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Mathematics

Assertion (A): ₹ 1,00,000 amounts to ₹ 1,21,000 in 1 year and to ₹1,46,410 in 2 years, compounded annually.

Reason (R): When the interest is compounded annually, then, we have A = P×(1+R100)nP \times \Big(1 + \dfrac{R}{100} \Big)^n

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Compound Interest

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Answer

Given,

P = ₹ 1,00,000

A = ₹ 1,21,000

n = 1 year

By formula,

A = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒121000=100000×(1+r100)1⇒121000100000=1+r100⇒r100=1.21−1⇒r=0.21×100⇒r=21%\Rightarrow 121000 = 100000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{121000}{100000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = 1.21 - 1 \\[1em] \Rightarrow r = 0.21 \times 100 \\[1em] \Rightarrow r = 21\%

Lets calculate the amount for 2 years with this rate of interest.

By formula,

A = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒A=100000×(1+21100)2⇒A=100000×(100+21100)2⇒A=100000×(121100)2⇒A=100000×(1.21)2⇒A=100000×1.4641⇒A=146410.\Rightarrow A = 100000 \times \Big(1 + \dfrac{21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\dfrac{100 + 21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\dfrac{121}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times (1.21)^2 \\[1em] \Rightarrow A = 100000 \times 1.4641 \\[1em] \Rightarrow A = 146410.

Assertion (A) is true.

Formula,

A = P×(1+R100)nP \times \Big(1 + \dfrac{R}{100} \Big)^n

Reason is true, as it is the formula to calculate the amount for compound interest when the interest is compounded annually.

Hence, option 3 is correct option.

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