Question

Ulculeu.
If the ratio of the compound interest (compounded annually) to the simple interest earned
on a principal at the same rate of interest per annum for 3 years is 331:300, what is the
rate of interest per annum?





with full explanation​

Answer 1

Let the Principal be Rs. P and same rate be R

Time = 3 years

• Compound interest = P{ ( 1 + R/100 )^T - 1 }
• Simple interest = PTR / 100

Ratio of CI and SI = 331 : 300

$\sf \Rightarrow \dfrac{P \bigg(\bigg(1 + \dfrac{R}{100} \bigg) ^{T} - 1 \bigg)}{ \dfrac{PTR}{100}} = \dfrac{331}{300}$

$\sf \Rightarrow \dfrac{ \bigg(1 + \dfrac{R}{100} \bigg) ^{3} - 1 }{ \dfrac{3R}{100}} = \dfrac{331}{300}$

$\sf \Rightarrow \bigg(1 + \dfrac{R}{100} \bigg) ^{3} - 1 = \dfrac{331}{300} \times \dfrac{3R}{100}$

$\sf \Rightarrow \bigg(1 + \dfrac{R}{100} \bigg) ^{3} - 1 = \dfrac{331R}{10000}$

Since a³ - b³ = (a - b)(a² + b² + ab)

$\sf \Rightarrow \bigg(1 + \dfrac{R}{100} - 1 \bigg) \bigg( \bigg(1 + \dfrac{R}{100} \bigg) ^{2} + 1 + 1 + \dfrac{R}{100} \bigg) = \dfrac{331R}{10000}$

$\sf \Rightarrow \dfrac{R}{100} \bigg( 1+ \bigg(\dfrac{R}{100} \bigg) ^{2} + \dfrac{2R}{10} + 1 + 1 + \dfrac{R}{100} \bigg) = \dfrac{331R}{10000}$

$\sf \Rightarrow \bigg(\dfrac{R}{100} \bigg) ^{2} + \dfrac{3R}{10} + 3 = \dfrac{331R}{10000} \times \dfrac{100}{R}$

$\sf \Rightarrow \dfrac{R^{2} }{10000} + \dfrac{3R}{100} + 3 = \dfrac{331}{100}$

Multiplying every term with 10000

$\sf \Rightarrow R^{2} + 300 R + 30000 = 33100$

$\sf \Rightarrow R^{2} + 300 R - 3100 = 0$

$\sf \Rightarrow R^{2} + 310R -1 0 R - 3100 = 0$

$\sf \Rightarrow R(R + 310) -1 0 (R + 310) = 0$

$\sf \Rightarrow (R + 310)(R + 10) = 0$

$\sf \Rightarrow R + 310 = 0 \qquad OR \qquad R - 10 = 0$

$\sf \Rightarrow R = - 310 \qquad OR \qquad R = 10$

Since Rate of interest cannot be negative

$\Rightarrow{ \boxed{ \sf R = 10 }}$

Hence Rate of interest is 10 %