Quantitative Aptitude

General Questions

Quantitative Aptitude Exercise Mode

General Questions

Practice and master this topic with our carefully crafted questions.

10 Questions
15 Minutes
0% Completed
QUEST ? ✓ !
Question 1
How many words can be formed by using letters of the word 'DELHI'?

A
50
B
72
C
85
D
120
Correct Answer: Option D

The word ‘DELHI’ contains 5 letters

Therefore, required number of words = 5P5 = 5! = (5 × 4 × 3 × 2 × 1) = 120

120 words can be formed by using letters of the word ‘DELHI’

Question 2
In how many ways can the letters of the word 'LEADER' be arranged?

A
72
B
144
C
360
D
720
Correct Answer: Option C

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways = 6!
(1!)(2!)(1!)(1!)(1!)
   
      = 360.

Question 3
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A
360
B
480
C
720
D
5040
Correct Answer: Option C

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Question 4
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
E
None of these
Correct Answer: Option D

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.

Question 5
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
E
None of these
Correct Answer: Option D

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 + 30
2 x 1 2 x 1 3 x 2 x 1 2
= (24 + 90 + 80 + 15)
= 209.

Question 6
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
E
None of these
Correct Answer: Option C

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.
Question 7
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

A
32
B
48
C
36
D
60
E
120
Correct Answer: Option C

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.

Question 8
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
E
None of these
Correct Answer: Option C

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.

Question 9
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

A
810
B
1440
C
2880
D
50400
E
5760
Correct Answer: Option D

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters = 7!
2!
 = 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in 5! = 20 ways.
3!

Required number of ways = (2520 x 20) = 50400.

Question 10
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A
63
B
90
C
126
D
45
E
135
Correct Answer: Option A

Required number of ways

= (7C5 x 3C2) = (7C2 x 3C1)

= 7 x 6 x 3
2 x 1

= 63.

1 2 3 Next
Page 1 of 3