5 Marks Questions

Question 15 Marks

Find the equations of two straight lines passing through $(1,2)$ and making an angle of $60^{\circ}$ with the line $x+y=$ [5]0 . Find also the area of the triangle formed by the three lines.

Answer

Suppose $A(1,2)$ be the vertex of the triangle $A B C$ and $x+y=0$ be the equation of $B C$.

Now, we have to find the equations of sides $A B$ and $A C$, each of which makes an angle $60^{\circ}$ with the line $x+y=0$
We know the equations of two lines passing through a point ( $\mathrm{x}_{1}, \mathrm{y}_{1}$ ) and making an angle $\alpha$ with the line whose slope is m .
$\Longrightarrow \mathrm{y}-\mathrm{y}_{1}=\frac{m \pm \tan \alpha}{1 \mp m \tan \alpha}\left(\mathrm{x}-\mathrm{x}_{1}\right) \ldots$ (i)
Here,we have
$\mathrm{x}_{1}=1, \mathrm{y}_{1}=2, \alpha=60^{\circ}, \mathrm{m}=-1$,substituting in (i)
Therefore, the equations of the required sides are
$\Longrightarrow \mathrm{y}-2=\frac{-1+\tan 60^{\circ}}{1+\tan 60^{\circ}}(\mathrm{x}-1)$ and $\mathrm{y}-2=\frac{-1-\tan 60^{\circ}}{1-\tan 60^{\circ}}(\mathrm{x}-1)$
$\Rightarrow \mathrm{y}-2=\frac{\sqrt{3}-1}{\sqrt{3}+1}(\mathrm{x}-1)$ and $\mathrm{y}-2=\frac{\sqrt{3}+1}{\sqrt{3}-1}(\mathrm{x}-1)$
$\Rightarrow \mathrm{y}-2=(2-\sqrt{3})(\mathrm{x}-1)$ and $\mathrm{y}-2=(2+\sqrt{3})(\mathrm{x}-1)$
Solving $x+y=0$ and $y-2=(2-\sqrt{3})(x-1)$, we obtain
$x=-\frac{\sqrt{3}+1}{2}, y=\frac{\sqrt{3}+1}{{ }^{2}}$
$\therefore B \equiv\left(-\frac{\sqrt{3}+1}{2}, \frac{\sqrt{3}+1}{2}\right)$ or $C \equiv\left(\frac{\sqrt{3}-1}{2},-\frac{\sqrt{3}-1}{2}\right)$
$\mathrm{AB}=\mathrm{BC}=\mathrm{AD}=\sqrt{6}$ units
$\therefore$ Area of the required triangles $=\frac{\sqrt{3} \times(\sqrt{6})^{2}}{4}=\frac{3 \sqrt{3}}{2}$ square units.

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Question 25 Marks

Find the mean deviation about the mean for the data

Income per day in ₹	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Number of persons	4	8	9	10	7	5	4	3

Answer

Income per day	Mid values $X _{ i }$	$f _{ i }$	$f _{ i } x _{ i }$	$x_i-358 \mid$	$f _{ j }\left\| x _{ i }-358\right\|$
0-100	50	4	200	308	1232
100-200	150	8	1200	208	1664
200-300	250	9	2250	108	972
300-400	350	10	3500	8	80
400-500	450	7	3150	92	644
500-600	550	5	2750	192	960
600-700	650	4	2600	292	1168
700-800	750	3	2250	392	1176
		50	17900		7896

Mean $\bar{x}=\frac{1}{N} \sum f_{i} x_{i}=\frac{1}{50} \times 17900=358$
Mean deviation about mean $=\frac{1}{N} \sum_{i=1}^{n} f_{i}\left|x_{i}-\bar{x}\right|$
$=\frac{1}{50} \times 7896=157.92$

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Question 35 Marks

Calculate Karl Pearson's coefficient of skewness for the following data:

CI	0-10	10-20	20-30	30-40	40-50	50-60
r	3	6	9	7	4	2

Answer

Let $\mathrm{A}=25$ and $\mathrm{h}=10$

Mean $\mathrm{A}+\frac{\sum_{i=1}^{n} f_{i} d_{i}}{\sum_{i=1}^{n} f_{i}} \times h=25+\frac{9}{31} \times 10=25+2.90=27.90$
For median: $\frac{N}{2}=\frac{31}{2}=15.5$
Median class: 20-30
Median $=1+\frac{\frac{\mathrm{N}}{2}-c}{f} \times h=20+\frac{15.5-9}{9} \times 10=20+\frac{65}{9}=20+7.22=27.22$
Standard deviation $(\sigma)=10 \sqrt{\frac{\sum_{i=1}^{n} f_{i} d_{i}^{2}}{\mathrm{~N}}-\left(\frac{\sum_{i=1}^{n} f_{i} d_{i}}{\mathrm{~N}}\right)^{2}}=10 \sqrt{\frac{59}{31}-\left(\frac{9}{31}\right)^{2}}$
$\sigma=10 \sqrt{1.90-(0.29)^{2}}=10 \sqrt{1.90-0.08}=10 \sqrt{1.82}$
$=10 \times 1.349=13.49$
$\therefore \mathrm{S}_{\mathrm{kp}}=\frac{\text { Mean }- \text { Mode }}{\sigma}=\frac{3(\text { Mean }- \text { Median })}{\sigma}=\frac{3(27.90-27.22)}{13.49}$
$=\frac{3 \times 0.68}{13.49}=\frac{2.04}{13.49}=0.15$

