5 Marks Questions

Question 15 Marks

Find the equations of the lines through the point $(3,2)$ which make an angle of $45^{\circ}$ with the line $\mathrm{x}-2 \mathrm{y}=3$.

Answer

The given line is $x-2 y-3=0$
$\Rightarrow y=\frac{x}{2}-\frac{3}{2}$
$\therefore$ Slope, $\mathrm{m}_{1}=\frac{1}{2}$
Let ' $m_{2}$ ' be the slope of a line $A B$ which passes through $(3,2)$.
Since the angle between the two line is $60^{\circ}$
$\therefore \tan 45^{\circ}= \pm \frac{m_{2}-m_{1}}{1+m_{1} m_{2}}$
$\Rightarrow 1= \pm \frac{m_{2}-\frac{1}{2}}{1+\frac{1}{2} m_{2}}$
$\Rightarrow 1= \pm \frac{2 m_{2}-1}{m_{2}+2}$
$\therefore \frac{2 m_{2}-1}{m_{2}+2}=1$ or $\frac{2 m_{2}-1}{m_{2}+2}=-1$
$\Rightarrow \mathrm{m}_{2}=3$ or $\mathrm{m}_{2}=-\frac{1}{3}$
$\therefore$ Equation of AB is
$\mathrm{y}-2=3(\mathrm{x}-3)$
$\Rightarrow 3 \mathrm{x}-\mathrm{y}=7\left(\mathrm{~m}_{2}=3\right)$
or, $\mathrm{y}-2=-\frac{1}{3}(\mathrm{x}-3)\left(m_{2}=-\frac{1}{3}\right)$
$\Rightarrow \mathrm{x}+3 \mathrm{y}=9$

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Question 25 Marks

Find the mean deviation from the median of the following frequency distribution:

x_i	10	11	12	13	14	15	16
f_i	3	8	14	19	7	6	3

Answer

Here, $\mathrm{N}=\sum f_{i}=60$, which is even
$\therefore$ Median (M) $=\frac{\left(\frac{N}{2}\right) \text { th term }+\left(\frac{N}{2}+1\right) \text { th term }}{2}=\frac{\left(\frac{60}{2}\right) \text { th term }+\left(\frac{60}{2}+1\right) \text { term }}{2}=\frac{30 \text { th term }+31 \text { st term }}{2}$
$=\frac{13+13}{2}=13$
We make the table from the given data:

Age (in yr) x_i	f_i	cf	\|x_i - M\|	f_i\|x_i - M\|
10	3	3	3	9
11	8	11	2	16
12	14	25	1	14
13	19	44	0	0
14	7	51	1	7
15	6	57	2	12
16	3	60	3	9
Total	N=60			$\sum f _{ i }\left\| x _{ i }- M \right\|=67$

Mean deviation from median, MD $=\frac{\sum f_{i}\left|x_{i}-M\right|}{N}=\frac{67}{60}=1.12 \mathrm{yr}$

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Question 35 Marks

Find the mean deviation about the mean for the following data:

Marks obtained	10-20	20-30	30-40	40-50	50-60	160-70
Number of students	8	6	12	5	2	7

Answer

Here length of class interval, $\mathrm{h}=10$. Let the assumed mean be $\mathrm{A}=35$.
Then, we prepare the following table by using assumed mean as given.

Marks obtained	Number of students {f_i)	Midpoint x_i	$d_i=\frac{x_i-35}{10}$	f_id_i	$\left\|x_i-\bar{x}\right\|$	$f_i \times\left\|x_i-\bar{x}\right\|$
10-20	8	15	-2	-16	22	176
20-30	6	25	-1	-6	12	72
30-40	12	35 = A	0	0	2	21
40-50	5	45	1	5	8	40
50-60	2	55	2	4	18	36
60-70	7	65	3	21	28	196
	$\begin{aligned} N & =\Sigma f_i \\ & =40\end{aligned}$			$\Sigma f_i d_i=8$		$\begin{aligned} \sum f_i & \times\left\|x_i-\bar{x}\right\| \\ & =544\end{aligned}$

$N=\Sigma f_{i}=40, \bar{x}=A+\left\{\frac{\Sigma f_{i} d_{i}}{N} \times h\right\}=35+\left\{\frac{8}{40} \times 10\right\}=37$
$\therefore \mathrm{MD}(\bar{x})=\frac{\sum f_{i} \times\left|x_{i}-\bar{x}\right|}{N}=\frac{544}{40}=\frac{136}{10}=13.6$.

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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

In a school, there are 1000 students, out of which 430 are girls. It is known that out of $430,10 \%$ of the girls study in class XII, what is the probability that a student chosen randomly studies in class XII, given that the chosen student is a girl?

Answer

Let ' A ' be the event that the chosen student studies in class XII and B be the event that the chosen student is a girl.
There are 430 girls out of 1000 students
So, $\mathrm{P}(\mathrm{B})=\mathrm{P}($ Chosen student is girl $)=\frac{430}{1000}=\frac{43}{100}$
Since, $10 \%$ of the girls studies in class XII
So, total number of girls studies in class XII
$=\frac{10}{100} \times 430=43$
Then, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}$ (Chosen student is a girl of class XII)
$=\frac{43}{1000}$
$\therefore$ Required probability $=\mathrm{P}(\mathrm{A} / \mathrm{B})$
$=\frac{P(A \cap B)}{P(B)} \quad\left[\because P(A / B)=\frac{P(A \cap B)}{P(B)}\right]$
$=\frac{43 / 1000}{43 / 100}=\frac{1}{10}$

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Question 65 Marks

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that one of them is black and other is red.

Answer

Given: A box containing 10 black and 8 red balls.
Total number of balls in box $=18$
Probability of getting a black ball in first draw $=\frac{10}{18}=\frac{5}{9}$
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw $=\frac{8}{18}=\frac{4}{9}$
Now, Probability of getting first ball is black and second is red $=\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$
Probability of getting a red ball in first draw $=\frac{8}{18}=\frac{4}{9}$
As the ball is replaced after first throw,
Hence, Probability of getting a black ball in second draw $=\frac{10}{18}=\frac{5}{9}$
Now, Probability of getting first ball is red and second is black $=\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$
Therefore, Probability of getting one of them is black and other is red :
= Probability of getting first ball is black and second is red + Probability of getting first ball is red and second is black
$=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}$

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Applied Maths 5 Marks Questions

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