Download App

Mathematics Case study (4 Marks)

Model Paper 3 · CBSE STD 12 Science (CBSE - English Medium). 3 questions.

Questions

Case study (4 Marks)

🎯

Test yourself on this topic

3 questions · timed · auto-graded

Question 14 Marks
Answer
i
Corner PointsValue of z = 4x - 6y
(0,3)4 x 0 - 6 x 3 = - 18
(5,0)4 x 5 - 6 x 0 = 20
(6,8)4 x 6 - 6 x 8 = - 24
(0,8)4 x 0 - 6 x 8 = - 48
Minimum value of Z is - 48 which occurs at (0, 8).
ii
Corner PointsValue of z = 4x - 6y
(0,3)4 x 0 - 6 x 3 = - 18
(5,0)4 x 5 - 6 x 0 = 20
(6,8)4 x 6 - 6 x 8 = - 24
(0,8)4 x 0 - 6 x 8 = - 48
Maximum value of Z is 20, which occurs at (5, 0).
iii
Corner PointsValue of z = 4x - 6y
(0,3)4 x 0 - 6 x 3 = - 18
(5,0)4 x 5 - 6 x 0 = 20
(6,8)4 x 6 - 6 x 8 = - 24
(0,8)4 x 0 - 6 x 8 = - 48
Maximum of Z-Minimum of Z = 20 - (- 48) = 20 + 48 = 68
OR
The corner points of the feasible region are U(0, 0) A(3, 0) B(3, 2) C(2,3) D(0, 3)
View full question & answer→
Question 24 Marks
Read the following text carefully and answer the questions that follow:
Two motorcycles $A$ and $B$ are running at the speed more than allowed speed on the road along the lines
$\vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}+\mu(2 \hat{i}+\hat{j}+\hat{k})$, respectively.
Image
$i$. Find the cartesian equation of the line along which motorcycle $A$ is running. $(1)$
$ii$. Find the direction cosines of line along which motorcycle $A$ is running. $(1)$
$iii$. Find the direction ratios of line along which motorcycle $B$ is running. $(2)$
OR
Find the shortest distance between the given lines. $(2)$
Answer
$i$. The line along which motorcycle $A$ is running, $\vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k}),$ which can be rewritten as
$(x \hat{i}+y \hat{j}+z \hat{k})=\lambda \hat{i}+2 \lambda \hat{j}-\lambda \hat{k}$
$\Rightarrow x =\lambda, y =2 \lambda, z =-\lambda$
$ \Rightarrow \frac{x}{1}=\lambda, \frac{y}{2}=\lambda, \frac{z}{-1}=\lambda$
$ii$. Clearly, $D.R$. of the required line are $<1,2,-1\rangle$
$\therefore D.C$. are
$\left(\frac{1}{\sqrt{1^2+2^2+(-1)^2}}, \frac{2}{\sqrt{1^2+2^2+(-1)^2}}, \frac{-1}{\sqrt{1^2+2^2+(-1)^2}}\right)$
$\text { i.e., }\left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right)$
$iii.$ The line along which motorcycle $B$ is running, is $\vec{r}=(3 \hat{i}+3 \hat{j})+\mu(2 \hat{i}+\hat{j}+\hat{k})$, which is parallel to the vector $2 \hat{i}+\hat{j}+\hat{k}$
$\therefore D.R.$ of the required line are $(2,1,1)$.
OR
Here, $\vec{a}_1=0 \hat{i}+0 \hat{j}+0 \hat{k}, \vec{a}_2$
$=3 \hat{i}+3 \hat{j}, \vec{b}_1=\hat{i}+2 \hat{j}-\hat{k}, \vec{b}_2=2 \hat{i}+\hat{j}+\hat{k}$
$\therefore \vec{a}_2-\vec{a}_1=3 \hat{i}+3 \hat{j}$ and
$\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 1\end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}$
Now,
$\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=(3 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}-3 \hat{j}-3 \hat{k})$
$=9-9=0$
Hence, shortest distance between the given lines is $0$ .
View full question & answer→
Question 34 Marks
Read the following text carefully and answer the questions that follow:
To hire a marketing manager, it's important to find a way to properly assess candidates who can bring radical
changes and has leadership experience.
