Question

Solve It .....!

1 - Find the compound interest on Rs. 6250 for 1 year at 8% per annum , interest being compounded half Yearly.

2- What sum Will amount to Rs. 13310 in 2 years at 10% p.a. of compounded interest, interest being compounded Annually?​

Answer 1

1 . $\huge{\bold{Given:-}}$

Principal = 6250

Rate = 8 % per annum

Time , n = 1 year

Since , it is given compounded half yearly

$\huge{\bold{ Solution }}$

Therefore,

$Amount \: = P {[1 + \frac{ \frac{R}{2} }{100}]}^{2n} \\ \\ A = 6250[ {1 + \frac{ \frac{8}{2} }{100}] }^{2 \times 1} \\ \\ A= 6250[ {1 + \frac{4}{100}] }^{2} \\ \\ A= 6250[{1 + \frac{1}{25}] }^{2} \\ \\ A= 6250[ { \frac{26}{25}] }^{2} \\ \\ A= 6250 \times \frac{26}{25} \times \frac{26}{25} \\ \\ A = 10 \times 26 \times 26 \\ \\ A= 6760 \\ \\ {\purple{\boxed{\large{\bold{Amount \: = Rs.6760}}}}} \\ \\ {\red{\boxed{\large{\bold{Compound \: interest = A - P}}}}} \\ \\ C.I= 6760 - 6250 \\ \\ CI = 510 \\ \\ {\blue{\boxed{\large{\bold{C.I= Rs.510}}}}}$

Hope its helpful ❤

Answer 2

Answer:

$\Large{\underline{\bf{Solution}}}}$

$\fbox{Question 1}$

Solution :-

$\huge\red{Given -}$

Principal Value - ₹6250

Time - 1 year

Rate - 8%

$\green{Compounded Half yearly = ? }$

$\fbox{solution}$

$amount = {p(1 + \frac{r}{200}})^{2}$

$a = 6250 \times {(1 + \frac{8}{200} })^{2}$

$a = 6250 \times \frac{26}{25} \times \frac{26}{25}$

$= 6700$

$\Blue{Compound Interest =A - P}$

Compounded Interest ⇒Rs.6760 - Rs.6250

⇒Rs. 510

$\fbox{Question 2}$

Solution:-

$\green{Let the required sum be Rs. P. Then ,}$

$13310 = p \times( {1 + \frac{10}{100} })^{2}$

Or,

$13310 = p \times \frac{11}{10} \times \frac{11}{10}$

⇒₹ 11,000