Question
Find the difference between the compound interest compounded yearly and half-yearly for the following:
Rs 20,000 for $1 \frac{1}{2}$ years at $16 \%$ p.a.
Rs 20,000 for $1 \frac{1}{2}$ years at $16 \%$ p.a.
✓
Answer
$P=\operatorname{Rs} 20,000 ; t=1 \frac{1}{2}$ years
When compounded yearly: $r=16 \%$ p.a.
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$A=R s 20000\left(1+\frac{16}{100}\right)\left(1+\frac{16}{100}\right)^{\frac{1}{2}}$
$=$ Rs $20000 \times 1.16 \times\left(1+\frac{1}{2} \times \frac{16}{100}\right)$
$=$ Rs $20,000 \times 1.16 \times 1.08$
$=$ Rs 25,056
C.I. $= A - P$
$=\operatorname{Rs}(25,056-20,000)$
$=$ Rs 5,056
When compounded half-yearly:
$A = P \left(1+\frac{ r }{100}\right)^{ n }$
$A=\operatorname{Rs} 20000\left(1+\frac{8}{100}\right)^3$
$=$ Rs $20,000 \times 1.08 \times 1.08 \times 1.08$ $=$ Rs $25,194.24$
C.I. $= A - P$
$=\operatorname{Rs}(25,194.24-20,000)$
$=$ Rs 5,194.24
Hence the difference in the interest=Rs $(5,194.24$ - $ 5,056)=\operatorname{Rs} 138.24 $
When compounded yearly: $r=16 \%$ p.a.
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$A=R s 20000\left(1+\frac{16}{100}\right)\left(1+\frac{16}{100}\right)^{\frac{1}{2}}$
$=$ Rs $20000 \times 1.16 \times\left(1+\frac{1}{2} \times \frac{16}{100}\right)$
$=$ Rs $20,000 \times 1.16 \times 1.08$
$=$ Rs 25,056
C.I. $= A - P$
$=\operatorname{Rs}(25,056-20,000)$
$=$ Rs 5,056
When compounded half-yearly:
$A = P \left(1+\frac{ r }{100}\right)^{ n }$
$A=\operatorname{Rs} 20000\left(1+\frac{8}{100}\right)^3$
$=$ Rs $20,000 \times 1.08 \times 1.08 \times 1.08$ $=$ Rs $25,194.24$
C.I. $= A - P$
$=\operatorname{Rs}(25,194.24-20,000)$
$=$ Rs 5,194.24
Hence the difference in the interest=Rs $(5,194.24$ - $ 5,056)=\operatorname{Rs} 138.24 $
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