Question

The difference between the simple and compound
interest on 12,500 for 3 years at 4% is​

Answer 1

Answer:-

The difference between the compound interest and the simple interest is Rs. 381.60

Explanation:-

Given :

Principle = Rs. 12,500

Time = 3 years

Rate = 4%

To find :

Difference between the Simple interest and Compound interest.

Solution :

We know that,

$s.i = \dfrac{p \times r \times t}{100}$

$= > s.i = rs. \: \dfrac{12500 \times 3 \times 4}{100}$

$= > s.i = rs. \: 1500$

Hence, The S.I is Rs. 1500

Now, Compound interest:-

Principle for the second year => Rs. (12500 + 1500) = Rs. 14000

Time = 3 years

Rate = 4 %

$= > s.i \: = rs. \: \dfrac{14000 \times 3 \times 4}{100}$

$= > s.i \: = rs. \: 1680$

Therefore, Principle for second year => Rs. (14000+1680) = Rs. 15680

Now, C.I for third year :-

$= rs. \: \dfrac{15680 \times 3 \times 4}{100}$

$= rs. \: 1881.60$

Hence, the C.I is Rs. 1881.60

Now, The difference = (C.I - S.I)

= Rs. (1881.60 - 1500.00)

= Rs. 381.60

Hence, the difference is Rs. 381.60

Answer 2

Answer =

₹60.8

Given:-

P = ₹ 12,500

R = 4%

T = 3 years

To find :-

The difference between simple interest and compound interest.

Solution :-

Let first calculate the S. I

$\boxed{\sf{S.I = \dfrac {p × r ×t }{100}}}$

Put the given value,

$S. I = \dfrac { 12,500 × 4 × 3 }{100}$

$S. I. = 1500$

Now, Calculate C. I

$\boxed{\sf{C.I = p (1 + \dfrac{R}{100})^t-p}}$

put the given value.

$C. I = 12,500 (1 + \dfrac{4}{100})^3-p$

$C. I= 12,500 (1 + \dfrac {1}{25})^3-p$

$C. I = 12,500 (\dfrac{25+1}{25})^3-p$

$C. I = 12,500 (\dfrac{26}{25})^3-p$

$C. I = 12,500 × \dfrac {17,576}{15,625}-p$

$C. I = 12,500 × 1.124-p$

$C. I = 14,060.8-p$

C. I = 14,060.8 - 12,500

C. I = 1560.8

Now,

According to question :-

= C. I - S. I

= 1560.8 - 1500

= ₹ 60.8