Ajay, Ramesh and Ravi attend the interview for the post of a marketing manager. Ajay, Ramesh and Ravi
chances of being selected as the manager of a firm are in the ratio $4:1:2$ respectively. The respective probabilities for them to introduce a radical change in marketing strategy are $0.3, 0.8,$ and $0.5$. If the change does take place.
Image
$i.$ Find the probability that it is due to the appointment of Ajay $(A). (1)$
$ii.$ Find the probability that it is due to the appointment of Ramesh $(B). (1)$
$iii$. Find the probability that it is due to the appointment of Ravi $(C). (2)$
OR
Find the probability that it is due to the appointment of Ramesh or Ravi. $(2)$
Answer
$i$. Let
$E_1: A$ jay $(A)$ is selected,
$E_2:$ Ramesh $(B)$ is selected,
$E_3:$ Ravi $(C)$ is selected
Let $A$ be the event of making a change
$P\left(E_1\right)=\frac{4}{7}, P\left(E_2\right)=\frac{1}{7}, P\left(E_3\right)=\frac{2}{7}$
$P \left( A / E _1\right)=0.3, P \left( A / E _2\right)=0.8, P \left( A / E_3\right)=0.5$
$P \left( E _1 / A \right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)+P\left(E_3\right) P\left(A / E_3\right)}$
$=\frac{\frac{4}{7} \times 0.3}{\frac{4}{7} \times 0.3+\frac{1}{7} \times 0.8+\frac{2}{7} \times 0.5}=\frac{\frac{1.2}{7}}{\frac{1.2}{7}+\frac{0.8}{7}+\frac{1}{7}}=\frac{\frac{1.2}{7}}{\frac{3}{7}}$
$=\frac{1.2}{3}=\frac{12}{30}=\frac{2}{5}$
$ii$. Let $E_1: \operatorname{Ajay}( A )$ is selected, $E _2:$ Ramesh $(B)$ is selected, $E _3:$ Ravi $( C )$ is selected
Let $A$ be the event of making a change
$P\left(E_1\right)=\frac{4}{7}, P\left(E_2\right)=\frac{1}{7}, P\left(E_3\right)=\frac{2}{7}$
$P \left( A / E _1\right)=0.3, P \left( A / E _2\right)=0.8, P \left( A / E _3\right)=0.5$
$P \left( E _2 / A \right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)+P\left(E_3\right) P\left(A / E_3\right)}$
$=\frac{\frac{1}{7} \times 0.8}{\frac{4}{7} \times 0.3+\frac{1}{7} \times 0.8+\frac{2}{7} \times 0.5}=\frac{\frac{0.8}{7}}{\frac{1.2}{7}+\frac{0.8}{7}+\frac{1}{7}}=\frac{\frac{0.8}{7}}{\frac{3}{7}}$
$=\frac{0.8}{3}=\frac{8}{30}=\frac{4}{15}$
$iii$. Let $E _1$ : Ajay $(A)$ is selected, $E _2$ : Ramesh $(B)$ is selected, $E _3$ : Ravi $(C)$ is selected
Let A be the event of making a change
$P\left(E_1\right)=\frac{4}{7}, P\left(E_2\right)=\frac{1}{7}, P\left(E_3\right)=\frac{2}{7}$
$P \left( A / E _1\right)=0.3, P \left( A / E _2\right)=0.8, P \left( A / E _3\right)=0.5$
$ P \left( E _3 / A \right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)+P\left(E_3\right) P\left(A / E_3\right)}$
$=\frac{\frac{2}{7} \times 0.5}{\frac{4}{7} \times 0.3+\frac{1}{7} \times 0.8+\frac{2}{7} \times 0.5}=\frac{\frac{1}{7}}{\frac{1.2}{7}+\frac{0.8}{7}+\frac{1}{7}}=\frac{1}{3}$
OR
Let $E_1$ : Ajay $(A)$ is selected, $E_2$ : Ramesh $(B)$ is selected, $E_3:$ Ravi $(C)$ is selected
Let $A$ be the event of making a change
$P\left(E_1\right)=\frac{4}{7}, P\left(E_2\right)=\frac{1}{7}, P\left(E_3\right)=\frac{2}{7}$
$P \left( A / E _1\right)=0.3, P \left( A / E _2\right)=0.8, P \left( A / E_3\right)=0.5$
Ramesh or Ravi
$\Rightarrow P \left( E _2 / A \right)+ P \left( E _3 / A \right)=\frac{4}{15}+\frac{1}{3}=\frac{9}{15}=\frac{3}{5}$
View full question & answer→
Generate Paper FreeAll Model Paper 3 topics
Case study (4 Marks) - Mathematics STD 12 Science Questions - Vidyadip