Question

sameera has taken a loan of 12500 at a rate of 12 p.c.p.a 3 years if the interest is compounded annually then how many rupees should he pay to clear his loan​

Answer 1

Answer:

The Answer here is 17,501.6

Step-by-step explanation:

So,

We know that the Principal Amount here is 12,500.

The rate of Interest is 12% Per Annum (one year)

And the Time period is 3 years.

So, this is the formula that I was taught:

                                 Principal (1 + Rate of interest/100) 3 (This is a degree)

So, 12,500 (1 + 12/100) 3

     12,500 (100/100 + 12/100) 3

The reason why "1" became "100/100" is because of the denominators. The denominators must be same.

Continuing:

                     

                       = 12,500 (112/100) 3

                = 12,500 * 112/100 * 112/100 * 112/100 (112/100 multiplies thrice because of degree)

                = 125 * 112 * 112 * 112 / 100 * 100 (12,500 became 125 due to cancellation.)

               = 17,50,16,000 / 10000

Again, cancellation:

              = 175016 / 10

Now using decimals:

                            17,501.6

We got it in decimals because "by 10" means the we move one decimal point.

Here's your answer!

Hope it helps!

Answer 2

Answer :-

• Sameera will have to pay Rs 17561.6 to clear her loan.

Given :-

• Sameera has taken a loan of Rs 12500 at a rate of 12 p.c.p.a for 3 years.
• The interest is compounded annually.

To find :-

• The amount he should pay to clear his loan.

Step-by-step explanation :-

• Here, the principal, rate and time have been given to us. It has also been given that the interest is compounded annually, so we have to find the amount and that will be the number of rupees Sameera will have to pay to clear his loan.

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Let's find the amount using the required formula!

We know that :-

$\underline{\boxed{\sf Amount = Principal \Bigg(1 + \dfrac{Rate}{100} \Bigg)^{Time}}}$

Here,

• Principal = Rs 12500.
• Rate = 12 p.c.p.a.
• Time = 3 years.

$\underline{\mathfrak{Substituting \:given \:values \:in \:the\: formula,}}$

$\rm Amount = 12500 \Bigg(1 + \dfrac{12}{100} \Bigg)^3$

Making 1 a fraction by taking 1 as the denominator,

$\rm Amount = 12500 \Bigg(\dfrac{1}{1} + \dfrac{12}{100} \Bigg)^3$

The LCM of 1 and 100 is 100, so adding the fractions using their denominators,

$\rm Amount = 12500 \Bigg(\dfrac{1 \times 100 + 12 \times 1}{100} \Bigg)^3$

On simplifying,

$\rm Amount = 12500 \Bigg(\dfrac{100 + 12}{100} \Bigg)^3$

Adding 12 to 100,

$\rm Amount = 12500 \Bigg(\dfrac{112}{100} \Bigg)^3$

Here the power is 3, so removing the brackets and multiplying 112/100 with itself 3 times,

$\rm Amount = 12500 \times \dfrac{112}{100} \times \dfrac{112}{100} \times \dfrac{112}{100}$

Let's multiply 112/100 with itself 3 times first.

$\rm Amount = 12500 \times \dfrac{112 \times 112 \times 112}{100 \times 100 \times 100}$

On multiplying,

$\rm Amount = 12500 \times \dfrac{1404928}{1000000}$

Cutting off the zeroes,

$\rm Amount = 125 \times \dfrac{1404928}{10000}$

Multiplying 125 with 1404928,

$\rm Amount = \dfrac{175616000}{10000}$

Dividing 175616000 by 10000,

$\overline{\boxed{\rm Amount = Rs \: 17561.6}}$

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• Hence, Sameera will have to pay Rs 17561.6 to clear his loan.