admin

Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

by
Question:
Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

[A].

1 hour

[B].

2 hours

[C].

6 hours

[D].

8 hours

Answer: Option C

Explanation:

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

1 + 1 = 1
x (x + 6) 4
x + 6 + x = 1
x(x + 6) 4

x2 – 2x – 24 = 0

(x -6)(x + 4) = 0

x = 6.     [neglecting the negative value of x]

A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:

by
Question: A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
[A].

4 1 hours
3

[B].

7 hours

[C].

8 hours

[D].

14 hours

Answer: Option D

Explanation:

Work done by the leak in 1 hour = 1 – 3 = 1 .
2 7 14

Leak will empty the tank in 14 hrs.

Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?

by
Question:
Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?

[A].

5
11

[B].

6
11

[C].

7
11

[D].

8
11

Answer: Option B

Explanation:

Part filled by (A + B + C) in 3 minutes = 3 1 + 1 + 1 = 3 x 11 = 11 .
30 20 10 60 20
Part filled by C in 3 minutes = 3 .
10
Required ratio = 3 x 20 = 6 .
10 11 11

Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

by
Question:
Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

[A].

1 13 hours
17

[B].

2 8 hours
11

[C].

3 9 hours
17

[D].

4 1 hours
2

Answer: Option C

Explanation:

Net part filled in 1 hour 1 + 1 – 1 = 17 .
5 6 12 60
The tank will be full in 60 hours i.e., 3 9 hours.
17 17

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

by
Question: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
[A].

210

[B].

1050

[C].

25200

[D].

21400

Answer: Option C

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

      = (7C3 x 4C2)
= 7 x 6 x 5 x 4 x 3
3 x 2 x 1 2 x 1
= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging
5 letters among themselves		 = 5!
= 5 x 4 x 3 x 2 x 1
= 120.

Required number of ways = (210 x 120) = 25200.

Video Explanation: https://youtu.be/dm-8T8Si5lg

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

by
Question:
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

[A].

5

[B].

10

[C].

15

[D].

20

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

by
Question:
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

[A].

32

[B].

48

[C].

64

[D].

96

Answer: Option C

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= 3 x 6 x 5 + 3 x 2 x 6 + 1
2 x 1 2 x 1
= (45 + 18 + 1)
= 64.

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

by
Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
[A].

159

[B].

194

[C].

205

[D].

209

Answer: Option D

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number
of ways		 = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5
2 x 1 2 x 1 3 x 2 x 1 2 x 1
= (24 + 90 + 80 + 15)
= 209.

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

by
Question:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

[A].

564

[B].

645

[C].

735

[D].

756

Answer: Option D

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2)
3 x 2 x 1 2 x 1
= 525 + 7 x 6 x 5 x 6 + 7 x 6
3 x 2 x 1 2 x 1
= (525 + 210 + 21)
= 756.

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

by
Question:
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

[A].

266

[B].

5040

[C].

11760

[D].

86400

Answer: Option C

Explanation:

Required number of ways = (8C5 x 10C6)
= (8C3 x 10C4)
= 8 x 7 x 6 x 10 x 9 x 8 x 7
3 x 2 x 1 4 x 3 x 2 x 1
= 11760.