Case study (4 Marks)

Questions

Question 14 Marks

Class	Frequency
0-10	6
10-20	7
20-30	15
30-40	16
40-50	4
50-60	2

i. Find the mean deviation about median. (1)
ii. Find the Median. (1)
iii. Write the formula to calculate the Mean deviation about median? (2)
OR
Write the formula to calculate median? (2)

Answer

Class	$f _{ i }$	cf	Mid-point $\left( x _{ j }\right)$	$\left\| x _{ i }- M \right\|$	$f _{ i }\left\| x _{ i }- M \right\|$
0-10	6	6	5	23	138
10-20	7	13	15	13	91
20-30	15	28	25	3	45
30-40	16	44	35	7	112
40-50	4	48	45	17	68
50-60	2	50	55	27	54
	50				508

Here, $\frac{N}{2}=\frac{29}{2}=25$
Here, 25th item lies in the class 20-30. Therefore, 20-30 is the median class.
Here, l = 20, cf = 13, f = 15, b = 10 and N = 50
$\because$ Median, $M =l+\frac{\frac{N}{2}-c f}{f} \times b$
$\Rightarrow M=20+\frac{25-13}{15} \times 10=20+8=28$
Thus, mean deviation about median is given by
$MD ( M )=\frac{1}{N} \sum_{i=1}^6 f_i\left|x_i-M\right|=\frac{1}{50} \times 508=10.16$
Hence, mean deviation about median is 10.16.
ii. Here, l = 20, cf = 13, f = 15, b = 10 and N = 50
$\begin{array}{l}\because \text { Median, } M=l+\frac{\frac{N}{2}-cf}{f} \times b \\ \Rightarrow M =20+\frac{25-13}{15} \times 10=20+8=28\end{array}$
$\text { iii. } MD=\frac{\Sigma f_4\left|x_1-M\right|^{15}}{N}$
OR
$M =1+\frac{\frac{ N}{2}-e f}{f} \times h$

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Question 24 Marks

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.

i. Find the total number of arrangements so that Raju and Ravi are at extreme positions? (1)
ii. Find the number of arrangements so that Joseph is sitting in the middle. (1)
iii. Find the number of arrangements so that three girls are together. (2)
OR
Find the number of arrangements so that Aman and Ravi are not together? (2)

Answer

i.Given Raju and Ravi are at the extreme positions
Case 1 Raju ______ ______ ______ ______ ______ Ravi
Case 2 Ravi ______ ______ ______ ______ ______ Raju
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
ii. ______ ______ ______ Joseph ______ ______ ______
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 $=5!\times 3!=120 \times 6=720$
OR
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 $\ldots$ (i)
Total number of sitting arrangements of all 7 students without restriction
______ ______ ______ ______ ______ ______ ______
All seven students can fill seven seats in 7! Ways
$71=7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=5040$ ways $\ldots$ (ii)
But here we need the number of arrangements so that Aman and Ravi are not together = Total number of sitring arrangements of all 7 students without restriction - Number of arrangements so that Aman and Ravi are togethes.
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

A farmer wishes to install 2 handpumps in his field for watering.

The farmer moves in the field while watering in such a way that the sum of distances between the farmer and each handpump is always 26m. Also, the distance between the hand pumps is 10 m.

i. Name the curve traced by farmer and hence find the foci of curve. (1)
ii. Find the equation of curve traced by farmer. (1)
iii. Find the length of major axis, minor axis and eccentricity of curve along which farmer moves. (2)
OR
iv. Find the length of latus rectum. (2)

Answer

i. The curve traced by farmer is ellipse. Because An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.
Two positions of hand pumps are foci Distance between two foci $=2 c=10$ Hence $c=5$ Here foci lie on $x$ axis \& coordinates of foci $=( \pm c, 0)$
Hence coordinates of foci $=( \pm 5,0)$
ii. $\frac{x^2}{169}+\frac{y^2}{144}=1$
Sum of distances from the foci = 2a
Sum of distances between the farmer and each hand pump is = 26 = 2a
$\Rightarrow 2 a=26 \Rightarrow a=13 m$
Distance between the handpump = 10m = 2c
$\begin{array}{l}\Rightarrow c=5 m \\
c^2=a^2-b^2 \\
\Rightarrow 25=169-b^2 \\
\Rightarrow b^2=144\end{array}$
Equation is $\frac{x^2}{169}+\frac{y^2}{144}=1$
iii. Equation of ellipse is $\frac{x^2}{169}+\frac{y^2}{144}=1$ comparing with standard equation of ellipse $a =13, b=12$ and $c=5$ (given)
Length of major axis $=2 a =2 \times 13=26$
Length of minor axis $=2 b=2 \times 12=24$
eccentricity $e=\frac{\in}{a}=\frac{5}{13}$
OR
Equation of the ellipse is $\frac{x^2}{169}+\frac{y^2}{144}=1$ hence $a =13$ and $b =12$
length of latus rectum of ellipse is given by $\frac{2 b^2}{ a }=\frac{2 \times 144}{13}$

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MATHS Case study (4 Marks)

Case study (4 Marks)

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