Question

A sum of ₹4000 is invested for 3 years at 5% CI annually. Calculate the CI earned in 3 years​

Answer 1

★ Solution :-

First, we should find the amount.

Amount :-

$\longrightarrow \sf Amount = Principle \bigg(1 + \dfrac{Rate}{100} \bigg)^{Time}$

$\longrightarrow \sf 4000 \bigg(1 + \dfrac{5}{100} \bigg)^{3}$

$\longrightarrow \sf 4000 \bigg(1 + \dfrac{1}{20} \bigg)^{3}$

$\longrightarrow \sf 4000 \bigg(\dfrac{20 + 1}{20} \bigg)^{3}$

$\longrightarrow \sf 4000 \bigg(\dfrac{21}{20} \bigg)^{3}$

$\longrightarrow \sf 4000 \bigg(\dfrac{21^3}{20^3} \bigg)$

$\longrightarrow \sf 4000 \bigg(\dfrac{9261}{8000} \bigg)$

$\longrightarrow \sf 1 \bigg(\dfrac{9261}{2} \bigg)$

$\longrightarrow \sf \cancel \dfrac{9261}{2} = 4630.5$

Now, we can find the compound interest.

Compound interest :-

$\longrightarrow \sf CI = Amount - Principle$

$\longrightarrow \sf 4630.5 - 4000$

$\longrightarrow \sf Rs.630.5$

Therefore, the compound interest is ₹630.5.

Answer 2

Answer:

Given :-

• A sum of Rs 4000 is invested for 3 years at 5% per annum in compound interest annually.

To Find :-

• What is the compound interest in 3 years.

Formula Used :-

$\bigstar$ Amount Formula :

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

where,

• A = Amount
• P = Principal
• r = Rate of Interest
• n = Time Taken

$\bigstar$ Compound Interest or C.I Formula :

$\longrightarrow \sf\boxed{\bold{\pink{C.I =\: A - P}}}$

where,

• C.I = Compound Interest
• A = Amount
• P = Principal

Solution :-

At, first we have to find the amount :

Given :

• Principal = Rs 4000
• Rate of Interest = 5% per annum
• Time = 3 years

According to the question by using the formula we get,

$\dashrightarrow \sf A =\: 4000\bigg[1 + \dfrac{5}{100}\bigg]^3$

$\dashrightarrow \sf A =\: 4000\bigg[\dfrac{100 + 5}{100}\bigg]^3$

$\dashrightarrow \sf A =\: 4000\bigg[\dfrac{105}{100}\bigg]^3$

$\dashrightarrow \sf A =\: 4000\bigg[\dfrac{105}{100} \times \dfrac{105}{100} \times \dfrac{105}{100}\bigg]$

$\dashrightarrow \sf A =\: 4000\bigg[\dfrac{105 \times 105 \times 105}{100 \times 100 \times 100}\bigg]$

$\dashrightarrow \sf A =\: 4000\bigg[\dfrac{1157625}{1000000}\bigg]$

$\dashrightarrow \sf A =\: \bigg[4\cancel{000} \times \dfrac{1157625}{1000\cancel{000}}\bigg]$

$\dashrightarrow \sf A =\: \bigg[4 \times \dfrac{1157625}{1000}\bigg]$

$\dashrightarrow \sf A =\: \bigg[\dfrac{4 \times 1157625}{1000}\bigg]$

$\dashrightarrow \sf A =\: \bigg[\dfrac{46305\cancel{00}}{10\cancel{00}}\bigg]$

$\dashrightarrow \sf A =\: \bigg[\dfrac{\cancel{46305}}{\cancel{10}}\bigg]$

$\dashrightarrow \sf\bold{\purple{A =\: Rs\: 4630.5}}$

Hence, the amount is Rs 4630.5 .

Now, we have to find the compound interest :

Given :

• Amount (A) = Rs 4630.5
• Principal (P) = Rs 4000

According to the question by using the formula we get,

$\leadsto \sf C.I =\: Rs\: 4630.5 - Rs\: 4000$

$\leadsto \sf\bold{\red{C.I =\: Rs\: 630.5}}$

$\therefore$ The compound interest in 3 years is Rs 630.5 .