Exercise : 3


1. Parameshwaran invested an amount of 12,000 at the simple interest rate of 10 pcpa and another amount at the simple interest rate of 20 pcpa. The total interest earned at the end of one year on the total amount invested became 14 pcpa. Find the total amount invested.
(a) 22,000
(b) 25,000
(c) 20,000
(d) 24,000
(e) None of these
Ans.c

Let the amount invested at 20% rate be x. According to the question,

12000 × 10100 + x × 20100 = (12000 + x) × 14100

or, 1200 + x5 = 1680 + 750x

or, x5 - 750x = 480

or, 350x = 480

\ x = 8000

\ Total amount invested = (12000 + 8000)= 20000

2. Raviraj invested an amount of 10,000 at compound interest rate of 10 pcpa for a period of three years. How much amount will Raviraj get after three years?
(a) 12,310
(b) 13,210
(c) 13,320
(d) 13,120
(e) None of these
Ans.e

Amount = 10000(1 + 10100)3

= 10000 × 1110 × 1110 × 1110

= 13310

3. Nelson borrowed some money at the rate of 6 p.c.p.a. for the first three years, 9 p.c.p.a. for the next five years and 13 p.c.p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is 8,160, how much money did he borrow?
(a) 12,000
(b) 10,000
(c) 8,000
(d) Data inadequate
(e) None of these
Ans.c

Let ‘x’ be the amount borrowed by Nelson

and \(\frac{x \times 6 \times 3}{100} + \frac{x \times 9 \times 5}{100} + \frac{x \times 13 \times 3}{100} = 8160\)

102100x = 8,160

∴ x = \(\frac{8,160 \times 100}{102} = 8000\)

4. Amal borrowed a sum of money with simple interest as per the following rate structure:

(a) 6 p.c. p.a. for the first three years
(b) 8 p.c. p.a. for the next five years
(c) 12 p.c. p.a. for the next eight years

If he paid a total of 5,040 as interest at the end of twelve years, how much money did he borrow?

(a) 8,000
(b) 10,000
(c) 12,000
(d) 6,000
(e) None of these
Ans.e

Let x be the amount Amal borrowed.

∴ 18% of x + 40% of x + 48% of x = 5040

or, 106% of x = 5040

∴ x = 5040106 × 100 = 4754.71

5. The simple interest in 14 months on a certain sum at the rate of 6 per cent per annum is 250 more than the interest on the same sum at the rate of 8 per cent in 8 months. How much amount was borrowed?
(a) 15000
(b) 25000
(c) 7500
(d) 14500
(e) None of these
Ans.a

Let the amount be x.

From the question, \(\frac{x \times 14 \times 6}{1200} - \frac{x \times 8 \times 8}{1200}\) = 250

∴ x = 15000

6. On retirement, a person gets 1.53 lakhs of his provident fund which he invests in a scheme at 20% p.a. His monthly income from this scheme will be
(a) 2,450
(b) 2,500
(c) 2,550
(d) 2,600
(e) None of these
Ans.b

Let S.I. = x

= \(\frac{1.53 \times 10^{5} \times 20}{100}\)

= 2,500

7. A sum was put at simple interest at a certain rate for 4 years Had it been put at 2% higher rate, it would have fetched 56 more. Find the sum.
(a) 500
(b) 600
(c) 700
(d) 800
(e) None of these
Ans.c

Difference in S.I. = \(\frac{P \times T}{100} \left(R_{1} - R_{2} \right)\)

⇒ 56 = \(\frac{P \times 4 \times 2}{100}\) (∵ R1 - R2 = 2)

⇒ P = \(\frac{56 \times 100}{4 \times 2} = 700\)

8. Simple interest on a certain sum is 16 over 25 of the sum. Find the rate per cent and time, if both are equal.
(a) 8% and 8 years
(b) 6% and 6 years
(c) 10% and 10 years
(d) 12 % and 12 years
(e) None of these
Ans.a

1625P = \(\frac{P \times R \times R}{100}\)

⇒ R2 = 160025 ⇒ R = 405 = 8%

Also, time = 8 years

9. The simple interest on 200 for 7 months at 5 paise per rupee per month is
(a) 70
(b) 7
(c) 35
(d) 30.50
(e) None of these
Ans.a

∵ Rate = 5 paise per rupee = 5%

∴ S.I. = \(\frac{200 \times 5 \times 7}{100} = 70\)

10. A father left a will of 68,000 to be divided between his two sons aged 10 years and 12 years such that they may get equal amount when each attains the age of 18 years If the money is reckoned at 10% p.a., find how much each gets at the time of the will.
(a) 30,000, 38,000
(b) 28,000, 40,000
(c) 32,000, 36,000
(d) Cannot be determined.
(e) None of these
Ans.c

Let one gets = x

then, second gets = (68,000 – x)

Given : A1 = A2

\(x + \frac{x \times 10 \times 8}{100} = \left(68,000 - x \right) + \frac{\left(68,000 - x \right) \times 10 \times 6}{100}\)

⇒ x[100 + 80] = (68,000 - x)[100 + 60]

180160x = 68,000 - x

⇒ 34x = 68000 × 16 ⇒ x = 32000

∴ second gets = 36,000