Case study (4 Marks)

Question 14 Marks

Ashish is writing examination. He is reading question paper during reading time. He reads instructions carefully. While reading instructions, he observed that the question paper consists of $15$ questions divided in to two parts part $I$ containing $8$ questions and part $II$ containing $7$ questions.

$i$. If Ashish is required to attempt $8$ questions in all selecting at least $3$ from each part, then in how many ways can he select these questions $(1)$
$ii$. If Ashish is required to attempt $8$ questions in all selecting $3$ from $I$ part, then in how many ways can he select these questions $(1)$
$iii.$ If Ashish is required to attempt $8$ questions in all selecting $4$ from part $I$ and $4$ from part $II,$ then in how many ways can he select these questions $(2)$
OR
If Ashish is required to attempt $8$ questions in all selecting $6$ from one section and remaining from another section, then in how many ways can he select these questions $(2)$

Answer

$i.$ Since, at least $3$ questions from each part have to be selected

Part $I$	Part $II$
$3$	$5$
$4$	$4$
$3$	$5$

So number of ways are
$3 $ questions from part $I$ and $5$ questions from part $II$ can be selected in $n^8 C_3 \times{ }^7 C_5$ ways
$4$ questions from part $I$ and $4$ questions from part $II$ can be selected in ${ }^8 C_4 \times{ }^7 C_4$ ways
$5$ questions from part $I$ and $3$ questions from part $II$ can be selected in ${ }^8 C_5 \times{ }^7 C_3$ ways
So required number of ways are
${ }^8 C_3 \times{ }^7 C_5+{ }^8 C_4 \times{ }^7 C_4+{ }^8 C_5 \times{ }^7 C_3$
$\Rightarrow \frac{8!}{5!\times 3!} \times \frac{7!}{5!\times 2!}+\frac{8!}{4!\times 4!} \times \frac{7!}{4!\times 3!}+\frac{8!}{5!\times 3!} \times \frac{7!}{4!\times 3!}$
$\Rightarrow \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1}+\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}+\frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}$
$\Rightarrow 56 \times 21+70 \times 35+56 \times 35$
$\Rightarrow 1176+2450+1960$
$\Rightarrow 5586$
$ii$. Ashish is selecting $3$ questions from part $I$ so he has to select remaining $5$ questions from part $II$
The number of ways of selection is $3$ questions from part $I$ and $5$ questions from part $II$ can be selected in ${ }^8 C_3 \times{ }^7 C_5$ ways
$\Rightarrow{ }^8 C_3 \times{ }^7 C_5$
$\Rightarrow \frac{8!}{5!\times 31} \times \frac{7!}{5!\times 2!}$
$\Rightarrow \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1}$
$\Rightarrow 56 \times 21$
$\Rightarrow 1176$
$iii.\ 4$ questions from part $I$ and $4$ questions from part $II$ can be selected
${ }^8 C_4 \times{ }^7 C_4$
$\Rightarrow \frac{8!}{4 \times 4!} \times \frac{7}{4!3!}$
$\Rightarrow \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}$
$\Rightarrow 70 \times 35$
$\Rightarrow 2450$
OR
$6$ questions from part $I$ and $2$ questions from part $II$ can be selected or $2$ questions from part $I$ and $6$ questions from part $II$ can be selected
$\Rightarrow{ }^8 C_6 \times{ }^7 C_2+{ }^8 C_2 \times{ }^7 C_6$
$\Rightarrow \frac{8!}{6!\times 2!} \times \frac{7!}{2!\times 5!}+\frac{8!}{6!\times 2!} \times \frac{7!}{1!\times 6!}$
$\Rightarrow \frac{8 \times 7}{2 \times 1} \times \frac{7 \times 6}{2 \times 1}+\frac{8 \times 7}{2 \times 1} \times 7$
$\Rightarrow 28 \times 21+28 \times 7$
$\Rightarrow 588+196=784$

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Question 24 Marks

For a group of $200 $ candidates, the mean and the standard deviation of scores were found to be $40$ and $15 ,$ respectively. Later on it was discovered that the scores of $43$ and $35$ were misread as $34$ and $53,$ respectively.

