Question

$\sf\large\underbrace{Question:-}$
A sum of money is invested at half yearly compound interest rate 5% If the difference of principals at the end of 6 and 12 months is Rs126.
Find (i) the sum of money invested
(ii)Amount at end of 1(1/2)years(one whole one by two years)​

Answer 1

i) Let

the principal be P. Rate of interest per half year = 5% Amount at the end of 6 months = P(1 + 5/100)

Amount at the end of 1 year = P(1 + 5/100)2

Difference between the amounts = Formula = P(1+R) power time

Rs 126 P(1 + 5/100)2 - P(1 + 5/100)

= 126 P(21/20)2 - P(21/20)

= 126 P(441/400) - P(21/20) = 126 (441P - 420P) / 400 = 126 21P/400 = 126 P = 2400

So, the sum of money invested is Rs 2400 .

ii) Amount at the end of 1 and half year =

2400(1 + 5/100)3

= 2400 (21/20)3

= 2400 (9261/8000)

= Rs 2778.30

Hope my answer help you

Answer 2

Let the amount invested be P.

The rate is given 5% healf yearly, that means the amount invested gets added up by its 5% after 6 months or 1/2 years. It is equivalent to the amount getting invested at 10% rate per annum.

• $r=10\%$

• $R=\dfrac{r}{2}=5\%$

Let P₁ be the amount after 6 months or 1/2 years. Here,

• $t_1=\dfrac{1}{2}$

• $T_1=2t_1=1$

Then,

$\longrightarrow P_1=P\left[1+\dfrac{R}{100}\right]^{T_1}$

$\longrightarrow P_1=P\left[1+\dfrac{5}{100}\right]^1$

$\longrightarrow P_1=\dfrac{21P}{20}$

Let P₂ be the amount after 12 months or 1 year. Here,

• $t_2=1$

• $T_2=2t_2=2$

Then,

$\longrightarrow P_2=P\left[1+\dfrac{R}{100}\right]^{T_2}$

$\longrightarrow P_2=P\left[1+\dfrac{5}{100}\right]^2$

$\longrightarrow P_2=\dfrac{441P}{400}$

Given that,

$\longrightarrow P_2-P_1=Rs.\,126.00$

$\longrightarrow\dfrac{441P}{400}-\dfrac{21P}{20}=Rs.\,126.00$

$\longrightarrow\dfrac{21P}{400}=Rs.\,126.00$

$\longrightarrow P=Rs.\,126.00\times\dfrac{400}{21}$

$\longrightarrow\underline{\underline{P=Rs.\,2400.00}}$

Hence the sum of money invested is Rs. 2400.00.

Let P₃ be the amount after 3/2 years. Here,

• $t_3=\dfrac{3}{2}$

• $T_3=2t_3=3$

Then,

$\longrightarrow P_3=P\left[1+\dfrac{R}{100}\right]^{T_3}$

$\longrightarrow P_3=2400\left[1+\dfrac{5}{100}\right]^3$

$\longrightarrow P_3=Rs.\,2400.00\times\dfrac{9261}{8000}$

$\longrightarrow\underline{\underline{P_3=Rs.\,2778.30}}$

Hence the amound at the end of 3/2 years is Rs. 2778.30.