Question

a sum of money invested at compound interest amount to 16500 in 1 year and to 19965 in 3 year find the rate per cent and the original sum​

Answer 1

Answer:

A sum of money, invested at compound interest, amounts to Rs. 16,500 in 1 year and to Rs. 19,965 in 3 years. Find the rate per cent and the original sum of money invested

Answer 2

$\sf \green{\underline {Question :-}}$

• a sum of money invested at compound interest amount to 16500 in 1 year and to 19965 in 3 year find the rate per cent and the original sum.

$\sf \orange{\underline <u>{</u><u>Given</u> :-}}$

• The amount after 1 year = ₹16,500.
• The amount after 3 year = ₹ 19,965.

$\sf \pink {\underline{ <u>To</u> <u>\</u><u>:</u><u> </u><u>find</u> :- }}$

• The rate percentage and the original sum.

$\sf \purple{\underline{ <u>Solu</u><u>tion</u> :- }}$

• Let's assume that the Principal = ₹(P)
• And, Rate = r%

By using the formula of C.I. :-

$\bf {\underline{\boxed {\bf A(amount)= P(1 + \frac{r\%}{100} {)}^{n}}}}$

For the 1 year :-

where,

• P = ₹16,500.
• Time (n) = 1 year.

$\bf \implies \: 16500 = p(1 + \frac{r\%}{100} {)}^{1} ........(i)$

For the 3 year :-

where,

• P = ₹19,965.
• Time (n) = 3 years.

$\bf \implies \: 19965 = p(1 + \frac{r\%}{100} {)}^{3} ........(ii)$

Now, Dividing eq.(i) by eq.(ii)

$\bf \implies \: \frac{\cancel{19965}}{ \cancel{16500}} = \cancel \frac{p( 1+ \frac{r}{100} {)}^{3}}{p(1 + \frac{r}{100} {)}^{1} } \\ \bf \implies \: \frac{121}{100} = (1 + \frac{r}{100} {)}^{2} \\ ( \bf \: on \: √ \: both \: side) \: \\ \bf \implies \: \frac{11}{10} = (1 + \frac{r}{100} ) \\ \bf \implies \: \frac{11}{10} = \frac{100 + r}{100} \\ \bf \implies \: \frac{11 \times 10 \cancel0}{ \cancel{10}} = 100 + r \\ \bf \implies \: 110 = 100 + r \\ \bf \implies\: r \: = 110 - 100 \\ \bf \therefore{ \pink \bigstar{\underline{\boxed{ \bf \: r = 10\%}}}}{ \pink \bigstar}$

Putting the value of r in the eq.(i).

$\bf \implies 16500 = p(1 + \frac{10}{100} {)}^{1} \\ \bf \implies \: 16500 = p \times \frac{11}{10} \\ \bf \implies \: \frac{ \cancel{16500} \times 10}{ \cancel{11}} = p \\ \bf \implies { \green \bigstar}{ \underline{ \boxed{\bf p = 15000}}}{ \green \bigstar}$

Hence,

• Rate = 10% ☞ ans.
• Original Principal = ₹15,000 ☞ ans.