2 Marks Questions

Question 12 Marks

Solve: $12 x \equiv 44(\bmod 59)$

Answer

We know that $a \equiv b (\bmod m ) \Rightarrow \frac{a}{x} \equiv \frac{b}{x}\left(\bmod \frac{m}{d}\right)$, where $d =( x , m )$.
$
\begin{array}{l}
\therefore 12 x \equiv 44(\bmod 59) \\
\Rightarrow 3 x \equiv 11(\bmod 59)[\because(4,59)=1]
\end{array}
$
We find that $(3,59)=1$, hence it has unique solution $(\bmod 59)$. Using division algorithm, we obtain
$
\begin{array}{l}
59=19 \times 3+2 \\
3=2 \times 1+1
\end{array}
$
Using back substitution, we obtain
$
\begin{array}{l}
1=3-2 \times 1 \\
\Rightarrow 1=3-(59-19 \times 3) \times 1 \\
\Rightarrow 1=59 \times(-1)+20 \times 3
\end{array}
$
The coefficient of 3 i.e. 20 is the inverse of $3(\bmod 59)$
Now,
$
\begin{array}{l}
3 x \equiv 11(\bmod 59) \\
\Rightarrow 20 \times 3 x \equiv 20 \times 11(\bmod 59) \text { [Multiplying throughout by inverse of } 3 \text { i.e. } 20 \text { ] } \\
\Rightarrow(20 \times 3) x \equiv 220(\bmod 59) \\
\Rightarrow x \equiv 43(\bmod 59)
\end{array}
$
Hence $x \equiv 43(\bmod 59)$ is the solution of the given linear congruence.

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Question 22 Marks

Find the declared rate of return compounded semiannually which is equivalent to $6 \%$ effective rate of return [Use
$
\left.(1.06)^{\frac{1}{2}}=1.0296\right]
$

Answer

Let declared rate of interest be r \% p.a. compounded half yearly.
Given effective rate of return (per rupee) $=\frac{6}{100}=0.06$ (per-rupee), $p =2$ half years.
$
\begin{array}{l}
\therefore 0.06=\left(1+\frac{r}{200}\right)^2-1 \\
\Rightarrow\left(1+\frac{r}{200}\right)^2=1.06 \Rightarrow 1+\frac{r}{200}=(1.06)^{\frac{1}{2}} \\
\Rightarrow 1+\frac{r}{200}=1.0296 \Rightarrow \frac{r}{200}=0.0296 \\
\Rightarrow r=0.0296 \times 200 \Rightarrow r=5.92
\end{array}
$
Hence, the declared rate of return $=5.92 \%$

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Question 32 Marks

An interviewer gives the following graph on a client's sales in the last 7 years to candidate and said find the CAGR. Given that $\left(\frac{9}{4}\right)^{\frac{1}{7}}=1.1228$.

Answer

Sales in 2006 were 0.8 crores (beginning value). In 2013, after 7 years, sales increased to 1.8 crores.
$
\begin{array}{l}
CAGR=\left(\frac{\text { End valve }}{\text { Beginning value }}\right)^{\frac{1}{n}}-1 \\
=\left(\frac{1.8}{0.8}\right)^{\frac{1}{7}}-1=\left(\frac{9}{4}\right)^{\frac{1}{7}}-1 \\
=0.1228
\end{array}
$
$
\text { CAGR % }=12.28 \%
$

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Question 42 Marks

By using property of definite integrals, evaluate $\int_0^2 x \sqrt{2-x} d x$

Answer

$\begin{array}{l}\int_0^2 x \sqrt{2-x} d x=\int_0^2(2-x)(2-(2-x))^{\frac{1}{2}} d x \text { (by property } P _4 \text { ) } \\ =\int_0^2(2-x) x^{\frac{1}{2}} d x=\int_0^2\left(2 x^{\frac{1}{2}}-x^{\frac{3}{2}}\right) d x \\ =\left[\frac{4}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}\right]_0^2=\frac{4}{3} \cdot 2 \sqrt{2}-\frac{2}{5} \cdot 4 \sqrt{2}=\frac{16}{15} \sqrt{2}\end{array}$

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Question 52 Marks

A person has set up a sinking fund so that he can accumulate ₹ 100000 in 10 years for his children's higher education.
How much amount should he deposit every six months if interest is 5% per annum compounded semi-annually?

Answer

Given A = ₹ 100000,$r=\frac{5}{2} \%$ per half year $\Rightarrow i=\frac{2.5}{100}=0.025$ and $n=20$ half year
Using formula
$
\begin{array}{l}
A=R\left[\frac{(1+i)^n-1}{i}\right] \Rightarrow 100,000=R\left[\frac{(1.025)^{20}-1}{0.025}\right] \\
\Rightarrow R=\frac{100000 \times 0.025}{(1.025)^{20}-1} \\
\Rightarrow R=\frac{2500}{1.637-1}=\frac{2500}{0.637}
\end{array}
$
R = ₹ 3924.64
Let $x =(1.025)^{20}$
Taking logarithm on both sides, we get
$
\begin{array}{l}
\log x=20 \log 1.025 \\
\Rightarrow \log x=20 \times 0.0107 \\
\Rightarrow \log x=0.2140 \\
\Rightarrow x=\text { antilog } 0.2140 \\
\Rightarrow x=1.637
\end{array}
$

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Question 62 Marks

Mr. Y has two investment options - either at 10% per annum compounded semi-annually or 9.5 % per annum
compounded continuously. Which option is preferable and why?

Answer

When compounded semi-annually we have $r =0.10, m=2$ Now, $r_{e f f}=\left(1+\frac{r}{m}\right)^m-1$
$
\begin{array}{l}
=\left(1+\frac{0.10}{2}\right)^2-1 \\
=0.1025 \text { or } 10.25 \%
\end{array}
$
when compounded continuously
$
\begin{array}{l}
r_{e f f}=e^r-1=e^{0.095}-1 \\
=0.0996=9.96 \%
\end{array}
$
Thus, the first investment is preferable.

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Question 72 Marks

Assuming a four yearly cycle, calculate the trend by the method of moving averages from the following data:

Year	1984	1985	1986	1987	1988	1989	1990	1991	1992	1993
Value	12	25	39	54	70	87	105	100	82	65

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