3 Marks Question

Question 13 Marks

₹ 1000 is invested every 3-months at 4.8% compounded quarterly. How much will the annuity be worth in 2 years?

Answer

$FV .=\frac{ C \cdot F \cdot\left[(1+i)^n-1\right]}{i}$
Here,$\quad$$\quad$$i=4.8 \%$ quarterly
$=\frac{0.048}{4}=0.012$
C.F. = ₹1000
$n=2 \times 4=8$
$\therefore$ $FV .=\frac{1000 \times\left[(1+0.012)^8-1\right]}{0.012}$
$=\frac{1000 \times\left[(1.012)^8-1\right]}{0.012}$
$=\frac{1000 \times(1.1001-1)}{0.012}$
$=\frac{1000 \times 0.1001}{0.012}$
$=\frac{100.13}{0.012}$
= ₹ 8344.18
Hence, in two years the annuity worth will be ₹ 8344.18.

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Question 23 Marks

₹ 5000 is paid every year for three years to pay off a loan. What is the loan amount if interest rate be 14% per annum compounded annually ?

Answer

Since, C.F. $=\frac{P V}{P(n, i)}$ where $P(n, i)=\frac{(1+i)^n-1}{i(1+i)^n}$
Here, $\quad$$\quad$ C.F = ₹ 5000
$ \begin{aligned} n & =3, i=14 \%=0.14 \\ P(n, i) & =\frac{(1+0.14)^3-1}{0.14(1+0.14)^3} \\ & =\frac{(1.14)^3-1}{0.14(1.14)^3} \\ & =\frac{1.4815-1}{0.14 \times 1.4815} \\ & =\frac{0.4815}{0.2074}=2.3216 \end{aligned} $
Therefore,
$ \begin{aligned} \qquad P V & =\text { C.E. } P(n, i) \\ & =5000 \times 2.3216 \end{aligned} $
= ₹ 11, 608

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Question 33 Marks

What is the monthly equivalent interest rate to a quarterly interest 2.5% ?

Answer

We have to find i mon knowing that i quart = 2.5% According to the relation of rate equivalence, the equality
$\left(1+i_{\text {ousart }}\right)^4=\left(1+i_{\text {mon }}\right)^{12}$
must be satisfied

$\therefore$ $\left(1+i_{\operatorname{mon}}\right)^{12}=\left(1+\frac{2.5}{100}\right)^4$
$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\left[\because\right.$ give $\left.i_{\text {quart }}=2.5 \%=\frac{2.5}{100}\right]$
$\Rightarrow$$\left(1+i_{\operatorname{mon}}\right)^{12}=(1.025)^4$
$\Rightarrow$$\left[\left(1+i_{\text {mon }}\right)^{12}\right]^{1 / 4}=\left[(1.025)^4\right]^{1 / 4}$
$\Rightarrow$$1+i_{\text {mon }}=(1.025)^{1 / 3}$
$\Rightarrow$$i_{\text {mon }}=(1.025)^{1 / 3}-1$
$\Rightarrow$$i_{\text {mon }}=1.008265-1$
$\Rightarrow$$i_{\text {mon }}=0.008265 \approx 0.8265 \%$

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Question 43 Marks

₹ 5000 is invested in a Term Deposit Scheme that fetches interest 6% per annum compounded quarterly. What will be interest after one year ? What is effective rate of interest ?

Answer

We know that,
Compound interest, $I=\left[P(1+i)^n-1\right]$
P = ₹ 7 5000, i = 6% = 0.06 p.a
$=0.06 \times \frac{1}{4}$ per quarter
= 0.015 per quarter
$\Rightarrow$$I$ = 5000[(1 + 0.015)⁴ - 1]
= 5000{(1.015)⁴ - 1}
= 5000 (1.0613-1)
= 5000 × 0.0613
= 306.82
effective rate of interest using I = PEI,
We get $\quad$$\quad$306.82 = 5000 × E × 1
E = $\frac{306.82}{5000}$ = 0.0613 or 6.13%
Note: We may arrive at the same result by using
E = (1 + i)ⁿ - 1
= (1 + 0.015)⁴- 1
= (1.015)⁴ - 1
= 1.0613-1
= 0.0613
= 6.13%

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Question 53 Marks

₹ 16000 invested at 10% p.a. compounded semi-annually amounts to ₹ 18522. Find the time period of investment.

Answer

Here,
P = ₹ 16000
A_n = ₹ 18522
$i=10 \times \frac{1}{2} \%=5 \%=0.05$
we have, $A_{i n}=P(1+i)^n$
$\Rightarrow 18522=16000(1+0.05)^n$
$\begin{aligned} \Rightarrow & \frac{18522}{16000} & =(1.05)^n \\ \Rightarrow & 1.157625 & =(1.05)^n \\ \Rightarrow & (1.05)^3 & =(1.05)^n \\ \Rightarrow & n =3\end{aligned}$
Therefore, time period of investment is three half years i.e.,$1 \frac{1}{2}$ years.

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