Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Consider the following diagram, where the forces in the cable are given.

i. What is the cartesian equation of line along $EA$? $(1)$
$\rightarrow$
$ii.$ The vector $ED$ is $(1)$
$iii.$ The length of the cable $EB$ is $(2)$
$OR$
What is the result of adding up all the vectors along the cables? $(2)$

Answer

$i.$ Clearly, the coordinates of $A$ are $(8, -6, 0)$ and that of $E$ are $(0, 0, 24)$.
Also, cartesian equation of line along $EA$ is given by
$\frac{x-0}{8-0}=\frac{y-0}{-6-0}=\frac{z-24}{0-24}$
$\Rightarrow \frac{x}{8}=\frac{y}{-6}=\frac{z-24}{-24} $
$\Rightarrow \frac{x}{-4}=\frac{y}{3}=\frac{z-24}{12}$
$ii.$ Clearly, the coordinates of $D$ are $(-8, -6, 0)$ and that of $E$ are $(0, 0, 24)$
$\therefore$ Vector $\overrightarrow{E D}$ is $(-8-0) \hat{i}+(-6-0) \hat{j}+(0-24) \hat{k}$, i.e., $-8 \hat{i}-6 \hat{j}-24 \hat{k}$.
$iii.$ Since, the coordinates of $B$ are $(8, 6, 0)$ and that of $E$ are $(0, 0, 24),$ therefore length of cable
$ EB =\sqrt{(8-0)^2+(6-0)^2+(0-24)^2}$
$=\sqrt{64+36+576}=\sqrt{676}=26 \text { units }$
$OR$
Sum of all vectors along the cables
$=\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$
$=(8 \hat{i}-6 \hat{j}-24 \hat{k})+(8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}-6\hat{j}-24 \hat{k})$
$=-96 \hat{k}$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Akash and Prakash appeared for first round of an interview for two vacancies. The probability of Nisha's selection is $\frac{1}{3}$ and that of Ayushi's selection is $\frac{1}{2}$.

$i.$ Find the probability that both of them are selected. $(1)$
$ii.$ The probability that none of them is selected. $(1)$
$iii.$ Find the probability that only one of them is selected.$(2)$
$OR$
Find the probability that atleast one of them is selected. $(2)$

Answer

$\text { i. } P(A)=\frac{1}{3}, P\left(A^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3}$
$P(B)=\frac{1}{2}, P\left(b^{\prime}\right)=1-\frac{1}{3}=\frac{1}{2}$
$P($Both are selected$) =P(A \cap B)=P(A) \cdot P(B)=\frac{1}{3} \cdot \frac{1}{2}$
$P($Both are selected$) =\frac{1}{6}$
$ii.$
$ P ( A )=\frac{1}{3}, P \left( A ^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3}$
$P ( B )=\frac{1}{2}, P \left( B ^{\prime}\right)=1-\frac{1}{3}=\frac{1}{2}$
$P($Both are selected$)=P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)=\frac{2}{3} \cdot \frac{1}{2}$
$P($Both are selected$) =\frac{1}{3}$
$iii.$
$ P ( A )=\frac{1}{3}, P \left( A ^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3}$
$P ( B )=\frac{1}{2}, P \left( B ^{\prime}\right)=1-\frac{1}{3}=\frac{1}{2}$
$OR$
$ P ( A )=\frac{1}{3}, P \left( A ^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3}$
$P ( B )=\frac{1}{2}, P \left( b ^{\prime}\right)=1-\frac{1}{3}=\frac{1}{2}$
$P($none of them selected$) =P\left(A^{\prime}\right) \cdot P(B)+P(A) \cdot P\left(B^{\prime}\right)=\frac{2}{3} \cdot \frac{1}{2}+\frac{1}{3} \cdot \frac{1}{2}$
$P($Both are selected$)=\frac{3}{6}=\frac{1}{2}$

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
Dinesh is having a jewelry shop at Green Park, normally he does not sit on the shop as he remains busy in political meetings. The manager Lisa takes care of jewelry shop where she sells earrings and necklaces. She gains profit of $₹30$ on pair of earrings $₹40$ on neckless. It takes $30$ minutes to make a pair of earrings and $1$ hour to make a necklace, and there are $10$ hours a week to make jewelry. In addition, there are only enough materials to make $15$ total of jewelry items per week..

$i.$ Formulate the above information mathematically. $(1)$
$ii.$ Graphically represent the given data. $(1)$
$iii.$ To obtain maximum profit how many pair of earing and neckleses should be sold? $(2)$
$OR$
What would be the profit if 5 pairs of earrings and 5 necklaces are made? $(2)$

Answer

$i.$ Let number of pairs of earing $= x$ and number of Necklaces $= y$
As per the given information
$x, y \geq 0$
$0.5 x+y \leq 10$
$x+y \leq 15$
Profit function = Z = 30x + 40y
$ii.$ Let number of pairs of earing $= x$ and number of Necklaces $= y$
As per the given information
$x, y \geq 0$
$0.5 x+y \leq 10$
$x+y \leq 15$
Profit function $=Z= 30x + 40y$

iii. From graph corner points are $(0, 0), (0, 10), (10, 5)$ and $(15, 0).$

corner points	maximum profit $= Z = 30x + 40y$
$(0,0)$	$Z = 0$
$(0,10)$	$Z = ₹400$
$(10,5)$	$Z = ₹500$
$(15,0)$	$Z = ₹450$

Hence profit is maximum when $x =$ number of pair of Earings $= 10$ and $y =$ Number of Neckleses
$OR$
When $x = 5$ and $y = 5$
$Z = 30x + 40y = 150 + 200 = ₹350$

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MATHS Case study (4 Marks)

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