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Question 45 Marks

Evaluate: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$

Answer

$\lim _{x \rightarrow 0} \frac{1+x-1+x}{x[\sqrt{1+x}+\sqrt{1-x}]}$ [By rationalising]
$=\lim _{x \rightarrow 0} \frac{2}{\sqrt{1+x}+\sqrt{1-x}}=\frac{2}{1+1}=1$

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Question 55 Marks

$60 \%$ students read Hindi newspaper, $40 \%$ students read Tamil newspaper and $20 \%$ students read both Hindi and Tamil newspaper. Find the probability that a student selected at random reads
i. Tamil newspaper given that he has already read Hindi newspaper.
ii. Hindi newspaper given that he has already read Tamil newspaper.
iii. neither Hindi nor Tamil newspaper.

Answer

Let H denote the event that the student reads Hindi newspaper and T denotes the event that the student reads Tamil newspaper.
Given $\mathrm{P}(\mathrm{H})=\frac{60}{100}, \mathrm{P}(\mathrm{T})=\frac{40}{100}$ and $\mathrm{P}(\mathrm{H} \cap \mathrm{T})=\frac{20}{100}$.
i. $\mathrm{P}(\mathrm{T} \mid \mathrm{H})=\frac{P(T \cap H)}{P(H)}=\frac{\frac{20}{100}}{\frac{60}{100}}=\frac{1}{3}$.
ii. $\mathrm{P}(\mathrm{H} \mid \mathrm{T})=\frac{P(H \cap T)}{P(T)}=\frac{\frac{20}{100}}{\frac{40}{100}}=\frac{1}{2}$.
iii. $\mathrm{P}\left(\mathrm{H}^{\prime} \cap \mathrm{T}^{\prime}\right)=1-\mathrm{P}(\mathrm{H} \cup \mathrm{T})=1-(\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{T})-\mathrm{P}(\mathrm{H} \cap \mathrm{T}))$
$
=1-\left(\frac{60}{100}+\frac{40}{100}-\frac{20}{100}\right) \cdot=\frac{20}{100}=\frac{1}{5}
$

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Question 65 Marks

Three persons A, B and C apply for a job of Manager in a private company. Chances of their selection (A, B and C are in the ratio 1:2:4. The probabilities that $\mathrm{A}, \mathrm{B}$ and C can introduce changes to improve profits of the company are $0.8,0.5$ and 0.3 , respectively. If the change does not take place, find the probability that it is due to the appointment of C .

Answer

Let us define the following events
$\mathrm{A}=$ selecting person A
$B=$ selecting person $B$
$\mathrm{C}=$ selecting person C
$\mathrm{P}(\mathrm{A})=\frac{1}{1+2+4}, \mathrm{P}(\mathrm{B})=\frac{2}{1+2+4}$
and $\mathrm{P}(\mathrm{C})=\frac{4}{1+2+4}$
$\mathrm{P}(\mathrm{A})=\frac{1}{7}, \mathrm{P}(\mathrm{B})=\frac{2}{7}$
and $\mathrm{P}(\mathrm{C})=\frac{4}{7}$
Let $\mathrm{E}=$ Event to introduce the changes in their profit.
Also given $P\left(\frac{E}{A}\right)=0.8, P\left(\frac{E}{B}\right)=0.5$ and $P\left(\frac{E}{C}\right)=0.3$
$\Rightarrow P\left(\frac{\bar{E}}{A}\right)=1-0.8=0.2, P\left(\frac{\bar{E}}{B}\right)=1-0.5=0.5$
and $P\left(\frac{\bar{E}}{C}\right)=1-0.3=0.7$
The probability that change does not take place by the appointment of C ,
$P\left(\frac{C}{E}\right)=\frac{P(C) \cdot P\left(\frac{\bar{E}}{C}\right)}{P(A) \times P\left(\frac{\bar{E}}{A}\right)+P(B) \times P\left(\frac{\bar{E}}{B}\right)+P(C) \times P\left(\frac{\bar{E}}{C}\right)}$
$=\frac{\frac{4}{7} \times 0.7}{\frac{1}{7} \times 0.2+\frac{2}{7} \times 0.5+\frac{4}{7} \times 0.7}$
$=\frac{2.8}{0.2+1.0+2.8}=\frac{2.8}{4}=0.7$

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Applied Maths 5 Marks Questions

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