Sawaal
16
Q:

Find the value of 'n' for which the nth term of two AP'S:

15,12,9.... and -15,-13,-11...... are equal?

A) n = 2 B) n = 5
C) n = 29/5 D) n = 1

Answer:   C) n = 29/5



Explanation:

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1) 

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

 

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

Subject: Permutations and Combinations
Job Role: Analyst
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In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together

A) 50000 B) 40500
C) 5040 D) 50400
 
Answer & Explanation Answer: D) 50400

Explanation:

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 3!/5!= 20 ways.

 

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

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Q:

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) 720 B) 520
C) 700 D) 750
 
Answer & Explanation Answer: A) 720

Explanation:

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) 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.

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Q:

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) 735
C) 756 D) 657
 
Answer & Explanation Answer: C) 756

Explanation:

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

 

Required number of ways= 7C3*6C2+7C4*6C1+7C5 = 756.

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11 17119
Q:

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A) 1260 B) 210
C) 210 x 6! D) 1512
 
Answer & Explanation Answer: A) 1260

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

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