Question

If the compound interest on Rs 500 at the rate of 10% per on Rs 105, find the time.​

Answer 1

GIVEN :–

• Compound interest = 105

• Principal amount = 500

• Rate = 10 %

TO FIND :–

• Time = ?

SOLUTION :–

• We know that –

$\\ \: \: \: \: \longrightarrow \: \: \pink{ \large\boxed{ \bf \: A=P \bigg(1+ \dfrac{R}{100} \bigg)^n}}$

• Here –

$\\ \: \: \: \longrightarrow \: \: \bf A = Total \: amount$

$\\ \: \: \: \longrightarrow \: \: \bf P = Principal \: amount$

$\\ \: \: \: \longrightarrow \: \: \bf R = Rate$

$\\ \: \: \: \longrightarrow \: \: \bf n = Time$

• Now put the values –

$\\ \implies \bf (500 + 105)=500 \times \bigg(1+ \dfrac{10}{100} \bigg)^n$

$\\\implies \bf (500 + 105)=500 \times \bigg(1+ \dfrac{10}{100} \bigg)^n$

$\\\implies \bf 605=500 \times(1+ 0.1)^n$

$\\\implies \bf \dfrac{605}{500} = (1+ 0.1)^n$

$\\\implies \bf 1.21 = (1.1)^n$

$\\\implies \bf (1.1)^{2} = (1.1)^n$

$\\\implies \bf n = 2$

• Hence , Time is 2 year.

Answer 2

Answer:

Given :-

• The compound interest on Rs 500 at the rate of 10% per annum on Rs 105.

To Find :-

• What is the time.

Formula Used :-

$\clubsuit$ Amount Formula :

$\longrightarrow \sf\boxed{\bold{\pink{Amount =\: Principal + Interest}}}$

$\clubsuit$ Amount Of Compound Interest Formula :

$\longrightarrow \sf\boxed{\bold{\pink{A =\: \bigg(1 + \dfrac{r}{100}\bigg)^n}}}$

where,

• A = Amount
• r = Rate of Interest
• n = Time

Solution :-

First, we have to find the amount :

Given :

$\bigstar$ Principal = Rs 500

$\bigstar$ Interest = Rs 105

According to the question by using the formula we get,

$\implies \sf Amount =\: Rs\: 500 + Rs\: 105$

$\implies \sf\bold{\purple{Amount =\: Rs\: 605}}$

Hence, the amount is Rs 605.

Now, we have to find the time :

Let,

$\mapsto \sf\bold{\green{Time =\: n\: years}}$

Given :

$\bigstar$ Amount = Rs 605

$\bigstar$ Principal = Rs 500

$\bigstar$ Rate of Interest = 10%

According to the question by using the formula we get,

$\longrightarrow \sf 605 =\: 500\bigg(1 + \dfrac{1\cancel{0}}{10\cancel{0}}\bigg)^n$

$\longrightarrow \sf 605 =\: 500\bigg(1 + \dfrac{1}{10}\bigg)^n$

$\longrightarrow \sf 605 =\: 500\bigg(\dfrac{10 + 1}{10}\bigg)^n$

$\longrightarrow \sf 605 =\: 500\bigg(\dfrac{11}{10}\bigg)^n$

$\longrightarrow \sf \dfrac{\cancel{605}}{\cancel{500}} =\: \bigg(\dfrac{11}{10}\bigg)^n$

$\longrightarrow \sf \dfrac{121}{100} =\: \bigg(\dfrac{11}{10}\bigg)^n$

$\longrightarrow \sf \bigg(\dfrac{11}{10}\bigg)^2 =\: \bigg(\dfrac{11}{10}\bigg)^n$

$\longrightarrow \sf \bigg(\cancel{\dfrac{11}{10}}\bigg)^2 =\: \bigg(\cancel{\dfrac{11}{10}}\bigg)^n$

$\longrightarrow \sf 2 =\: n$

$\longrightarrow \sf\bold{\red{n =\: 2\: years}}$

$\therefore$ The time is 2 years.