5 Marks Questions

Question 15 Marks

Find the equation of a circle whose centre is a point $(1,-2)$ and which passes through the centre of the circle $2 x^{2}$$+2 y^{2}+4 y=5$.

Answer

Given circle is $2 x^{2}+2 y^{2}+4 y=5$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{y}-\frac{5}{2}=0$
Centre is $(0,-1)$.
radius $=\sqrt{(1-0)^{2}+(-2+1)^{2}}=\sqrt{2}$
$\therefore$ circle is $(x-1)^{2}+(y+2)^{2}=2$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+\mathrm{Ay}+3=0$

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Question 25 Marks

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

Income per day	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

We construct the following table. (5th and 6th columns are filled after calculating the mean.)

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

Here $\mathrm{n}=\Sigma f_{i}=50, \Sigma f_{i} x_{i}=17900$
$\therefore$ Mean $=\bar{x}=\frac{1}{n} \Sigma f_{i} x_{i}=\frac{17900}{50}=358$
M.D. $(\bar{x})=\frac{1}{n} \Sigma f_{i}\left|x_{i}-\bar{x}\right|=\frac{7896}{50}=157.92$

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Question 35 Marks

Compute the moment coefficient of skewness $\beta_{1}$ for the following distribution:

Marks obtained	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
Frequency	6	12	22	24	16	12	8

Answer

We construct the following table:

class	Frequency $f _{ i }$	Class mark ( $x _{ i }$)	$f _{ i } x _{ i }$	$\left( x _{ i }-\bar{x}\right)$	$\left( x _{ i }-\bar{x}\right)^2$	$\left( x _{ i }-\bar{x}\right)^3$	$f _{ i }\left( x _{ i }-\bar{x}\right)^2$	$f _{ i }\left( x _{ i }-\bar{x}\right)^3$
0-10	6	5	30	-30	900	-27000	5400	-162000
10-20	12	15	180	-20	400	-8000	4800	-96000
20-30	22	25	550	-10	100	-1000	2200	-22000
30-40	24	35	840	0	0	0	0	0
40-50	16	45	720	10	100	1000	1600	16000
50-60	12	55	660	20	400	8000	48000	96000
60-70	8	65	520	30	900	27000	7200	216000
	100		3500				26000	48000

Here, $\bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{3500}{100}=35$
$\therefore \mu_{2}=\frac{\Sigma f_{i}\left(x_{i}-\bar{x}\right)^{2}}{\Sigma f_{i}}=\frac{26000}{100}=260$
$\mu_{3}=\frac{\Sigma f_{i}\left(x_{i}-\bar{x}\right)^{3}}{\Sigma f_{i}}=\frac{48000}{100}=480$
$\therefore \beta_{1}=\frac{\mu_{3}^{2}}{\mu_{2}^{3}}=\frac{(480)^{2}}{(260)^{3}}=0.013$

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Question 45 Marks

Suppose $f(x)=\left\{\begin{array}{cc}a+b x, & x<1 \\ 4, & x=1 \\ b-a x, & x>1\end{array}\right.$ and if $\lim _{x \rightarrow 1} \mathrm{f}(\mathrm{x})=\mathrm{f}(1)$, then what are the possible values of a and b ?

Answer

We have,
$f(x)=\left\{\begin{array}{cc}a+b x, & x<1 \\ 4, & x=1 \\ b-a x, x>1 & \end{array}\right.$
Now, LHL $=\lim _{x \rightarrow 1^{-}} f(x)$
$=\lim _{x \rightarrow 1^{-}}(a+b x)=\lim _{h \rightarrow 0}[a+b(1-h)]$ [putting $\mathrm{x}=1-\mathrm{h}$ as $\mathrm{x} \rightarrow 1$, then $\mathrm{h} \rightarrow 0$ ]
$=\mathrm{a}+\mathrm{b}$
$\mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)$
$=\lim _{x \rightarrow 1^{+}}(b-a x)=\lim _{h \rightarrow 0}[b-a(1+h)]$ [putting $\mathrm{x}=1+\mathrm{h}$ as $\mathrm{x} \rightarrow 1$, then $\mathrm{h} \rightarrow 0$ ]
$=\mathrm{b}-\mathrm{a}$
Since, $\lim _{x \rightarrow 1} \mathrm{f}(\mathrm{x})=\mathrm{f}(1)$
$\therefore \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(1)$
$\Rightarrow \mathrm{a}+\mathrm{b}=\mathrm{b}-\mathrm{a}=4[\because \mathrm{f}(1)=4$, given $]$
$\Rightarrow \mathrm{a}+\mathrm{b}=4$..(i) and $\mathrm{b}-\mathrm{a}=4$..(ii)
On solving (i) and (ii), we get
$\mathrm{a}=0, \mathrm{~b}=4$

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Question 55 Marks

A die has two faces with number 1, three faces each with number 2 and one face with number 3 . If die is ruled once, find (i) P (3) (ii) $\mathrm{P}(1$ or 2$)$ (iii) $\mathrm{P}(2)$

Answer

(i) Out of 6 faces, one face marked with number 3
$\therefore P(3)=\frac{1}{6}$
(ii) Out of 6 faces, two faces marked with number 1 and three faces marked with number 2
$\therefore P(1$ or 2$)=\frac{5}{6}$
(iii) Out of 6 faces, three faces marked with number 2
$\therefore P(2)=\frac{3}{6}=\frac{1}{2}$

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Question 65 Marks

In a factory which manufactures bolts, machines A, B and C manufacture respectively $30 \%, 50 \%$ and $20 \%$ of the bolts. Of their outputs 3, 4 and 1 per cent respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B.

Answer

$\mathrm{P}(\mathrm{A})=\frac{30}{100}=\frac{3}{10}, \mathrm{P}(\mathrm{B})=\frac{50}{100}=\frac{5}{10}, \mathrm{P}(\mathrm{C})=\frac{20}{100}=\frac{2}{10}$
E : bolt is defective
$\mathrm{P}\left(\frac{E}{A}\right)=\frac{3}{100}, \mathrm{P}\left(\frac{E}{B}\right)=\frac{4}{100}, \mathrm{P}\left(\frac{E}{C}\right)=\frac{1}{100}$
Using Bayes' Theorem probability that defective bolt is not manufactured by machine B.
$1-\mathrm{P}\left(\frac{E}{B}\right)=1-\frac{\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\frac{E}{B}\right)}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\frac{E}{A}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\frac{E}{B}\right)+\mathrm{P}(\mathrm{C}) \cdot \mathrm{P}\left(\frac{E}{C}\right)}$
$=1-\frac{\frac{5}{10} \times \frac{4}{100}}{\frac{3}{10} \times \frac{3}{100}+\frac{5}{10} \times \frac{4}{100}+\frac{2}{10} \times \frac{1}{100}}$
$=1-\frac{20}{9+20+2}=1-\frac{20}{31}=\frac{11}{31}$

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Applied Maths 5 Marks Questions

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