Question

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Radha invested ₹1600 on compound interest for 2 years. She received ₹1764 after the specified period. Find the rate of interest per annum.

A-7%
B-6%
C-5%
D-4%
​

Answer 1

Step-by-step explanation:

Given,

A = 1764, P = 1600, n = 2 yrs.,

R =?

For compound interest

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

1764 = 1600(1 + R/100)^2

1764/1600 = (1 + R/100)^2

(1 + R/100) = 42/40

R/100 = (42 - 40)/40

R = (2/40)*100

R = 5%

Therefore, rate of interest = 5%

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Answer 2

Answer:

Given :-

• Radha invested ₹1600 on compound interest for 2 years. She received ₹1764 after the specified period.

To Find :-

• What is the rate of interest per annum.

Formula Used :-

$\clubsuit$ Amount Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}\: \: \: \bigstar$

where,

• A = Amount
• P = Principal
• r = Rate of Interest
• n = Time Period

Solution :-

Given :

• Principal (P) = ₹1600
• Amount (A) = ₹1764
• Time Period (n) = 2 years

According to the question by using the formula we get,

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

$\implies \sf 1764 =\: 1600\bigg(1 + \dfrac{r}{100}\bigg)^2$

$\implies \sf \dfrac{1764}{1600} =\: \bigg(1 + \dfrac{r}{100}\bigg)^2$

$\implies \sf \bigg(\dfrac{42}{40}\bigg)^2 =\: \bigg(1 + \dfrac{r}{100}\bigg)^2$

$\implies \sf \dfrac{42}{40} =\: 1 + \dfrac{r}{100}$

$\implies \sf \dfrac{42}{40} - 1 =\: \dfrac{r}{100}$

$\implies \sf \dfrac{42 - 40}{40} =\: \dfrac{r}{100}$

$\implies \sf \dfrac{2}{40} =\: \dfrac{r}{100}$

By doing cross multiplication we get,

$\implies \sf r \times 40 =\: 2(100)$

$\implies \sf 40r =\: 2 \times 100$

$\implies \sf 40r =\: 200$

$\implies \sf r =\: \dfrac{\cancel{200}}{\cancel{40}}$

$\implies \sf r =\: \dfrac{5}{1}$

$\implies \sf\bold{\red{r =\: 5\%}}$

$\sf\bold{\purple{\underline{\therefore\: The\: rate\: of\: interest\: per\: annum\: is\: 5\%\: .}}}$

Hence, the correct options is option no (C) 5% .

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