5 Marks Questions

Question 15 Marks

(i) At what rate of interest will the present value of a perpetuity of ₹ 300 payable at the end of each quarter be ₹ 24000 ?
(ii) What sum of money invested now could establish a scholarship of ₹ 5000 which is to be awarded at the end of every year forever, if money is worth $8 \%$ per annum.
(iii) Rehaan invested ₹ 7000 in a Term Deposit Scheme that fetches interest $7.5 \%$ per annum compounded semi-annually. What will be the interest after 3 years? Given $(1.0375)^6=1.2472$

Answer

(i) Let the rate of investment be $r \%$ per annum, then
$
i=\frac{r}{4 \times 100}=\frac{r}{400}
$
Given, perpetuity, R=₹ 300
and payment, P=₹ 24000
$
\begin{array}{rlrl}
\text { Using, } & P & =\frac{R}{i} \\
\Rightarrow & & i & =\frac{R}{P} \\
\Rightarrow & \frac{r}{400} & =\frac{300}{24000} \\
\Rightarrow & & r & =\frac{300 \times 400}{24000}=5 \%
\end{array}
$
Hence, rate of interest $=5 \%$ per annum.
(ii) It is a flat perpetuity, so Present value of perpetuity
$
\begin{array}{l}
=\frac{\text { Cash flow }}{\text { Interest rate }} \\
=\frac{5000}{\frac{8}{100}}
\end{array}
$
$=\frac{5000 \times 100}{8}$
=₹ 62500
(iii) We know that, Compound interest
$
\begin{aligned}
I & =P\left[(1+i)^n-1\right] \\
i & =7.5 \% \text { p.a. } \\
& =0.075 \text { p.a. }
\end{aligned}
$
$\begin{aligned} & =0.075 \times \frac{1}{2} \text { per six months } \\ & =0.0375 \text { per six months } 1 \\ I & =7000\left[(1+0.0375)^6-1\right] \\ & =7000\left[(1.0375)^6-1\right] \\ & =7000[1.2472-1] \\ & =7000 \times 0.2472\end{aligned}$
= ₹ 1730.4

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Question 25 Marks

A person amortizes a loan of ₹ 1500000 for renovation of his house by 8 years mortgage at the rate of $1 2 \%$ p.a. compounded monthly. Find
(i) the equated monthly instalment
(ii) the principal outstanding at the beginning of $40^{\text {th }}$ month.
(iii) the interest paid in $40^{\text {th }}$ payment. $\left[\right.$ Given $(1.01)^{96}=2.5993,(1.01)^{57}=1.7633$ ]

Answer

Given, P= ₹ 15,00,000, $i=\frac{12}{12 \times 100}=\frac{1}{100}=0.01$ and $n=8 \times 12=96$
(i)
$
\begin{aligned}
E M I & =\frac{15,00,000 \times 0.01 \times(1.01)^{96}}{(1.01)^{96}-1} \\
& =\frac{15,00,000 \times 0.01 \times 2.5993}{2.5993-1} \\
& =\frac{15,00,000 \times 0.01 \times 2.5993}{1.5993}
\end{aligned}
$
= ₹ 24,379.10
(ii) Principal outstanding at the beginning of $40^{\text {th }}$ month
$
\begin{array}{l}
=\frac{E M I\left[(1+i)^{96-40+1}-1\right]}{i(1+i)^{96-40+1}} \\
=\frac{24379.10 \times\left[(1.01)^{57}-1\right]}{0.01(1.01)^{57}} \\
=\frac{24379.10 \times(1.7633-1)}{0.01 \times 1.7633} \\
=\frac{24379.10 \times 0.7633}{0.017633}
\end{array}
$
= ₹ 1,055,326.20
(iii) Interest paid in $40^{\text {th }}$ payment
$
\begin{array}{l}
=\frac{E M I\left[(1+i)^{96-40+1}-1\right]}{(1+i)^{96-40+1}} \\
=\frac{24379.10\left[(1.01)^{57}-1\right]}{(1.01)^{57}} \\
=\frac{24379.10 \times 0.7633}{1.7633}
\end{array}
$
= ₹ 10553.26

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Question 35 Marks

A machine costing ₹ 200000 has effective life 7 years and its scrap value is ₹ 30000. What amount should the company put into a sinking fund earning $5 \%$ per annum, so that it can replace the machine after its useful life ? Assume that a new machine will cost ₹ 300000 after 7 years. [Given $\log (1.05)=0.0212$ and antilog $(0.1484)=$ 1.407]

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Question 45 Marks

The cost of a TV depreciates by ₹ 800 during the second year and by ₹ 700 during the third year. Calculate:
(i) the rate of depreciation per annum.
(ii) the original cost of the machine.
(ii) the value of the TV at the end of third year.

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Applied Maths 5 Marks Questions

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