Question

1. Find the difference between compound interest and simple interest if P = ₹.4000, R= 5 p.c.p.a and N= 2 yrs.​

Answer 1

Answer:

0.0185

Step-by-step explanation:

SIMPLE INTEREST = PRT / 100

= 7.4 × 5 × 2 / 100 = 0.74

COMPOUND INTEREST, A = P(1+R/100)^n

= 7.4 (1+5/100)^2 = 7.4 ×105/100 ×105/100 = 8.1585

CI = A-P = 8.1585-7.4 = 0.7585

DIFFERENCE

= 0.7585-0.74

=0.0185

hope it helps. mark as brainliest if my answer is best.

Answer 2

Before, finding the answer. Let's find out on how we can find the answer.

• First, we must find the Compound interest by the formula of

$\sf{ Amount = P \left( 1 + \dfrac{R}{100} \right) ^n}$

Compound Interest = Amount - Principal

• Next, we must find the Simple Interest by the formula of

$\sf \dfrac{p \times r \times t}{100}$

• At last, we must Subtract the Compound Interest and Simple Interest.

_________________________

Given :

• Principal = Rs. 4000
• Rate = 5%
• Time = 2 years

To find :

• Difference between compound interest and simple interest.

Solution :

We know that,

$\sf{ Amount = P \left( 1 + \dfrac{R}{100} \right) ^n}$

$\sf = \sf{ 4000 \left( 1 + \dfrac{ \cancel5}{ \cancel1 \cancel0 \cancel0} \right) ^2}$

$\sf{ = 4000 \left( 1 + \dfrac{1}{20} \right) ^2}$

$\sf{ = 4000 \left( \dfrac{20 + 1}{100} \right) ^2}$

$\sf{ = 4000 \left( \dfrac{21}{20} \right) ^2}$

$\sf{ Amount = 4000 \left( \dfrac{21}{20} \times \dfrac{21}{20} \right) ^2}$

$= 10 \times 21 \times 21$

$\sf = 4410$

Compound Interest = Principal - Amount

= Rs. 4000 - 4410

= Rs. 410

Hence, Compound Interest is Rs. 410.

$\sf \: Simple \: Interest = \dfrac{p \times r \times t}{100}$

$\sf \: = \dfrac{4000 \times5 \times 2 }{ 100 }$

$\sf = \dfrac{40000}{100}$

$= 400$

Hence, Simple Interest is Rs.400

Difference = Compound Interest - Simple Interest

= Rs. 4410 - 400

= Rs. 4010

Hence, Difference is Rs. 4010.