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Case study (4 Marks)

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Question 14 Marks
Answer
i.Since, at least 3 questions from each part have to be selected
Part IPart II
35
44
35
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 
${ }^8 C_3 \times{ }^7 C_5+{ }^8 C_4 \times{ }^7 C_4+{ }^8 C_5 \times{ }^7 C_3$
$\begin{array}{l}
\Rightarrow \frac{8!}{5!\times 3!} \times \frac{7!}{5!\times 2!}+\frac{8!}{4!\times 4!} \times \frac{7!}{4!\times 3!}+\frac{8!}{5!\times 3!} \times \frac{7!}{4!\times 3!} \\
\Rightarrow \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1}+\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}+\frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} \\
\Rightarrow 56 \times 21+70 \times 35+56 \times 35 \\
\Rightarrow 1176+2450+1960 \\
\Rightarrow 5586\end{array}$
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
$\begin{array}{l}\Rightarrow{ }^8 C_3 \times{ }^7 C_5 \\ \Rightarrow \frac{8!}{5!\times 31} \times \frac{7!}{5!\times 2!} \\ \Rightarrow \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1} \\ \Rightarrow 56 \times 21 \\ \Rightarrow 1176\end{array}$
iii. 4 questions from part I and 4 questions from part II can be selected 
$\begin{array}{l}{ }^8 C_4 \times{ }^7 C_4 \\ \Rightarrow \frac{8!}{4 \times 4!} \times \frac{7}{4!3!} \\ \Rightarrow \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \\ \Rightarrow 70 \times 35 \\ \Rightarrow 2450\end{array}$
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 
$\begin{array}{l}\Rightarrow{ }^8 C_6 \times{ }^7 C_2+{ }^8 C_2 \times{ }^7 C_6 \\ \Rightarrow \frac{8!}{6!\times 2!} \times \frac{7!}{2!\times 5!}+\frac{8!}{6!\times 2!} \times \frac{7!}{1!\times 6!}\end{array}$
$\begin{array}{l}\Rightarrow \frac{8 \times 7}{2 \times 1} \times \frac{7 \times 6}{2 \times 1}+\frac{8 \times 7}{2 \times 1} \times 7 \\ \Rightarrow 28 \times 21+28 \times 7 \\ \Rightarrow 588+196=784\end{array}$

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Question 24 Marks
Answer
i. $SD =\sigma=15$
$\Rightarrow$ Variance $=15^2=225$
According to the formula,
$\begin{array}{l}\text { Variance }=\left(\frac{1}{n} \sum x_i^2\right)-\left(\frac{1}{n} \sum x_i\right)^2 \\ \therefore \frac{1}{200} \sum x_i^2-(40)^2=225 \\ \Rightarrow \frac{1}{200} \sum\left(x_i\right)^2-1600=225 \\ \Rightarrow \sum\left(x_i\right)^2=200 \times 1825=365000\end{array}$
This is an incorrect reading. 
$\begin{array}{l}\therefore \text { Corrected } \sum\left(x_i\right)^2=365000-34^2-53^2+43^2+35^2 \\ =365000-1156-2809+1849+1225 \\ =364109\end{array}$
$\begin{array}{l}\text { Corrected variance }=\left(\frac{1}{n} \times \text { Corrected } \sum x_i\right)-(\text { Corrected mean })^2 \\ =\left(\frac{1}{200} \times 364109\right)-(39.955)^2 \\ =1820.545-1596.402 \\ =224.14\end{array}$
ii. The formula of variance is $\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}{n}$.
iii. Corrected mean $=\frac{\text { Corrected } \sum x_1}{200}$
$\begin{array}{l}=\frac{7993}{200} \\
=39.955\end{array}$
OR
We have: 
$\begin{array}{l} n =200, \bar{X}=40, \sigma=15 \\ \frac{1}{n} \sum x_{ i }=\bar{X} \\ \therefore \frac{1}{200} \sum x_i=40 \\ \Rightarrow \sum x_i=40 \times 200=8000\end{array}$
Since the score was misread, this sum is incorrect. 
$\begin{array}{l}\Rightarrow \text { Corrected } \sum x_i=8000-34-53+43+35 \\ =8000-7 \\ =7993\end{array}$
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Question 34 Marks
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 $x^2=-4 a y$
$\begin{array}{l}\Rightarrow-4 a=-16 \\ \Rightarrow a=4\end{array}$
ii. compare $x^2=-16 y$ with $x^2=-4 a y$
$\begin{array}{l}\Rightarrow-4 a=-16 \\ \Rightarrow a=4\end{array}$
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) 
$\begin{array}{l}\Rightarrow 25=4 a \times 2 \\ \Rightarrow 4 a=\frac{25}{2}\end{array}$
hence required equation of parabola is 
$\begin{array}{l}x^2=\frac{25}{2} y \\ \Rightarrow 2 x ^2=25 y\end{array}$
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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Case study (4 Marks) - MATHS STD 11 Science Questions - Vidyadip