Question

What is the difference between amounts for 2 years and 3 years for Rs. 500 at 10% compound interest?

1️⃣ ₹.665.50
2️⃣ ₹ 60
3️⃣ ₹ 60.50
4️⃣ ₹ 605​

Answer 1

Formula of CI and A:-

${\pmb {\red{CI = P( 1 + \frac{r}{100} )^n - P }}}$

${ \pmb { \red{Amount = P ( 1 + \frac{r}{100} )^n }}}$

Given,

$\sf \: P = ₹ 500 \\ \sf R = 10 \: \%$

$\sf \: T_1 = 2 \: years$

$\sf \: T_2 = 3 \: years$

$\rm \underline\blue{1st Amount = P ( 1 + \frac{r}{100} )^n}$

$\implies500 ( 1 + \frac{10}{100} )^2 \\ \\ \implies 500( \frac{100 + 10}{100} ) ^{2} \\ \\ \implies \: 500( \frac{110}{100} ) ^{2} \\ \\ \implies \: 5 \cancel{00} \times \frac{11 \cancel0}{1 \cancel0 \cancel0} \times \frac{11 \cancel0}{1 \cancel{00}} \\ \\ \implies5 \times 11 \times 11 \\ \\ \therefore \: ₹605$

$\rm \underline\blue{2nd Amount = P ( 1 + \frac{r}{100} )^n}$

$\implies500 ( 1 + \frac{10}{100} )^3 \\ \\ \implies 500( \frac{100 + 10}{100} ) ^{3} \\ \\ \implies \: 500( \frac{110}{100} ) ^{3} \\ \\ \implies \: 5 \cancel{00} \times \frac{11 \cancel0}{1 \cancel0 \cancel0} \times \frac{11 \cancel0}{1 \cancel{00}} \: \times \: \frac{11 \cancel0}{10 \cancel0} \\ \\ \implies5 \times 11 \times 11 \times \frac{11}{10} \\ \\ \therefore \: ₹ \frac{6655}{10} = ₹665.5$

Henceforth,

Difference = 2nd Amount - 1st Amount

⟹ ₹ 665.5 - ₹ 605

$\therefore$ ₹60.5

So, Option 3️⃣ ₹60.50 $\bf\red{ [Ans]}$

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Note :-

• CI = Compound Interest
• A = Amount

$\large \fbox \red {Hope \: it \: helps \: you}$

Answer 2

Answer:

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