Question

A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate the interest of the first year.

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Answer 1

Answer:

Simple interest (I) = Rs. 20,000

Hence, for first year

»» P = Rs. 20,000

»» N = 1 year

»» R = 8%

So, Intrest in the end of 1 year I =

»» 20,000 × 1 × 8 / 100

»» Rs. 1,600

hope it helps you.

Answer 2

Given: $P=Rs.20,000, t=2yrs. and R=8$%.

We have to calculate the interest of the first year.

We are solving in the following way:

As we know that the formula to calculate the simple interest is:

$\mathrm{I}=\frac{\mathrm{P} \times \mathrm{N} \times \mathrm{R}}{100}$

Here for the first year,

Principal, $P= Rs 20,000$

Number of years, $N=1$year

Rate, $R=8$%

So, Interest at the end of $1$st year will be:

$\mathrm{I}=\frac{20,000 \times 1 \times 8}{100}\\\\=\mathrm{Rs} 1,600$

So, for the second year,

P = $Rs 20,000+ Rs 1,600= Rs 21,600$

$N=1$year

Rate, $R=8$%

Therefore,

$\mathrm{I}=\frac{21,600 \times 1 \times 8}{100}\\\\=I =\mathrm{Rs} 1,728$

Hence, the interest of the first year will be $\text { Rs } 1,600$and for the second year will be $\text { Rs } 1,728$