Question

A man invests 10000 for 3 years at a certain rate of interest, compounded annually.

At the end of one year it amounts to 10800. Calculate : (i) the rate of interest per
annum, (ii) the interest accrued in the second year, (iii) the amount at the end of the third
year (to the nearest rupee).​

Answer 1

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It is given that

Principal(P) = 10,000

Period (T)= 1 year

Sum amount (A)= 11200

Rate of interest =?

(i) We know that

Interest (I)= 11200- 10000= 1200

So the rate of interest

R= (1×100)/(P×T)

Substituting the values

R= (1200×100)/(1000×1)

So we get

R= 12% p.a

Therefore, the rate of interest per annum is 12% p.a

(ii) We know that

Period (T)= 2 years

Rate of interest (R) = 12% p.a.

Here

A =P(1+R/100)

1

Substituting the values

A = 10000(1+12/100)

2

By further calculation

A= 10000(28/25)

2

We can write it as

A= 10000×28/25×28/25

So we get

A=16×28×28

A=12544