Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Mrs. Maya is the owner of a high-rise residential society having $50$ apartments. When he set rent at $₹ 10000$ month, all apartments are rented. If he increases rent by $₹ 250 $ month, one fewer apartment is rented.
The maintenance cost for each occupied unit is $₹ 500$ month.

$i$. If $P$ is the rent price per apartment and $N$ is the number of rented apartments, then find the profit. $(1)$
$ii.$ If $x$ represents the number of apartments which are not rented, then express profit as a function of $x. (1)$
$iii.$ Find the number of apartments which are not rented so that profit is maximum. $(2)$
OR
Verify that profit is maximum at critical value of $x$ by second derivative test. $(2)$

Answer

$i$. If $P$ is the rent price per apartment and $N$ is the number of rented apartments, the profit is given by $N P-500 N= N \ ( P -500)$
$[\because . ₹ 500 $ month is the maintenance charge for each occupied unit$]$
$ii$. Let $R$ be the rent price per apartment and $N$ is the number of rented apartments.
Now, if $x$ be the number of non $-$ rented apartments, then $N(x)=50-x$ and $R(x)=10000+250 x$
Thus, profit $= P ( x )= NR =(50- x )(10000+250 x -500)$
$=(50-x)(9500+250 x)=250(50-x)(38+x)$
$iii$. We have, $P(x)=250(50-x)(38+x)$
Now, $P^{\prime}(x)=250[50-x-(38+x)]=250[12-2 x]$
For maxima/minima, put $P ^{\prime}( x )=0$
$\Rightarrow 12-2 x=0$
$ \Rightarrow x=6$
Number of apartments are $6$ .
OR
$P^{\prime}(x)=250(12-2 x)$
$P^{\prime \prime}(x)=-500<0$
$\Rightarrow P ( x )$ is maximum at $x =6$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Team $\text{P, Q, R}$ went for playing a tug of war game. .
Teams $\text{P, Q, R}$ have attached a rope to a metal ring and is trying to pull the ring into their own areas $($team areas when in the given figure below$)$.
Team $P$ pulls with force$F _1=4 \hat{i}+0 \hat{j} KN$
Team $Q$ pull with force $F _2=-2 \hat{i}+4 \hat{j} KN$
Team $R$ pulls with force $F _3=-3 \hat{i}-3 \hat{j} KN$

$i.$ What is the magnitude of the teams combined force? $(1)$
$ii.$ Find the magnitude of Team $B. (1)$
$iii.$ Which team will win the game? $(2)$
$OR$
Find the probability that she gets grade $A$ in at least one subject. $(2)$

Answer

$i.$ Let $F$ be the combined force,
$ \therefore \vec{F}=\vec{F}_1+\vec{F}_2+\vec{F}_3$
$=(4 \hat{i}+0 \hat{j})+(-2 \hat{i}+4 \hat{j})+(-3 \hat{i}-3 \hat{j})$
$=(4-2-3) \hat{i}+(0+4-3) \hat{j}$
$=-\hat{i}+\hat{j}$
$\therefore \vec{F}\left|=\left|\sqrt{(-1)^2+1^2}\right|\right.$
$=|\sqrt{2}| KN $
$ii.$ Magnitude of force of Team $B =$
$ \left|\vec{F}_2\right|=\left|\sqrt{(-2)^2+4^2}\right|=\sqrt{20} KN$
$=2 \sqrt{5} KN$
$iii.$ We have,$ \left|\vec{F}_1\right|=\left|\sqrt{(4)^2+0^2}\right|=4 KN $$\left|\vec{F}_2\right|=\left|\sqrt{(-2)^2+4^2}\right|=\sqrt{20} KN$
$\text { and }\left|\vec{F}_3\right|=\left|\sqrt{(-3)^2+(-3)^2}\right|=\sqrt{18} KN $
Here, the magnitude of force $F_2$ is greater, therefore team $Q$ will win the game.
$OR$
we have,
Combined force, $\vec{F}=-\hat{i}+\hat{j}$
$\therefore \theta \tan ^{-1}\left(\frac{F_y}{F_x}\right)=\tan ^{-1}\left(\frac{1}{-1}\right)$
$=\tan ^{-1}(1)$
$=\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$
$=\frac{3 \pi}{4} \text { radians }$

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
Shama is studying in class $XII$.
She wants do graduate in chemical engineering.
Her main subjects are mathematics, physics, and chemistry. In the examination, her probabilities of getting grade $A$ in these subjects are $0.2, 0.3,$ and $0.5$ respectively.

