Question

Find the compound interest on rupees 4000 for 2 1\2 years at 10% per annum.

with explanation of process.

Answer 1

Given Principal = 4000, Time n = (5/2) = 2.5 years, R = 10%.

We know that A = P(1 + r/100)^n

⇒ 4000(1 + 10/100)^2

⇒ 4000(11/10)^2

⇒ 4000(121/100)

⇒ 4840.

Interest after 2 years = 4840.

Interest after half year = 4840 * (10/100) * (1/2)

                                     = 242.

Amount After 2.5 years = 4840 + 242

                                       = 5082.

We know that C.I = A - P

                             = 5082 - 4000

                             = 1082.

Therefore, Compound Interest = 1082.

Hope it helps!

Answer 2

Questioñ:

Find the compound interest on rupees 4000 for 2 1\2 years at 10% per annum.

Solutioñ:

Given

• Principal(P) = Rs 4000

• Rate of Interest(r ) = 10%

• Time = 2.5 years Here first we take n = 2 years.

We know,

A = P( 1 + r /100)n Where A = Amount and n = number of years

So,

A = 4000 (11/10)²

A = 4000 ( 121 / 100 )

A = Rs 4840

• The Amount after two years = Rs 4840
• Now Principal = Rs 4840

By taking principal as ₹4840

Simple interest for last ½ year is = PRT/100

SI = (4840 × 10 × 1 / 2) / 100

SI = Rs.242

Amount after 2 ½ year = Rs 4840 + Rs 242

= Rs 5082

C.I = A - P

C.I = Rs 5082 - Rs 4000 = Rs1082

∴ The compound interest is Rs 1082.

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