Question

Find the compound interest on 40000 INR for 1.5 years, at a rate of 10% per annum, being compounded half-yearly.​

Answer 1

Answer:

• The compound interest is 630.50 INR.

Step-by-step explanation:

To Find:

• The compound interest on 40000 INR for 1.5 years, at a rate of 10% per annum, being compounded half-yearly.

Formula used:

In compound interest.

• C.I. = P(1 + R/100)ᵀ - P

Where,

• P = Principal
• R = Rate of interest
• T = Time period
• C.I. = Compound interest

We have:

• Principal = 40000 INR
• Being compounded half-yearly.
• Time period = 1.5 years = 3 half-year
• Rate = 10/2 % = 5%

Finding the compound interest:

⟶ C.I. = 4000(1 + 5/100)³ - 4000

⟶ C.I. = 4000(1 + 0.05)³ - 4000

⟶ C.I. = 4000(1.05)³ - 4000

⟶ C.I. = 4000(1.157625) - 4000

⟶ C.I. = 4630.50 - 4000

⟶ C.I. = 630.50

∴ Compound interest = 630.50 INR

Answer 2

Correct question:-

Find the compound interest on 4000 INR for 1.5 years, at a rate of 10% per annum, being compounded half-yearly.

Given:-

• Amount (P) = 4000 INR
• Rate (R) = 10%
• Time (T) = 1.5 years

To Find:-

• Compound interest (C.I.) which is Compounded half yearly.

Formula used:-

• C.I. = $P(1+\dfrac{R}{100})^{T}$ - Amount

Solution:-

As given in the question, that it is compounded half yearly. So,

• Time {T} = 3 years

• Rate {R} = 5%

Putting the values in the Formula of Compound Interest,

C.I.$=\:P(1+\dfrac{R}{100})^{T}$ - 4000

$=\:4000(1+\dfrac{5}{100})^{3}$ - 4000

$=\:4000(\dfrac{100+5}{100})^{3}$ - 4000

$=\:4000(\dfrac{105}{100})^{3}$ - 4000

$=\:4000×(1.05)^{3}$ - 4000

$=\:4000×1.157625$ - 4000

$=\:4630.5$ - 4000

$=\:630.5$

Hence, the Compound Interest on 4000 INR for 1.5 years, at a rate of 10% per annum, being compounded half-yearly is 630.5 INR.

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