3 Marks Question

Question 13 Marks

For a group of 60 boys students, the mean and S.D. of statistics marks are 45 and 2 respectively. The same figures for a group of 40 girls students are 55 and 3 respectively. What is the mean and S.D. of marks if the two groups are pooled together?

Answer

As given $\mathrm{n}_{1}=60, \bar{x}_{1}=45, \sigma_{1}=2, \mathrm{n}_{2}=40, \bar{x}_{2}=53, \sigma_{2}=3$
Thus, combined mean is given by
$\bar{x}=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$
$=\frac{60 \times 45+40 \times 55}{60+40}$
$=49$
Thus, $\mathrm{d}_{1}=\bar{x}_{1}-\bar{x}=45-49=-4$
$\mathrm{d}_{2}=\bar{x}_{2}-\bar{x}=55-49=6$
Combined S.D. $=\sqrt{\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}+n_{1} d_{1}^{2}+n_{2} d_{2}^{2}}{n_{1}+n_{2}}}$
$=\sqrt{\frac{60 \times(2)^{2}+40 \times(3)^{2}+60 \times(-4)^{2}+40 \times(6)^{2}}{60+40}}$
$=\sqrt{30}=5.48$.

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Question 23 Marks

Using logarithmic table, find the value of $\sqrt{\frac{41.32 \times 20.18}{12.69}}$.

Answer

Let $x=\sqrt{\frac{41.32 \times 20.18}{12.69}}$. Then,
$\log x=\log \left\{\frac{41.32 \times 20.18}{12.69}\right\}^{\frac{1}{2}}=\frac{1}{2} \log \left\{\frac{41.32 \times 20.18}{12.69}\right\}$
$\Rightarrow \log \mathrm{x}=\frac{1}{2}[\log (41.32 \times 20.18)-\log 12.69]=\frac{1}{2}[\log 41.32+\log 20.18-\log 12.69]$
$\Rightarrow \log \mathrm{x}=\frac{1}{2}[1.6162+1.3049-1.1035]=\frac{1}{2}(2.9211-1.1035)=\frac{1}{2}(1.8176)=0.9088$
$\therefore \mathrm{x}=\operatorname{antilog}(0.9088)=8.106$

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Question 33 Marks

a. In what time will ₹ 85000 amount to ₹ 157675 at $4.5 \%$ p.a?
b. A sum of ₹ 46875 was lent out at simple interest and at the end of 1 year 8 months the total amount was ₹ 50,000 . Find the rate of interest percent per annum.

Answer

a. We know,
$
\begin{aligned}
& \mathrm{A}=\mathrm{P}(1+\mathrm{it}) \\
& \Rightarrow 157675=85000\left(1+\frac{4.5}{100} \times t\right) \\
& \Rightarrow \frac{157675}{85000}=\frac{100+4.5 t}{100} \\
& \Rightarrow 4.5 \mathrm{t}=\left[\frac{157675}{85000} \times 100\right]-100 \\
& \Rightarrow 4.5 \mathrm{t}=\frac{85.5}{4.5}=19
\end{aligned}
$
$\therefore$ In 19 years ₹ 85000 will amount to ₹ 157675 at $4.5 \%$ p.a. simple interest rate.
b. We know,
$
\begin{aligned}
& \mathrm{A}=\mathrm{P}(1+\mathrm{it}) \\
& \Rightarrow 50000=46875\left(1+i 1 \frac{8}{12}\right) \\
& \Rightarrow \frac{50000}{46875}=1+\frac{5}{3} i \\
& \Rightarrow(1.067-1) \times \frac{3}{5}=\mathrm{i} \\
& \Rightarrow \mathrm{i}=0.04 \\
& \Rightarrow \mathrm{i}=4 \% .
\end{aligned}
$

