Aptitude, Aptitude Hack, Aptitude Tricks & Tips, Mock Test, Test Series,

Showing posts with label Words. Show all posts
Showing posts with label Words. Show all posts

Monday, 13 May 2019

Aptitude Hack#79(13-5-19)

Question:

How many words can be formed by using all letters of the word “Bihar”

   A )   110
   B )   120
   C )   130
   D )   140

Sunday, 12 May 2019

Aptitude Hack#79(13-5-19)

Question:

How many words can be formed by using all letters of the word “Bihar”

   A )   110
   B )   120
   C )   130
   D )   140

Answer: B) 120

Explanation:

The word Bihar contains 5 different letters.
Required number of words = 5p5 
= 5! = (5x4x3x2x1) 
= 120.

Sunday, 5 May 2019

Aptitude Hack#72(6-5-19)

Question:

How many words can be formed by using all letters of the word ‘daughter’ so that the vowels always come together?
 A )   4320 words
 B )   4220 words
 C )   4120 words
 D )   5320 words

Tuesday, 30 April 2019

Aptitude Hack#72(6-5-19)

Question:

How many words can be formed by using all letters of the word ‘daughter’ so that the vowels always come together?
 A )   4320 words
 B )   4220 words
 C )   4120 words
 D )   5320 words

Answer: A) 4320 words

Explanation:
 Given the word contains 8 different letters. 
 When the vowels aue are always together, 
we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are dgntr (aue).
Then 6 letters to be arranged in 6p6 = 6! = 720 ways.
The vowels in the group (aue) may be arranged in 3! = 6 ways.
Required number of words = (720×6) = 4320.