Question

14) If Rs 40000 amounts to Rs 46305 in 11/2 years, compound interest payable half yearly,
find the rate of interest per annum?​

Answer 1

★Given :  

• Amount (A) = Rs. 46,305
• Principal(P) = Rs. 40,000

•  $\sf Time = 1\dfrac{1}{2} yrs$

★To find :

• The rate of interest per annum.

★Solution :

$\sf Time = 1\dfrac{1}{2} yrs = \dfrac{3}{2} yrs \\\\\therefore 2n = 2\times \dfrac{3}{2} = 3yrs$

 

As the Interest is Payable Half Yearly,

 $\leadsto \underline{\boxed{\sf A = P\bigg(1+\frac{r}{200}\bigg)^{2n}}}$

$\implies \sf 46305 = 40000\bigg(1+\dfrac{r}{200}\bigg)^{3}\\\\\implies \sf \bigg(1+\dfrac{r}{200}\bigg)^{3} = \dfrac{46305}{40000} \\\\\implies \sf \bigg(1+\dfrac{r}{200}\bigg)^3=\bigg( \dfrac{21}{20} \bigg)^3\\\\\implies \sf 1+\dfrac{r}{200} = \dfrac{21}{20} \\ \\\implies \sf \dfrac{r}{200} =\dfrac{21}{20} \\\\\implies \sf \underline{\boxed{\bold{ r = 10\% }}}$

Hence the rate of interest is 10%.

_____________

Answer 2

Given

• Principal (P) = Rs 40000 /-
• Amount (A) = Rs 46305 /-
• Time period (n) = 1 ½ years

To find

• Rate of interest per annum

Solution

According to the question, compound interest is payable half yearly, so when it is half yearly, we have time period = 2n, n is time period.

• T = 1 ½ years
• 2n = 2 × ³⁄₂
• 2n = 3 years

The interest is payed half yearly, so :-

• A = P (1 + r/200)²ⁿ
• 46305 = 40000 (1 + r/200)³
• 46305/40000 = (1 + r/200)³
• 21³/20³ = (1 + r/200)³
• (21/20)³ = (1 + r/100)³
• 1 + r/100 = 21/20
• r/100 = 21/20
• r = 10

Hence, the rate of interest per annum (r) is 10 %