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Question

Answers

A. Rs $151200$

B. Rs $100000$

C. Rs $51200$

D. Rs $251200$

Answer
Verified

From the question:

Given that:

\[

P = 100000 \\

r = 20\% = \dfrac{{20}}{{100}} = 0.2 \\

n = 2{\text{ }}years{\text{ }}3{\text{ }}months \\

\]

Where $P = $ Principal

$r = $ Rate of interest

$n = $ Time period

Formula for compound interest:

$A = P{\left( {1 + \dfrac{r}{n}} \right)^{nt}}$

Convert the value of n in years.

We get:

$

n = 2 + \dfrac{1}{4} = \dfrac{{8 + 1}}{4} \\

\Rightarrow n = \dfrac{9}{4} = 2.25 \\

$

Now the time period is $2.25$ years.

Form the question:

Given that:

Interest is compounded annually:

For first year:

Interest$ = 100000 \times 0.2 \times 1 = 20000$

Thus for second year ${P_{new}} = 100000 + 20000 = 120000$

Interest for second year $120000 \times 0.2 \times 1 = 24000$

Thus for last year ${P_{new}} = 120000 + 24000 = 144000$

Thus we get the interest for complete two years.

Now interest for rest of $0.25$ years$ = $ $144000 \times 0.2 \times 0.25 = 7200$

Now the new amount is $ = 14400 + 7200 = 151200$

Thus the interest is $ = 151200 - 100000 = 51200$

This is the compound interest on Rs $100000$ at $20\% $ per annum for $2$ year’s $3$ months.

Hence

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