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Q:

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

A) 2580	B) 3687
C) 4320	D) 5460

Answer & Explanation Answer: C) 4320

Explanation:

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.

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Filed Under: Permutations and Combinations
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Q:

A work which could be finished in 11 days was finished 4 days earlier after 4 more men joined. The number of men employed was ?

A) 7	B) 8
C) 9	D) 10

Answer & Explanation Answer: A) 7

Explanation:

As the work is same
M1D1 = M2D2

Here Let the number of men employed was M
=> 11M = 7(M+4)
=> 11M - 7M = 28
=> M = 7

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Filed Under: Time and Work
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Q:

A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?

A) 47	B) 46
C) 37	D) 35

Answer & Explanation Answer: C) 37

Explanation:

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.

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Filed Under: HCF and LCM
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Q:

Ajay and Vijay have some marbles with them. Ajay told Vijay "if you give me 'x' marbles, both of us will have equal number of marbles". Vijay then told Ajay "if you give me twice as many marbles, I will have 30 more marbles than you would". Find 'x' ?

A) 4	B) 10
C) 8	D) 5

Answer & Explanation Answer: D) 5

Explanation:

If Vijay gives 'x' marbles to Ajay then Vijay and Ajay would have V - x and A + x marbles.
V - x = A + x --- (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A - 2x and V + 2x marbles.
V + 2x - (A - 2x) = 30 => V - A + 4x = 30 --- (2)
From (1) we have V - A = 2x
Substituting V - A = 2x in (2)
6x = 30 => x = 5.

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Filed Under: Arithmetical Reasoning
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Q:

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most ?

A) 215	B) 268
C) 254	D) 216

Answer & Explanation Answer: A) 215

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

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Filed Under: Permutations and Combinations
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Q:

The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is ?

A) 3 years	B) 4 years
C) 2.5 years	D) 2 years

Answer & Explanation Answer: B) 4 years

Explanation:

$P {(1 + \frac{20}{100})}^{n}$ > 2P

Now, (6/5 x 6/5 x 6/5 x 6/5) > 2.

So, n = 4 years.

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Filed Under: Compound Interest
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Q:

In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 172 votes. Find the total number of valid votes ?

A) 446	B) 415
C) 404	D) 430

Answer & Explanation Answer: D) 430

Explanation:

Let the total number of valid votes be 'V'.

70% of V = 70/100 x V = 7V/10

Number of votes secured by the other candidate = V - 7V/100 = 3V/10

Given, 7V/10 - 3V/10 = 172 => 4V/10 = 172

=> 4V = 1720 => V = 430.

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Filed Under: Percentage
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Q:

The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is ?

A) 2(6!)	B) 6! x 7
C) 6! x ⁷P₆	D) None

Answer & Explanation Answer: C) 6! x ⁷P₆

Explanation:

We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.

Hence required number of ways = 6! x ⁷P₆

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Filed Under: Permutations and Combinations
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