Question

Find the compound interest for P = z 4000, R-
5%, P.a and n = 2years.​

Answer 1

Given:

$\small\rightarrow$ Principal = Rs. 4000

$\small\rightarrow$ Rate of interest = 5 %

$\small\rightarrow$ Time = 2 years

To Find:

$\small\rightarrow$ Compound Interest = ?

Solution:

we have:

• Principal = Rs. 4000
• Rate = 5 %
• Time = 2 years

As we know that,

$\: \: \sf compound \: interest = p ( {(1 + \frac{r}{100} )}^{n} - 1)$

Now put on formula

$\: \: \: \sf \rightarrow \: c.i = 4000( {(1 + \frac{5}{100} )}^{2} - 1) \\ \\ \: \: \sf \rightarrow \: c.i = 4000( {(1 + \cancel\frac{5}{100}) }^{2} - 1) \\ \\ \: \: \: \sf \rightarrow \: c.i = 4000( {(1 + \frac{1}{20}) }^{2} - 1) \\ \\ \: \: \sf \rightarrow \: c.i = 4000( {( \frac{21}{20}) }^{2} - 1) \\ \\ \: \: \sf \rightarrow \: c .i = 4000( \frac{441}{400} - 1) \\ \\ \: \: \sf \rightarrow \: c.i = 4000 \times \frac{41} { 400} \\ \\ \: \: \sf \rightarrow \: c.i = 410 \: rupees$

Therefore, Compound Interest is 410 rupees.

Answer 2

Step-by-step explanation:

⤵️

ci=410 rs.

hope this helps you mate ✨