Question

Fabina borrows : 12,500 at 12% per annum for 3 years at simple interest and
Radha borrows the same amount for the same time period at 10% per annum,
compounded annually. Who pays more interest and by how much?​

Answer 1

Answer:

check solution in attachment

if any doubt...you can ask

Answer 2

Question :-⬇️

Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Given :-

Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually.

Find Out :-

Who pays more interest and by how much?

Solution :-

Let fabina pays x amount of interest and radha pays y amount of interest .

Now , we will see the amount paid by both after 3 years.

For finding amount paid by Fabina -

P = Rs. 12,500

N = 3 years.

R = 12%

✭ $\bf SI = \dfrac{P×R×N}{100}$ ✭

➙ $\sf SI =\: \dfrac{12500 \times 3 \times 12}{100}$

➙ $\sf SI = \: 125 \times 3 \times 12$

$\implies$ ${\small{\bold{\purple{\underline{SI =\: Rs \: 4500}}}}}$

For finding amount paid by Radha -

P = Rs. 12500

N = 3 years.

R = 10% which is compounded annually.

✭ $\bf A = \: P\bigg(1 + \dfrac{r}{100} \bigg)^{n}$ ✭

➙ $\sf A = \: 12500\bigg(1 + \dfrac{10}{100} \bigg) {}^{3}$

➙ $\sf A = \: 12500\bigg( \dfrac{11}{10} \bigg) {}^{3}$

➙ $\sf A = \: 12500 \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10}$

➙ $\sf A = \dfrac{125 \times 11 \times 11 \times 11}{10}$

$\implies$ ${\small{\bold{\purple{\underline{A =\: Rs \: 16637.5}}}}}$

∴ CI = Rs. 16637.5 - 12500

➙ Rs. 4137. 50

Therefore , amount paid by radha = Rs. 4137.50

So , difference of money

➙ Rs. 4500 - Rs. 4137.50

➙ Rs. 362.5

Henceforth, Fabina pays Rs. 362.5 more.

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\orange{\rule{45pt}{7pt}}\green{\rule{45pt}{7pt}}$