

**E) x = y or relationship cannot be determined**

Explanation:

x^{2}+ x – 12 = 0

Gives, x = -4, 3

2y^{2}– 9y + 10 = 0

2y^{2}– 9y – 5y + 10 = 0

Gives, y = 2, 5/2

Put on number line

-4 2 5/2 3

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