Question

Find the compound interest on Rs. 2,500 for 15 months at 8% compounded quarterly.

Answer 1

Answer:

Given :-

• A sum of Rs 2500 for 15 months at 8% compound quarterly.

To Find :-

• What is the compound interest.

Formula Used :-

$\clubsuit$ Amount formula when the interest is compounded quarterly :

$\bigstar \: \: \: \: \sf\boxed{\bold{\pink{A =\: P\Bigg[1 + \dfrac{\dfrac{r}{4}}{100}\Bigg]^{4n}}}}\: \: \: \: \: \bigstar$

where,

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

$\clubsuit$ Compound Interest Formula :

$\bigstar \: \: \: \: \sf\boxed{\bold{\pink{Compound\: Interest =\: A - P}}}\: \: \: \: \bigstar$

where,

• A = Amount
• P = Principal

Solution :-

Given :

✫ Principal = Rs 2500

✫ Time = $\sf 15\: months =\: \dfrac{15}{12}\: yr =\: \bf{\dfrac{5}{4}\: yr}$

✫ Rate of Interest = 8%

✰ First we have to find the amount :

$\mapsto \bf A =\: P\Bigg[1 + \dfrac{\dfrac{r}{4}}{100}\Bigg]^{4n}$

$\mapsto \sf A =\: 2500\Bigg[1 + \dfrac{\dfrac{8}{4}}{100}\Bigg]^{(\cancel{4} \times \frac{5}{\cancel{4}})}$

$\mapsto \sf A =\: 2500\bigg[1 + \dfrac{\cancel{8}}{\cancel{4}} \times \dfrac{1}{100}\bigg]^5$

$\mapsto \sf A =\: 2500\bigg[1 + \dfrac{2}{1} \times \dfrac{1}{100}\bigg]^5$

$\mapsto \sf A =\: 2500\bigg[1 + \dfrac{2}{100}\bigg]^5$

$\mapsto \sf A =\: 2500\bigg[\dfrac{100 + 2}{100}\bigg]^5$

$\mapsto \sf A =\: 2500\bigg[\dfrac{102}{100}\bigg]^5$

$\mapsto \sf A =\: 2500[1.02]^5$

$\mapsto \sf A =\: 2500[1.02 \times 1.02 \times 1.02 \times 1.02 \times 1.02]$

$\mapsto \sf A =\: 2500 \times 1.104$

$\mapsto \sf\bold{\purple{A =\: Rs\: 2760}}$

✰ Now, we have to find the compound interest :

Given :

• Amount = Rs 2760
• Principal = Rs 2500

According to the question by using the formula we get,

$\dashrightarrow \bf Compound\: Interest =\: A - P$

$\dashrightarrow \sf Compound\: Interest =\: Rs\: 2760 - Rs\: 2500$

$\dashrightarrow \sf\bold{\red{Compound\: Interest =\: Rs\: 260}}$

$\therefore$ The compound interest is Rs 260 .

Answer 2

$\sf\red{A= P {(1 + \frac{ \frac{r}{4} }{100} )}^{4n} }$

$\implies$$\sf{A=2500 {(1 + \frac{ \frac{8}{4} }{100} )}^{(4 \times \frac{5}{4} )} }$

$\implies$$\sf{A=2500{(1 + \frac{8}{4} \times \frac{1}{100}) }^{5} }$

$\implies$$\sf{A = 2500 {(1 + \frac{2}{1} \times \frac{1}{100} )}^{5} }$

$\implies$$\sf{A=2500 {(1 + \frac{2}{100}) }^{5} }$

$\implies$$\sf{A=2500 {( \frac{100 + 2}{100} )}^{5}}$

$\implies$$\sf{A=2500{( \frac{102}{100} )}^{5} }$

$\implies$$\sf{A=2500 {(1.02)}^{5} }$

$\implies$$\sf{A=2500(1.02×1.02×1.02×1.02×1.02)}$

$\implies$$\sf{2500×1.104}$

$\implies$$\sf{₹2,760}$

$\sf\red{Compound \: Interest = A-P}$

$\implies$$\sf{₹2,760-₹2,500}$

$\implies$$\sf{₹260}$