GRE Questions and Answers with Explanation

Q:

A certain sum of money amounts to Rs 1008 in 2 years and to Rs 1164 in 3 ½ years. Find the sum and the rate of interest.

A) 800, 14%	B) 800, 13%
C) 800, 12%	D) 800, 19%

Answer & Explanation Answer: B) 800, 13%

Explanation:

S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 .

S.I. for 2 years = Rs (156 x $\frac{2}{3}$ x 2)= Rs 208.

Therefore, Principal = Rs (1008 - 208) = Rs 800.

Now, P = 800, T= 2 and S.I. = 208.

Therefore, Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%

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Q:

At what rate percent per annum will a sum of money double in 8 years.

A) 12.5%	B) 13.5%
C) 11.5%	D) 14.5%

Answer & Explanation Answer: A) 12.5%

Explanation:

Let principal = P, Then, S.I.= P and Time = 8 years

We know that S.I. = PTR/100

Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.

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Q:

A sum at simple interest at $13 \frac{1}{2}$ % per annum amounts to Rs. 2502.50 after 4 years. Find the sum

A) Rs.1625	B) Rs1525
C) Rs.1425	D) Rs.1325

Answer & Explanation Answer: A) Rs.1625

Explanation:

Let sum be Rs.x then, S.I. =Rs. [ x * (27/2) * 4 * (1/100) ] = Rs. 27x/50

Amount =Rs [ x + (27x/50)] = Rs.77x/50

=> 77X/50 = 2502.50

=>X= (2502.50 * 50)/77 = 1625

Hence Sum = Rs.1625

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Q:

Find: S.l. on Rs 3000 at 18% per annum for the period from 4th

A) 118	B) 105
C) 108	D) 110

Answer & Explanation Answer: C) 108

Explanation:

(24 + 31 + 18) days = 73 days = 1/5 year .

P = Rs 3000 and R = 18 % p.a.

$S . I = (\frac{3000 * 18 * \frac{1}{5}}{100})$

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Q:

Find: S.l. on Rs 6250 at 14% per annum for 146 days.

A) Rs.350	B) Rs.450
C) Rs.550	D) Rs.650

Answer & Explanation Answer: A) Rs.350

Explanation:

P = Rs. 6250, R = 14 % & T = (146/365) years = 2/5 years .

S.I= $R s . [\frac{6250 * 14 * \frac{2}{5}}{100}] = R s . 350$

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Q:

Find: S.l on Rs 68000 at $16 \frac{2}{3}$ 1 % per annum for 9 months

A) 6500	B) 7500
C) 8500	D) 9500

Answer & Explanation Answer: C) 8500

Explanation:

P = 68000, R = $\frac{50}{3}$ % & T = 9 months (3/4 years)

$S . I = \frac{p \times r \times t}{100} = [\frac{(6800 \times \frac{50}{3} \times \frac{3}{4})}{100}] = R s . 8500$

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Q:

What is the sum which earned interest?
I. The total simple interest was Rs. 7000 after 7 years.
II. The total of sum and simple interest was double of the sum after 5 years.

A) I alone sufficient while II alone not sufficient to answer	B) II alone sufficient while I alone not sufficient to answer
C) Either I or II alone sufficient to answer	D) Both I and II are necessary to answer

Answer & Explanation Answer: D) Both I and II are necessary to answer

Explanation:

Let the sum be Rs. x.

I gives, S.I. = Rs. 7000 and T = 7 years.

II gives, Sum + S.I. for 5 years = 2 x Sum Sum = S.I. for 5 years.

Now, S.I. for 7 years = Rs. 7000.

therefore, S.I. for 1 year = Rs. 1000.

Thus, I and II both are needed to get the answer.

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Q:

1. The simple interest on a sum of money is Rs. 50. What is the sum?
I. The interest rate is 10% p.a.
II. The sum earned simple interest in 10 years

A) I alone sufficient while II alone not sufficient to answer	B) II alone sufficient while I alone not sufficient to answer
C) Either I or II alone sufficient to answer	D) Both I and II are necessary to answer

Answer & Explanation Answer: D) Both I and II are necessary to answer

Explanation:

Given : S.I. = Rs. 50.

I gives, R = 10% p.a.

II gives, T = 10 years.

Sum = (100 x S.I)/(t x r ) = Rs.(100 x 50)/(10 x 10) = Rs.50

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