Question

(b) 18,000 for 25 years at 10% per annum compounded annually.
​

Answer 1

Step-by-step explanation:

\begin{gathered}A = 18000 ( 1 + \frac{10}{100})^{2}\\= 18000 \times ( 1.1)^{2} \\= Rs\:21780\end{gathered}

A=18000(1+

100

10

)

2

=18000×(1.1)

2

=Rs21780

\begin{gathered}Compound \: Interest \:for \: 2 \: years\\ = A - P\\= Rs \: 21780 - Rs \: 18000 \\= Rs\: 3780 \: ---(1)\end{gathered}

CompoundInterestfor2years

=A−P

=Rs21780−Rs18000

=Rs3780−−−(1)

\begin{gathered}Now, principal \:for \: next \: 6\: months = A = Rs\:21780 , \\Time (T) = \frac{1}{2} \: years ,\\Rate \:of \: Interest (R) = 10\%\end{gathered}

Now,principalfornext6months=A=Rs21780,

Time(T)=

2

1

years,

RateofInterest(R)=10%

\begin{gathered}Simple \: Interest (I) = \frac{PTR}{100}\\= \frac{ 21780\times \frac{1}{2} \times 10}{100}\\= Rs\:1089 \: ---(2)\end{gathered}

SimpleInterest(I)=

100

PTR

=

100

21780×

2

1

×10

=Rs1089−−−(2)

\begin{gathered}Total \: Interest = Rs \: 3780 + Rs \: 1089\\= Rs\:4869\end{gathered}

TotalInterest=Rs3780+Rs1089

=Rs4869