Question

Find the amount and compound interest on ₹ 10,000 for 3 years if the rate of interest for successive years are $5\%$, $10\%$ and $20\%$ years respectively

• A. ₹ 33100 , ₹ 3310
• B. 13500 , ₹ 3500
• C. ₹ 13860, ₹ 3860
• D. ₹ 13980 , ₹ 3980

Answer 1

The correct option is C

₹ 13860, ₹ 3860

Since the successive rates are given , we have

A = P( 1 + $\frac{r_1}{100}$ ) ( 1 + $\frac{r_2}{100}$ ) ( 1 + $\frac{r_3}{100}$ )

We have P = ₹ 10,000 ; $r_1$ = 5% $r_2$ = 10 % , $r_3$ = 20 %

A = P( 1 + $\frac{5}{100}$ ) $\times$ ( 1 + $\frac{10}{100}$ ) $\times$ ( 1 + $\frac{20}{100}$)

= 10000 $\times$ $\frac{105}{100}$ $\times$ $\frac{110}{100}$ $\times$ $\frac{120}{100}$

= 10000 $\times$ $\frac{21}{20}$ $\times$ $\frac{11}{10}$ $\times$ $\frac{12}{10}$

= ₹ 13860

Compound Interest = ₹ ( 13860 - 10,000) = ₹ 3860

Alternate Solution

For the first year

P = 10000 , R = 5% , T = 1 Year

Interest for the first year = $\frac{P\times R\times T}{100}$ = $\frac{10,000\times 5\times 1}{100}$ = ₹ 500

For the second year

P = 10,000 + 500 = 10,500 , R = 10% , T = 1 year

Interest for the second year = $\frac{P\times R\times T}{100}$ = $\frac{10,500\times 10\times 1}{100}$ = ₹ 1050

For the third year

P = 10,500 + 1050 = 11,550 , R = 20% , T = 1 year

Interest for the third year = $\frac{P\times R\times T}{100}$ = $\frac{11,550\times 20\times 1}{100}$ =₹ 2310

Total interest for 3 years = 500 + 1050 + 2310 = ₹ 3860

Amount = 10,000 + 3860 = ₹ 13860