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Q:

A sum of money invested for a certain number of years at 8% p.a. simple interest grows to Rs.180. The same sum of money invested for the same number of years at 4% p.a. simple interest grows to Rs.120 only. For how many years was the sum invested?

A) 25years	B) 15years
C) 20years	D) 22years

Answer & Explanation Answer: A) 25years

Explanation:

From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.

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Q:

A man borrowed Rs 24000 from two money lenders. For one loan, he paid 15% per annum and for the other 18% per annum. At the end of one year, he paid Rs 4050. How much did he borrow at each rate ?

A) Rs.16000	B) Rs.12000
C) Rs.15000	D) Rs.13000

Answer & Explanation Answer: C) Rs.15000

Explanation:

Let the sum at 15% be Rs x and that at 18% be Rs (24000 - x).

{(x * 15 * 1)/100 } + { [(24000 – x) * 18 * 1]/100 } = 4050

or 15 x + 432000 - 18x = 405000 or x = 9000.

Money borrowed at 15% = Rs 9000 .

Money borrowed at 18% = Rs 15000.

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Q:

A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum.

A) Rs.4000	B) Rs.9000
C) Rs.5000	D) Rs.6000

Answer & Explanation Answer: D) Rs.6000

Explanation:

Let sum = P and original rate = R. Then

[(P * (R+2) * 3)/100] - [ (P * R * 3)/100] = 360

3P*(R+2) - 3PR = 36000

3PR + 6P - 3PR = 36000

6P = 36000

P = 6000

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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:

A father left a will of Rs.35 lakhs between his two daughters aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years. The original amount of Rs.35 lakhs has been instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at the time of the will?

A) 17.5 lakhs	B) 21 lakhs
C) 15 lakhs	D) 20 lakhs

Answer & Explanation Answer: B) 21 lakhs

Explanation:

Let Rs.x be the amount that the elder daughter got at the time of the will. Therefore, the younger daughter got (3,500,000 - x).

The elder daughter’s money earns interest for (21 - 16) = 5 years @ 10% p.a simple interest.

The younger daughter’s money earns interest for (21 - 8.5) = 12.5 years @ 10% p.a simple interest.

As the sum of money that each of the daughters get when they are 21 is the same,

$x + \frac{5 * 10 * x}{100} = (3, 500, 000 - x) + \frac{12.5 * 10 * (3, 500, 000 - x)}{100}$

$x + \frac{50}{x} = 3, 500, 000 - x + \frac{125}{100} * 3, 500, 000 - \frac{125 x}{100}$

$2 x + \frac{50 x}{100} + \frac{125 x}{100} = 3, 500, 000 * (1 + \frac{5}{4})$

$\frac{200 x + 50 x + 125 x}{100} = \frac{9}{4} * (3, 500, 000)$

=> $x = 2, 100, 000 = 21 l a k h s$

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