Question

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$\sf{Find \: \: the \: \: amount \: \: on \: \: 210000 \: \: at \: \: the} \\ \sf { rate \: \: of \: \: 3 \: \: percent \: \: for \: \: 9 \: \: years \: \: 2 \:} \\ \sf {month \: \: compounded \: \: annually.}$

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Answer 1

$\sf\large\underline\blue{Given:}$

$\rm{\implies Principal=Rs.210000}$

$\rm{\implies Rate=3\%}$

$\rm{\implies Time=9\: years\:and\:2\: months}$

$\sf\large\underline\blue{To\: Find:}$

$\rm{\implies Amount=?}$

$\sf\large\underline\blue{Solution:}$

• Here, the time is given in fraction form. So, we have to use a special formula to calculate the amount. At first, convert the time into a mixed fraction. Once converted, we then get the fraction as 9 (1/6)]

$\sf\large\underline\blue{Formula\:used:}$

$\tt{\implies A=P\bigg(1+\dfrac{r}{100}\bigg)^n\bigg(1+\dfrac{rF}{100}\bigg)}$

• F=1/6, And n=9

$\tt{\implies A=210000\bigg(1+\dfrac{3}{100}\bigg)^9\bigg(1+\dfrac{3*1}{6*100}\bigg)}$

$\tt{\implies A=210000\bigg(\dfrac{103}{100}\bigg)^9\bigg(1+\dfrac{3}{600}\bigg)}$

$\tt{\implies A=210000\bigg(\dfrac{103}{100}\bigg)^9\bigg(1+\dfrac{1}{200}\bigg)}$

$\tt{\implies A=210000*(1.03)^9\times\dfrac{201}{200}}$

$\tt{\implies A=210000*1.30477318383*\dfrac{201}{200}}$

$\tt{\implies A=274002.368604*\dfrac{201}{200}}$

$\tt{\implies A=\dfrac{55074476.1}{200}}$

$\tt{\implies A=275372.381}$

$\sf\large{Hence,}$

$\tt{\implies Amount=Rs.275372.381}$

Answer 2

Given:

Principal = Rs. 2,10,000

Rate = 3℅

Time = 9 year and 2 months

To Find:

Amount

Formula:

$A = P(1 + \frac{r}{100} ) {}^{n} (1 + \frac{rF}{100} )$

F = 1/6 and n = 9

$A = 210000 {(1 + \frac{3}{100}) }^{9} (1 + \frac{3 \times 1}{6 \times 100} ) \\ Amount = Rs.275372.381$

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