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Question 43 Marks

Find the domain and the range of the given function: $f(x)=\frac{1}{\sqrt{5-x}}$

Answer

Given $f(x)=\frac{1}{\sqrt{5-x}}$
For $\mathrm{D}_{\mathrm{f}}, \mathrm{f}(\mathrm{x})$ must be a real number
$\Rightarrow \frac{1}{\sqrt{5-x}}$ must be a real number
$\Rightarrow 5-\mathrm{x}>0 \Rightarrow 5>\mathrm{x} \Rightarrow \mathrm{x}<5$
$\Rightarrow \mathrm{D}_{\mathrm{f}}=(-\infty, 5)$
For $\mathrm{R}_{\mathrm{f}}$, let $\mathrm{y}=\frac{1}{\sqrt{5-x}}$
As $x<5,0<5-x$
$\Rightarrow 5-\mathrm{x}>0 \Rightarrow \sqrt{5-x}>0$
$\Rightarrow \frac{1}{\sqrt{5-x}}>0\left(\because \frac{1}{a}>0\right.$ if and only if a $\left.>0\right)$
$\Rightarrow \mathrm{y}>0$
$\Rightarrow \mathrm{R}_{\mathrm{f}}=(0, \infty)$

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Question 53 Marks

Find the equations of the lines for which $\tan \theta=\frac{1}{2}$, where $\theta$ is the inclination of the line and
i. $y$-intercept is $-\frac{3}{2}$
ii. $x$-intercept is 4

Answer

Here slope of the line $\mathrm{m}=\tan \theta=\frac{1}{2}$
i. y -intercept is $-\frac{3}{2}$ i.e. $\mathrm{c}=-\frac{3}{2}$
Using slope-intercept form case I, the equation of the line is
$
\begin{aligned}
& \mathrm{y}=\frac{1}{2} \mathrm{x}+\left(-\frac{3}{2}\right)[\because \mathrm{y}=\mathrm{mx}+\mathrm{c}] \\
& \Rightarrow 2 \mathrm{y}=\mathrm{x}-3 \\
& \Rightarrow 2 \mathrm{y}-\mathrm{x}+3=0
\end{aligned}
$
ii. $x$-intercept is 4 i.e. $d=4$
Using slope-intercept form case II, the equation of the line is
$y=\frac{1}{2}(x-4)[\because y=m(x-d)]$
$
\begin{aligned}
& \Rightarrow 2 y=x-4 \\
& \Rightarrow 2 y-x+4=0
\end{aligned}
$

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Question 63 Marks

Find three numbers in G.P. whose product is 216 and the sum of their products in pairs is 156.

Answer

Let three numbers in G.P. be $\frac{a}{r}$, a, ar
$\therefore$ Their product $=\frac{a}{r} \cdot \mathrm{a} \cdot \mathrm{ar}=216$ (given)
$\Rightarrow a^{3}=216=(6)^{3} \Rightarrow a=6$.
Also sum of their products in pairs $=156$ (given)
$\Rightarrow \frac{a}{r} \cdot \mathrm{a}+\mathrm{a} \cdot \mathrm{ar}+\mathrm{ar} \cdot \frac{a}{r}=156$
$\Rightarrow a^{2}\left(\frac{1}{r}+r+1\right)=156$
$\Rightarrow 6^{2} \cdot \frac{1+r^{2}+r}{r}=156$
$\Rightarrow 3 \cdot \frac{r^{2}+r+1}{r}=13$
$\Rightarrow 3 r^{2}+3 r+3=13 r$
$\Rightarrow 3 r^{2}-10 r+3=0$
$\Rightarrow(r-3)\left(r-\frac{1}{3}\right)=0 \Rightarrow r=3, \frac{1}{3}$
When $r=3$, numbers are $2,6,18$ and when $r=\frac{1}{3}$, numbers are $18,6,2$

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Question 73 Marks

If $A M$ and $G M$ are between two positive numbers $x$ and $y$ are 13 and 12 respectively, find the numbers.

Answer

Given: $\frac{x+y}{2}=13 \Rightarrow \mathrm{x}+\mathrm{y}=26 \ldots$ (i)
and $\sqrt{x} y=12 \Rightarrow \mathrm{xy}=144 \ldots$ (ii)
From (i) and (ii)
$x(26-x)=144 \Rightarrow 26 x-x^{2}=144$
$\Rightarrow x^{2}-26 x+144=0$
$\Rightarrow(\mathrm{x}-18)(\mathrm{x}-8)=0$
$\Rightarrow \mathrm{x}-18=0$ or $\mathrm{x}-8=0$
$\Rightarrow \mathrm{x}=18$ or 8
$\therefore$ Numbers are 18, 8 or 8,18 .

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Applied Maths 3 Marks Question

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