Question

find the amount and compound interest compounded annually on
20000at 9% p. a. for 2 years​

Answer 1

Answer:

Rs.23762

Step-by-step explanation:

$P(1+\frac{R}{100})^{2}\\P=20000\\R=9%\\T=2 years\\=20000(1+\frac{9}{100})^{2} \\=20000(\frac{109}{100})^{2} \\=20000* \frac{109}{100}*\frac{109}{100}\\=20000*\frac{11881}{10000}\\=2*11881\\=23762$

Answer 2

Given:-

• Principal = 20000
• Rate = 9%
• Time = 2 years

To Find:-

• Amount and Compound Interest after 2 years.

Solution:-

We know,

The formula to find Amount when the interest is compounded annually is:-

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

Hence,

$\sf{A = 20000\bigg(1+\dfrac{9}{100}\bigg)^2}$

= $\sf{A = 20000\bigg(\dfrac{100+9}{100}\bigg)^2}$

= $\sf{A = 20000\bigg(\dfrac{109}{100}\bigg)^2}$

= $\sf{A = 20000\bigg(\dfrac{109}{100}\bigg)\bigg(\dfrac{109}{100}\bigg)}$

= $\sf{A = 23762}$

Therefore, amount will be Rs.23762.

Now,

CI = Amount - Principal

CI = 23762 - 20000

CI = Rs.3762.

Therefore, CI after 2 years will be Rs.3762.

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Explore more!!!

Formula to calculate amount when the interest is compounded half-yearly is:-

• A = $\sf{P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}$

Formula to calculate amount when the interest is compounded quarterly is:-

• A = $\sf{P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}$

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Note:-

Here,

• A = Amount
• P = Principal
• R = Rate
• n = Time
• CI = Compound Interest.

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