Aptitude

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? Read More »

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:

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: Read More »

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?

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? Read More »

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].

[B].

[C].

[D].

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:

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: Read More »

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?

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

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?

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? Read More »

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].

[B].

[C].

[D].

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?

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? Read More »

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].

[B].

[C].

[D].

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?

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? Read More »

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

4 x 3

6 x 5 x 4

x 4

6 x 5

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?

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? Read More »

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.