6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

There are 44 students in a hostel, due to the administration, 15 new students has joined. The expense of the mess increase by Rs. 33 per day. While the average expenditure per head diminished by Rs. 3, what was the original expenditure of the mess ?

A) Rs. 404	B) Rs. 514
C) Rs. 340	D) Rs. 616

Answer & Explanation Answer: D) Rs. 616

Explanation:

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

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Filed Under: Average
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
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11 9521

Q:

A sum of money invested at simpleinterest triples itself in 8 years. How many times will it become in 20 years time?

A) 8 times	B) 7 times
C) 6 times	D) 9 times

Answer & Explanation Answer: C) 6 times

Explanation:

A = P(1 + rt/100)

So, A= 3P (it becomes triples)

3P = P(1 + 8r/100)→r = 25%

Now, when t = 20

⇒A = P(1 + 25*20/100)

⇒A = P(1+ 5)⇒A = 6P

So, it becomes 6 times.

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Filed Under: Simple Interest
Exam Prep: Bank Exams

12 9510

Q:

There are 10 Letters and 10 correspondingly 10 different Address. If the letter are put into envelope randomly, then find the Probability that Exactly 9 letters will at the Correct Address ?

A) 1/10	B) 1/9
C) 1	D) 0

Answer & Explanation Answer: D) 0

Explanation:

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address
P(E) = favorable outcomes /total outcomes
Here favorable outcomes are '0'.
So probability is '0'.

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Filed Under: Probability
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5 9504

Q:

If $\frac{a}{b} = \frac{4}{5}$ and $\frac{b}{c} = \frac{15}{16}$ , then $\frac{c^{2} - a^{2}}{c^{2} + a^{2}}$ is :

A) 7/5	B) 7/25
C) 7/125	D) 7

Answer & Explanation Answer: B) 7/25

Explanation:

$\frac{a}{b} \times \frac{b}{c} = \frac{4}{5} \times \frac{15}{16} = \frac{3}{4} = \frac{16 - 9}{16 + 9}$

= $\frac{7}{25}$

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Filed Under: Simplification

7 9501

Q:

A man said to his son, "I was two-thirds of your present age when you were born". IIf the present age of the man is 48 years, find the present age of the son.

Answer

Let the Present age of the son be P, which means he was born P years ago.At the time the son was born, the age of the man was : (48 - P). So, according to the statement made by the man, his age when the son was born should be equal to 2/3 of P.

Therefore, $(48-P)=\frac{2}{3}P$

Solving, we get P = 28.8 years; which is the present age of the son.

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Subject: Problems on Ages

10 9500

Q:

Find the cash realised by selling Rs. 2440, 9.5% stock at 4 discount (brokerage 1/4 %)

A) 2000	B) 2298
C) 2290	D) 2289

Answer & Explanation Answer: B) 2298

Explanation:

By selling Rs. 100 stock , cash realised = $R s . [(100 - 4) - \frac{1}{4}] = R s . \frac{383}{4}$

By selling Rs. 2400 stock, cash realised = $R s . (\frac{383}{4} * \frac{1}{100} * 2400)$ = Rs 2298.

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Filed Under: Stocks and Shares

12 9496

Q:

Sixty men can stitch 150 shirts in 30 days working 8 hours a day. In how many days can 35 men stitch 200 shirts working 6 hours a day ?

A) 87.5	B) 89
C) 91.5	D) 93.5

Answer & Explanation Answer: C) 91.5

Explanation:

We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 x 30 x 8)/ 150 = (35 x D2 x 6) / 200
D2 = (60 x 30 x 8 x 200) / (150 x 35 x 6) => D2 = 91.42 days =~ 91.5 days.

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Filed Under: Time and Work
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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5 9495

Q:

Find the least number with which 16200 should be multiplied, to make it a perfect cube ?

A) 45	B) 48
C) 360	D) 36

Answer & Explanation Answer: A) 45

Explanation:

16200 = $2^{3} \times 3^{4} \times 5^{2}$

A perfect cube has a property of having the indices of all its prime factors divisible by 3.

Required number = $3^{2} \times 5$ = 9x5 = 45.

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Filed Under: Problems on Numbers
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6 9472

Quantitative Aptitude - Arithmetic Ability Questions

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