Question

If the simple interest on a sum of money for 2 years at 5% per annum is RS. 40, what

will be the compound interest on the same sum at the same rate for the same time?k​

Answer 1

Answer:

Compound interest wil be Rs 41 on the same sum at the same rate for the same time

Step-by-step explanation:

Given that:

• Simple interest = Rs 40
• Time = 2 years
• Rate of interest = 5% per annum

Let us assume that the certain sum of money be x.

Formula to find simple interest:

S.I. = (p × r × t × 0.01)

Where,

• S.I. = Simple interest
• p = Principal/Sum of money
• r = Rate of interest
• t = Time
• 0.01 = 1/100

Finding the sum of money:

→ 40 = (x × 5 × 2 × 0.01)

→ 40 = 0.1x

→ x = 40/0.1

→ x = 400

∴ Principal = Rs 400

Formula to find Amount in compound interest:

A = p(1 + 0.01r)ᵗ

Where,

• A = Amount
• p = Principal/Sum of money
• r = Rate of interest
• t = Time
• 0.01 = 1/100

Finding the amount in compound interest:

→ A = 400(1 + 0.01 × 5)²

→ A = 400(1 + 0.05)²

→ A = 400(1.05)²

→ A = 400 × 1.05 × 1.05

→ A = 441

∴ Amount = Rs 441

Formula to find compound interest:

C.I. = A - p

Where,

• C.I. = Compound interest
• A = Amount
• p = Principal/Sum of money

Finding the compound interest:

→ C.I. = 441 - 400

→ C.I. = R41

∴ Compound interest = Rs 441

Answer 2

Given :-

• Simple Interest => ₹40
• Rate => 5%
• Time => 2 years

To Find :-

Compound Interest on same thing

Solution :-

As we know that

$\sf \: Principal = \dfrac{ SI \times 100}{Rate \times Time}$

$\sf \: Principal = \dfrac{40 \times 100}{5 \times 2}$

$\sf \: Principal = \dfrac{4000}{10}$

$\sf \: Principal = 400$

Now,

$\sf \: A = P \bigg(1 + \dfrac{R}{100} \bigg)^{n}$

$\sf \: A =400 \bigg(1 + \dfrac{5}{100} \bigg)^{2}$

$\sf \: A \: = 400 \bigg( \dfrac{100 + 5}{100} \bigg)^{2}$

$\sf \: A = 400 \times \dfrac{105}{100} \times \dfrac{105}{100}$

$\sf \: A = 400 \times \dfrac{21}{20} \times \dfrac{21}{20}$

$\sf \: A = \dfrac{176,400}{400}$

$\sf \: A = 441$

Now,

Finding

$\sf \: CI = Amount - Principal$

$\sf \: CI =441 - 400$

$\fbox{CI =41}$