Case study (4 Marks)

Question 14 Marks

Read the text carefully and answer the questions:
Two friends Pankaj and Pooja were playing cards. There were 52 cards in a deck.

(a) In how many ways Sunil can select all four cards from same suit?
(b) In how many ways Anita can select four cards from different suit?
(c) In how many ways Sunil can select all face cards?
(d) In how many ways Anita can select two cards of same colour?

Answer

Read the text carefully and answer the questions:
Two friends Pankaj and Pooja were playing cards. There were 52 cards in a deck.

(i) Sunil can choose four cards from same suit in $4 \times{ }^{13} \mathrm{C}_{4}$ ways
$=4 \times \frac{13!}{9!\times 4!}$
$=4715=2860$
(ii) Here one card to be selected from each suit therefore, he can select in ${ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}$ ways $=\left({ }^{13} \mathrm{C}_{1}\right)^{4}=28561$
(iii) There are 12 face cards and 4 are to be selected out of these 12 cards. This can be done in ${ }^{12} \mathrm{C}_{4}$ ways
$=\frac{12!}{8!4!} 495$
(iv) ${ }_{\text {Anita can select two cards of same colour in }}{ }^{26} \mathrm{C}_{2}+{ }^{26} \mathrm{C}_{2}$ ways $=325+325=650$

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Question 24 Marks

Read the text carefully and answer the questions:
The purpose of the student council is to give students an opportunity to develop leadership by organizing and carrying out school activities and service projects. Create an environment where every student can voice out their concern or need. Raju, Ravi Joseph, Sangeeta, Priya, Meena and Aman are members of student's council. There is a photo session in a school these 7 students are to be seated in a row for photo session.

(a) Find the total number of arrangements so that Raju and Ravi are at extreme positions?
(b) Find the number of arrangements so that Joseph is sitting in the middle.
(c) Find the number of arrangements so that three girls are together.
(d) Find the number of arrangements so that Aman and Ravi are not together?

Answer

(i) Given Raju and Ravi are at the extreme positions

So remaining 5 places are filled in 5 ! Ways in both cases
$5 !=5 \times 4 \times 3 \times 2 \times 1=120$
Hence total number of arrangements $=120 \times 2=240$ ways

So here middle place is occupied by Joseph remaining 6 places are filled by remaining 6 students in 6 ! Ways $6!=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$ ways
(iii) When all girls are together let's consider them as a single unit. So four 4 boys with single group of girls can be arranged in $4+1=5$ ! Ways

5 ! $=5 \times 4 \times 3 \times 2 \times 1=120$
But all the tree girls can be arranged in themselves in 3! Ways $=3 \times 2 \times 1=6$
Hence total number of ways $=5!\times 3!=120 \times 6=720$
(iv) When Aman and Ravi are together let's consider them as a single unit. So remaining 5 students with single group of Aman and Ravi can be arranged in $5+1=6$ ! Ways

$6!=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$
But Aman and Ravi can be arranged in themselves in 2! Ways $=2 \times 1=2$
Hence total number of ways $=6!\times 2!=720 \times 2=1440$ ways $\ldots\text{(i)}$
Total number of sitting arrangements of all 7 students without restriction

All seven students can fill seven seats in 7 ! Ways
$7!=7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=5040$ ways $\ldots\text{(ii)}$
But here we need the number of arrangements so that Aman and Ravi are not together $=$ Total number of sitting arrangements of all 7 students without restriction - Number of arrangements so that Aman and Ravi are together.
From (i) and (ii) we have
The number of arrangements so that Aman and Ravi are not together $=5040-1440=3600$

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Question 34 Marks

Read the text carefully and answer the questions:
In this age of competitions rank plays an important role. Let us try to answer some questions related to percentile rank with respect to the given data of marks of 10 students $13,52,42,22,44,105,45,88,90,76$.
(a) If score is 90 , then find the percentile rank?
(b) If score is 52 then find the percentile rank?
(c) If score is 105 then find the percentile rank?
OR
If percentile rank is 60 then find the score?

Answer

Read the text carefully and answer the questions:
In this age of competitions rank plays an important role. Let us try to answer some questions related to percentile rank with respect to the given data of marks of 10 students $13,52,42,22,44,105,45,88,90,76$.
(i) Percentile rank $=\frac{8}{10} \times 100=80$
(ii) Percentile rank $=\frac{5}{10} \times 100=50$
(iii) Percentile rank $=\frac{9}{10} \times 100=90$
OR
$60=\frac{\text { marks less than } a}{10} \times 100 \Rightarrow$ marks less than $a =6$
$a=76$

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Question 44 Marks

Read the text carefully and answer the questions:
A triangular park has two of its vertices as $B(-4,1)$ and $C(2,11)$. The third vertex $A$ is a point dividing the line joining the points $(3,1)$ and $(6,4)$ in the ratio $2: 1$.
(a) Find the coordinates of third vertex A.
(b) Find the equation of line passing through A and parallel to BC.
(c) Find the equation of the line passing through A and perpendicular to BC .
OR
Find the area of triangular field ABC .

Answer

Read the text carefully and answer the questions:
A triangular park has two of its vertices as $\mathrm{B}(-4,1)$ and $\mathrm{C}(2,11)$. The third vertex A is a point dividing the line joining the points $(3,1)$ and $(6,4)$ in the ratio 2 : 1 .

Coordinates of $A =\left(\frac{2 \times 6+1 \times 3}{2+1}, \frac{2 \times 4+1 \times 1}{2+1}\right)$ i.e. $(5,3)$
(ii) Slope of line $BC =\frac{11-1}{2+4}=\frac{5}{3}$
$\therefore$ Slope of line parallel to $B C=\frac{5}{3}$
Equation of line through $\mathrm{A}(5,3)$ and parallel to BC is
$y-3=\frac{5}{3}(x-5) \Rightarrow 3 y-9=5 x-25$
$\Rightarrow 5 \mathrm{x}-3 \mathrm{y}-16=0$
(iii) Slope of line perpendicular to $\mathrm{BC}=-\frac{3}{5}$
Equation of line through $\mathrm{A}(5,3)$ and perpendicular to BC is
$y-3=-\frac{3}{5}(x-5) \Rightarrow 5 y-15=-3 x+15$
$\Rightarrow 3 \mathrm{x}+5 \mathrm{y}-30=0$
OR
$\mathrm{A}(5,3), \mathrm{B}(-4,1), \mathrm{C}(2,11)$
Area of $\Delta \mathrm{ABC}=\frac{1}{2}|5(1-11)+(-4)(11-3)+2(3-1)|$ sq.units
$=\frac{1}{2}|-50-32+4|$ sq. units
$=\frac{1}{2} \times 78 s q$. units $=39$ sq.units

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Applied Maths Case study (4 Marks)

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