Question

sameerrao has taken a loan of rupees 12500 at a rate of 12 p.c.p.a for 3 years .if the interest is compounded annually then how many ruppes should he pay to clear his loan?
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Answer 1

Answer:

Principal= 12500

Rate of intrest = 12% p.a

N=3 year

therefore amount= P(1+R/100)^N

= 12500(1+12/100)^3

=12500(1+25+3/25)^3

=12500*28/25*28/25*28/25

=20*28*28*28/25

=4*28*28*28/5

=17561.6

Therefore he paid rupees 17561.6

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Answer 2

Answer:

Given :-

• Sameerrao has taken a loan of Rs 12500 at the rate of 12% per annum for 3 years.
• The interest is compounded annually.

To Find :-

• How much amount should he pay to clean his loan.

Formula Used :-

$\clubsuit$ Amount Formula :

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

where,

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

Solution :-

Given :

• Principal = Rs 12500
• Rate of Interest = 12% per annum
• Time Period = 3 years

According to the question by using the formula we get,

$\implies \sf\bold{\underline{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}$

$\implies \sf A =\: 12500\bigg(1 + \dfrac{12}{100}\bigg)^3$

$\implies \sf A =\: 12500\bigg(\dfrac{100 \times 1 + 12}{100}\bigg)^3$

$\implies \sf A =\: 12500\bigg(\dfrac{100 + 12}{100}\bigg)^3$

$\implies \sf A =\: 12500\bigg(\dfrac{112}{100}\bigg)^3$

$\implies \sf A =\: 12500\bigg(\dfrac{112}{100} \times \dfrac{112}{100} \times \dfrac{112}{100}\bigg)$

$\implies \sf A =\: 12500\bigg(\dfrac{1404928}{1000000}\bigg)$

$\implies \sf A =\: \dfrac{12500 \times 1404928}{1000000}$

$\implies \sf A =\: \dfrac{175616\cancel{00000}}{10\cancel{00000}}$

$\implies \sf A =\: \dfrac{175616}{10}$

$\implies \sf\bold{A =\: Rs\: 17561.60}$

$\therefore$ Sameerrao has to pay Rs 17561.60 to clear his loan.