Question

on what sum will the compound interest for 2 years at 4% p.a. be rs 5712?

Answer 1

Given CI = rs 5712
Time is 2 years
Rate is 4%

Now, applying formula CI = P((1+r/100)-1)

Let the principle be x

5712=x((1+4/100)^2)-1)

5712=x((26/25)^2-1)

5712=51/625 x

x = 5712×625/51

x= 70000

So the P is rs 70000

Answer 2

$\large\underline\pink{Given:-}$

• Compound Interest, C.I = Rs 5712
• Time, t = 2 year
• Rate, r = 4%

$\large\underline\pink{To find:-}$

• Fine the principal ....?

$\large\underline\pink{Solutions:-}$

• C.I = A - P

• $A \: \: = \: \: p \: {({1} \: + \: \frac{r}{100})}^{n}$

Compound Interest = Amount - Principal

• $\: \: \: \: \: \leadsto \: \: {5712} \: \: = \: \: A \: - \: P$

• $\: \: \: \: \: \leadsto \: \: A \: \: = \: \: {5712} \: + \: P$

• $\: \: \: \: \: \: \: A \: \: = \: \: p \: {({1} \: + \: \frac{r}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: + \: p \: \: = \: \: p \: {({1} \: + \: \frac{4}{100})}^{2}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: + \: p \: \: = \: \: p \: {({1} \: + \: \frac{1}{25})}^{2}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: + \: p \: \: = \: \: p \: {( \frac{{25} \: + \: {1}}{25})}^{2}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: + \: p \: \: = \: \: p \: {( \frac{26}{25})}^{2}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: + \: p \: \: = \: \: p \: {( \frac{676}{625})}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: + \: p \: \: = \: \: p \: \times \: \frac{676}{625}$

$\: \: \: \: \: \: \: \leadsto \: \: {625} \: {({5712} \: + \: p)} \: \: = \: \: {676p}$

$\: \: \: \: \: \: \: \leadsto \: \: {3570000} \: + \: {625p} \: \: = \: \: {676p}$

$\: \: \: \: \: \: \: \leadsto \: \: {3570000} \: \: = \: \: {676p} \: - \: {625p}$

$\: \: \: \: \: \: \: \leadsto \: \: {3570000} \: \: = \: \: {51p}$

$\: \: \: \: \: \: \: \leadsto \: \: {p} \: \: = \: \: \frac{3570000}{51}$

$\: \: \: \: \: \: \: {p} \: \: = \: \: Rs \: {70,000}$

Hence, sum will be Rs 70,000.