Student	English	Hindi	S.st	Science	Maths
Ramu	$39$	$59$	$84$	$80$	$41$
Rajitha	$79$	$92$	$68$	$38$	$75$
Komala	$41$	$60$	$38$	$71$	$82$
Patil	$77$	$77$	$87$	$75$	$42$
Pursi	$72$	$65$	$69$	$83$	$67$
Gayathri	$46$	$96$	$53$	$71$	$39$

$i.$ Find the correct variance. $(1)$
$ii$. What is the formula of variance. $(1)$
$iii.$ Find the correct mean. $(2)$
OR
Find the sum of correct scores.$ (2)$

Answer

$i.\ SD =\sigma=15$
$\Rightarrow$ Variance $=15^2=225$
According to the formula,
$\text { Variance }=\left(\frac{1}{n} \sum x_i^2\right)-\left(\frac{1}{n} \sum x_i\right)^2$
$\therefore \frac{1}{200} \sum x_i^2-(40)^2=225$
$\Rightarrow \frac{1}{200} \sum\left(x_i\right)^2-1600=225$
$\Rightarrow \sum\left(x_i\right)^2=200 \times 1825=365000$
This is an incorrect reading.
$\therefore \text { Corrected } \sum\left(x_i\right)^2=365000-34^2-53^2+43^2+35^2$
$=365000-1156-2809+1849+1225$
$=364109$
$\text { Corrected variance }=\left(\frac{1}{n} \times \text { Corrected } \sum x_i\right)-(\text { Corrected mean })^2$
$=\left(\frac{1}{200} \times 364109\right)-(39.955)^2$
$=1820.545-1596.402$
$=224.14$
$ii$. The formula of variance is $\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}{n}$.
$iii. $ Corrected mean $=\frac{\text { Corrected } \sum x_1}{200}$
$=\frac{7993}{200}$
$=39.955$
OR
We have:
$ n =200, \bar{X}=40, \sigma=15$
$\frac{1}{n} \sum x_{ i }=\bar{X}$
$\therefore \frac{1}{200} \sum x_i=40$
$\Rightarrow \sum x_i=40 \times 200=8000$
Since the score was misread, this sum is incorrect.
$\Rightarrow $ Corrected $ \sum x_i=8000-34-53+43+35$
$=8000-7$
$=7993$

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Question 34 Marks

Indian track and field athlete Neeraj Chopra, who competes in the Javelin throw, won a gold medal at Tokyo Olympics. He is the first track and field athlete to win a gold medal for India at the Olympics.

$i$. Name the shape of path followed by a javelin. If equation of such a curve is given by $x^2=-16 y,$ then find the coordinates of foci. $(1)$
$ii.$ Find the equation of directrix and length of latus rectum of parabola $x^2=-16 y. (1)$
$iii.$ Find the equation of parabola with Vertex $(0,0)$, passing through $(5,2)$ and symmetric with respect to $y-$ axis and also find equation of directrix.$ (2)$
OR
Find the equation of the parabola with focus $(2,0)$ and directrix $x=-2$ and also length of latus rectum. $(2)$

Answer

$i$.The path traced by Javelin is parabola.
A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point $($not on the line$)$ in the plane.
compare $x^2=-16 y$ with $x^2=-4 a y$
$\Rightarrow-4 a=-16$
$\Rightarrow a=4$
$ii.$ compare $x^2=-16 y$ with $x^2=-4 a y$
$\Rightarrow-4 a=-16$
$\Rightarrow a=4$
Equation of directrix for parabola $x^2=-4 a y$ is $y=a$
$\Rightarrow$ Equation of directrix for parabola $x^2=-16 y$ is $y=4$
Length of latus rectum is $4 a=4 \times 4=16$
$iii$. Equation of parabola with axis along $y -$ axis
$x^2=4 a y$
which passes through $(5, 2)$
$\Rightarrow 25=4 a \times 2$
$\Rightarrow 4 a=\frac{25}{2}$
hence required equation of parabola is
$x^2=\frac{25}{2} y$
$\Rightarrow 2 x ^2=25 y$
Equation of directrix is $y= -a$
Hence required equation of directrix is $8y + 25 = 0.$
OR
Since the focus $(2,0)$ lies on the $x-$ axis, the $x-$ axis itself is the axis of the parabola.
Hence the equation of the parabola is of the form either $y^2=4 a x$ or $y^2=-4 a x$.
Since the directrix is $x=-2$ and the focus is $(2,0),$
the parabola is to be of the form $y^2=4 a x$ with $a=2$.
Hence the required equation is $y^2=4(2) x=8 x$
length of latus rectum $=4 a=8$

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Maths Case study (4 Marks)

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