Exercise : 2


1. Difference between the compound interest and the simple interest accrued on an amount of 18000, in two years was 405. What was the rate of interest p.c.p.a?
(a) 16
(b) 12
(c) 15
(d) Cannot be determined
(e) None of these
Ans.c

Rate % = \(\sqrt{\frac{405 \times 100 \times 100}{18000}}\) = 15%

2. Anish borrowed 15000 at the rate of 12% and an other amount at the rate of 15% for two years. The total interest paid by him was 9000. How much did he borrow?
(a) 32,000
(b) 33,000
(c) 30,000
(d) 35,000
(e) None of these
Ans.b

Let x be the other amount

\ 3x10 + 3600 = 900 ⇒ x = 18000

\ total borrowed sum = 33000

3. The compound interest on any sum at the rate of 5% for two years is 512.50. Find the sum.
(a) 5200
(b) 4800
(c) 5000
(d) 5500
(e) None of these
Ans.c

Let the sum be x.

512.50 = \(x\left | 1 + \frac{5}{100} \right |^{2} - 1 = x\left | \frac{441 - 400}{400} \right |\)

\ x = \(\frac{512.50 \times 400}{41} = 5000\)

4. Mr Amin borrowed some money from Mr Vishwas. The rate of interest for first two years is 8% p.a., for the next three years is 11 % p.a. and for the period beyond 5 years 14% p.a. Mr Vishwas got an amount of 10920 as an interest at the end of eight years. Then what amount was borrowed by Mr Amin ?
(a) 12000
(b) 15000
(c) 1400
(d) Data inadequate
(e) None of these
Ans.a

Let ‘x’ be the amount borrowed by Mr Amin.

\(\frac{x \times 2 \times 8}{100} + \frac{x \times 3 \times 11}{100} + \frac{x \times 3 \times 14}{100} = 10920\)

or, 91100x = 10920 or x = \(\frac{10920 \times 100}{91} = 12000\)

5. Mr Sridharan invested money in two schemes A and B,offering compound interest @ 8 p.c.p.a. and 9 p.c.p.a.respectively. If the total amount of interest accured through the two schemes together in two years was 4818.30 and the total amount invested was 27,000, what was the amount invested in Scheme A?
(a) 15,000
(b) 13,500
(c) 12,000
(d) Cannot be determined
(e) None of these
Ans.c

Let, in scheme A, Sridharan invest x.

Then, his investment in scheme B = (27000 – x).

Now,

\(x\left ( 1 + \frac{8}{100} \right )^{2} + \left (2700 - x \right )\left ( 1 + \frac{9}{100} \right )^{2}\)

– 27000 = 4818.30

or, x(1.08)2 + (27000 – x)(1.09)2 = 31818.30

or, 1.1664x + 32078.7 – 1.1881x = 31818.30

or, 0.0217x = 260.4

or, x = 260.40.0217 = 12000

6. Seema invested an amount of 16000 for two years at compound interest and received an amount of 17640 on maturity. What is the rate of interest?
(a) 8 pcpa
(b) 5 pcpa
(c) 4 pcpa
(d) Data inadequate
(e) None of these
Ans.b

\(\frac{A}{P} = \left ( 1 + \frac{r}{100} \right )^{t} \;\; or, \; \frac{17640}{16000} = \left ( 1 + \frac{r}{100} \right )^{2}\)

2120 = 1 + r100

⇒ r = 5%

7. Amit Kumar invested an amount of 15,000 at compound interest rate of 10 pcpa for a period of two years. What amount will he receive at the end of two years?
(a) 18,000
(b) 18,500
(c) 17,000
(d) 17,500
(e) None of these
Ans.e

Amount = 15000 (1 + 10100)2

= 15000 × 1110 × 1110 = 18150

8. In a business A and C invested amounts in the ratio 2 : 1. Whereas the ratio between amounts invested by A and B was 3 : 2. If 1,57,300 was their profit, how much amount did B receive?
(a) 72,600
(b) 48,400
(c) 36,300
(d) 24,200
(e) None of these
Ans.b

Ratio A : B = 3 : 2 and A : C = 2 : 1

\ A : B : C = 6 : 4 : 3

Profit share of B = 4 × 1,57,300 = 48400

9. Mr. Sane invested a total amount of 16,500 for two years in two schemes A and B with rate of simple interest 10 p.c.p.a. and 12 p.c.p.a. respectively. If the total amount of interest earned was 3,620, what was the amount invested in scheme B?
(a) 8,000
(b) 8,600
(c) 8,150
(d) Data inadequate
(e) None of these
Ans.a

% interest on total amount per annum

= \(\frac{3620 \times 100}{16500 \times 2} = \frac{362}{33} \%\)

Now, use Alligation method.

5

Hence, ratio of amount invested in schemes A and B

= (12 - 36233) : (36233 - 10) = 17 : 16

Hence, amount invested in B = \(\frac{16 \times 16500}{\left (17 + 16 \right )} = 8000\)

10. The difference between the simple and the compound interest compounded every six months at the rate of 10% p.a. at the end of two years is 124.05. What is the sum?
(a) 10,000
(b) 6,000
(c) 12,000
(d) 8,000
(e) None of these
Ans.d

Let the sum be x.

Then, \(\left [ x\left ( 1 + \frac{5}{100} \right )^{4} - x \right ] - \left [ \frac{x \times 10 \times 2}{100} \right ] = 124.05\)

Solving the above eqns, we get x = 8,000.