$1.$ Find the probability that she gets grade $A$ in all subjects. $(1)$
$2.$ Find the probability that she gets grade $A$ in no subjects. $(1)$
$3.$ Find the probability that she gets grade $A$ in two subjects. $(2)$
$OR$
Find the probability that she gets grade $A$ in at least one subject. $(2)$

Answer

$i. P ($Grade $A$ in Maths$) = P(M) = 0.2$
$P ($Grade $A$ in Physics$) = P(P) = 0.3$
$P ($Grade $A$ in Chemistry$) = P(C) = 0.5$
$ P (\text { not A garde in Maths) }= P (\bar{M})=1-0.2=0.8$
$P (\text { not A garde in Physics })= P (\bar{P})=1-0.3=0.7$
$P (\text { not A garde in Chemistry })= P (\bar{C})=1-0.5=0.5$
$P (\text { getting grade } A \text { in all subjects })= P ( M \cap P \cap C )$
$= P ( M ) \times P ( P ) \times P ( C )$
$=0.2 \times 0.3 \times 0.5=0.03$
$ii. P ($Grade $A$ in Maths$) = P(M) = 0.2$
$P ($Grade $A$ in Physics$) = P(P) = 0.3$
$P ($Grade $A$ in Chemistry$) = P(C) = 0.5$
$ P (\text { not A garde in Maths })= P (\bar{M})=1-0.2=0.8$
$P (\text { not A garde in Physics })= P (\bar{P})=1-0.3=0.7$
$P (\text { not A garde in Chemistry })= P (\bar{C})=1-0.5=0.5$
$P (\text { getting grade } A \text { in on subjects })=P(\bar{M} \cap \bar{P} \cap \bar{C})$
$=P(\bar{M}) \times P(\bar{P}) \times P(\bar{C})$
$=0.8 \times 0.7 \times 0.5=0.280$
$P ($Grade $A$ in Maths $)= P(M) = 0.2$
$P ($Grade $A$ in Physics$) = P(P) = 0.3$
$P ($Grade $A$ in Chemistry$) I = P(C) = 0.5$
$ P (\text { not } A \text { garde in Maths })= P (\bar{M})=1-0.2=0.8$
$P (\text { not A garde in Physics })= P (\bar{P})=1-0.3=0.7$
$P (\text { not } A \text { garde in Chemistry })= P (\bar{C})=1-0.5=0.5$
P(getting grade A in 2 subjects)
$\Rightarrow P (\text { grade } A \text { in } M \text { and } P \text { not in } C )+ P (\text { grade } A \text { in } P \ C \text { not in } M )+ P (\text { grade } A \text { in } M \ C \text { not in } P )$
$\Rightarrow P (M \cap P \cap \bar{C})+ P (P \cap C \cap \bar{M})+ P (M \cap C \cap \bar{P})$
$\Rightarrow 0.2 \times 0.3 \times 0.5+0.3 \times 0.5 \times 0.8+0.2 \times 0.5 \times 0.7=0.03+0.12+0.07$
$P($getting grade $A$ in $2$ subjects$) = 0.22$
OR
$P ($Grade $A$ in Maths$) = P(M) = 0.2$
$P ($Grade $A$ in Physics$) = P(P) = 0.3$
$P ($Grade $A$ in Chemistry $)= P(C) = 0.5$
$ P (\text { not A garde in Maths })= P (\bar{M})=1-0.2=0.8$
$P (\text { not A garde in Physics })= P (\bar{P})=1-0.3=0.7$
$P (\text { not A garde in Chemistry })= P (\bar{C})=1-0.5=0.5$
$P($getting grade $A$ in $1$ subjects$)$
$\Rightarrow P (\text { grade } A \text { in } M \text { not in } P \text { and } C )+ P (\text { grade } A \text { in } P \text { not in } M \text { and } C )+ P (\text { grade } A \text { in } C \text { not in } P \text { and } M )$
$\Rightarrow P (M \cap \bar{P} \cap \bar{C})+ P (P \cap \bar{C} \cap \bar{M})+ P (C \cap \bar{M} \cap \bar{P})$
$\Rightarrow 0.2 \times 0.7 \times 0.5+0.3 \times 0.5 \times 0.8+0.5 \times 0.8 \times 0.7=0.07+0.12+0.028$
$P ($ getting grade $A$ in $1$ subjects$) =0.47$

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MATHS Case study (4 Marks)

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