Case study (4 Marks)

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Question 14 Marks

Arun is running in a racecourse note that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m.

i. Path traced by Arun represents which type of curve. Find the length of major axis? (1)
ii. Find the equation of the curve traced by Arun? (1)
iii. Find the eccentricity of path traced by Arun? (2)
OR
iv. Find the length of latus rectum for the path traced by Arun. (2)

Answer

i. 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. Hence path traced by Arun is ellipse.
Sum of the distances of the point moving point to the foci is equal to length of major axis =10m
ii. Given 2a = 10 & 2c = 8
$\begin{array}{l}\Rightarrow a=5 \& c=4 \\ c^2=a^2+b^2 \\ \Rightarrow 16=25+b^2 \\ \Rightarrow b^2=25-16=9\end{array}$
Equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Required equation is $\frac{x^2}{25}+\frac{y^2}{9}=1$
iii. equation is of given curve is $\frac{x^2}{25}+\frac{y^2}{9}=1$
$a=5, b=3$ and given $2 c=8$ hence $c=4$
Eccentricity $=\frac{c}{a}=\frac{4}{5}$
OR
$\frac{x^2}{25}+\frac{y^2}{9}=1$
Hence $a=5$ and $b=3$
Length of latus rectum of ellipse is given by $\frac{2 b^2}{a}=\frac{2 \times 9}{5}=\frac{18}{5}$

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Question 24 Marks

A teacher conducted a surprise test of Mathematics, Physics and Chemistry for class XI on Monday. The mean and standard deviation of marks obtained by 50 students of the class in three subjects are given below:

Subject	Mathematics	Physics	Chemistry
Mean	42	32	40.9
Standard deviation	12	15	20

i. Which of the three subjects shows the highest variability? (1)
ii. What is the coefficient of variation of marks obtained by the students in Chemistry? (1)
iii. What is the coefficient of variation of marks obtained by the students in Physics? (2)
OR
What is the coefficient of variation of marks obtained by the students in Mathematics? (2)

Answer

i. The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry.
ii. Standard deviation of Chemistry $=20$
C.V. (in Chemistry) $=\frac{20}{40.9} \times 100=48.89$
iii. Standard deviation of Physics $=15$\
The coefficient of variation, C.V. $=\frac{\text { Stardard deviation }}{\text { Mean }} \times 100$
C.V. (in Physics) $=\frac{15}{32} \times 100=46.87$
OR
Standard deviation of Mathematics $=12$
The coefficient of variation, C.V. $=\frac{\text { Standard deviation }}{\text { Mean }} \times 100$
C.V. (in Mathematics) $=\frac{12}{42} \times 100=28.57$

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Question 34 Marks

A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collections, in such a way that the order of the objects does not matter.

How many different words can be formed by using all the letters of the word ALLAHABAD?
i. In how many of them, vowels occupy the even position?
ii. In how many of them, both L do not come together

Answer

In a word ALLAHABAD, we have

Letters	A	L	H	B	D	Total
Number	4	2	1	1	1	9

So, the total number of words $=\frac{9!}{4!2!}=\frac{9 \times 8 \times 7 \times 6 \times 5}{2 \times 1}=7560$
i. There are 4 vowels and all are alike i.e., 4 A's.
Also, there are 4 even places which are 2nd, 4th, 6th and 8th. So, these 4 even places can be occupied by 4 vowels in $\frac{4!}{4!}=1$ way. Now, we are left with 5 places in which 5 letters, of which two are alike (2 L's) and other distinct, can be arranged in $\frac{5!}{2!}$ ways.
Hence, the total number of words in which vowels occupy the even places $=\frac{5!}{2!} \times \frac{4!}{4!}=\frac{5!}{2!}=60$
ii. Considering both L together and treating them as one letter. We have,

Letters	A	LL	H	B	D	Total
Number	4	1	1	1	1	8

Then, 8 letters can be arranged in $\frac{81}{4!}$ ways.
So, the number of words in which both $L$ come together $=\frac{8!}{4}=8 \times 7 \times 6 \times 5=1680$
Hence, the number of words in which both L do not come together
$=$ Total number of words - Number of words in which both $L$ come together
=7560-1680=5880
Hence, the total number of words in which both L do not come together is 5880

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