Question

Find the difference between si and ci on rs.2500 for 2 years at 4% p.A., compound interest, interest being compounded semi-annually.

Answer 1

S.I=P×R×T/100

=2500×4×2/100

=200

Now rate is reckoned as half yearly

So,r =2℅ and time got doubled i.e 2×2=4 years

C.I=P(1+r/100)^t-P

= 2500(1+2/100)^4-2500

=2706.08-2500

=206.08

difference=C.I-S.I

=206.08-200

=6.08

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Answer 2

$\sf{\underline{For\:simple\:interest:}}$

Principal (P) = Rs. 2,500

Rate (R) = $4\%$ p.a.

Time (T) = 2 years

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{SI= \frac{PRT}{100}}}$

$\sf{\underline{By\:substituting\:the\:above\:values:}}$

$\implies$ $\sf{SI= \frac{PRT}{100}}$

$\implies$ $\sf{SI = \frac{2500 \times 4 \times 2}{100}}$

$\implies$ $\sf{SI = \frac{20000}{100}}$

$\implies$ $\sf{SI = Rs.\:200}$

$\sf{\underline{Therefore:}}$ SI: Rs. 200

$\sf{\underline{For\:compound\:interest:}}$

Principal (P) = Rs. 2500

Rate (R) = $4\%$ p.a. = $2\%$ per half year

Time (n) = 2 yrs = 4 half yrs

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{A= P(1 + \frac{R}{100} ) ^{n}}}$

$\sf{\underline{By\:substituting\:the\:above\:values:}}$

$\implies$ $\sf{A= P(1 + \frac{R}{100} ) ^{n}}$

$\implies$ $\sf{A= 2500(1 + \frac{2}{100} ) ^{4}}$

$\implies$ $\sf{A = 2500 \times ( \frac{51}{50} ) ^{4} }$

$\implies$ $\sf{A= Rs.\:2706.0804}$

$\sf{\underline{Now:}}$

$\boxed{\sf{CI = A - P}}$

$\implies$ $\sf{CI= 2706.0804 - 2500}$

$\implies$ $\sf{CI = Rs.\:206.0804}$

$\sf{\underline{Therefore:}}$ CI: Rs. 206.0804

$\sf{\underline{Now:}}$

$\sf{\underline{In\:order\:to\:find\:the\:difference:}}$

$\boxed{\sf{Difference = CI - SI}}$

$\implies$ $\sf{D= 206.0804 - 200}$

$\implies$ $\sf{D= Rs.\:6.08}$

$\sf{\underline{Answer:}}$ Rs. 6.08