Question

13 A sum of 9600 is invested for 3 years at 10% per annum at compound interest.
(1) What is the sum due at the end of the first year?
(i) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in 2 years.
(iv) Find the difference between the answers in (ii) and (1) and find the interest on this
sum for one year.
(v) Hence, write down the compound interest for the third
year.​

Answer 1

Answer:

It is given that

Principal= 9600

Rate of interest = 10% p.a Period 3 years

We know that

Interest for the first year= Prt/100 Substituting the values

= (9600 × 10 × 1)/100

= 960

(i) Amount after one year = 9600 - 960=

10560

So principal for the second year

10560

Here the interest for the second year= (10560 x 10 x 1)/100

(ii) Amount after two years =10560 +

1056 11616 =

=1056

(iii) Compound interest earned in 2 years =

960+10560 = 2016

(iv) Difference between the answers in (ii) and (i) =11616 - 10560 1056

We know that

Interest on 1056 for 1 year at the rate of 10% p.a. (1056 × 10 × 1)/100

105.60

(v) Here Principal for the third year = 11616 So the interest for the third year = (11616 x

10 × 1)/100

=1161.60