The compound interest on a sum of money in
2 years and 4 years are Rs 5,460 and Rs 12,066.60
respectively. Find the difference between principal
and compound interest of 3 years.​

Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

[tex]C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P[/tex]

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

[tex]5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P[/tex]...(1)

[tex]12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P[/tex] ...(2)

Dividing equation (2) by (1), we have;

[tex]\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 }[/tex]

[tex]Let \ \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;[/tex]

[tex]\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 } = \dfrac{x^2 - 1}{x - 1}[/tex]

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x =  12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

[tex]\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}[/tex]

[tex]1 + \dfrac{r}{100} =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}[/tex]

We get

[tex]\dfrac{12,066.60}{5,460} =\dfrac{221}{100}[/tex]

[tex]\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}[/tex]

[tex]1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}[/tex]

[tex]\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1[/tex]

[tex]\dfrac{r}{100} = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1[/tex]

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

[tex]5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P[/tex]

[tex]5,460 = P \times \left ( 1 + 0.1\right ) ^{2} - P = P \times \left (\left ( 1 + 0.1\right ) ^{2} - 1\right) = P \times \dfrac{21}{100}[/tex]

[tex]P = \dfrac{100}{21} \times 5,460 = 26,000[/tex]

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

[tex]CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606[/tex]

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

The difference between the principal and the compound interest in three years, d = Rs 17,994.

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