Case study (4 Marks)

Question 14 Marks

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Particulars	Firm A	Firm B
No. of wage earners	586	648
Mean of monthly wages	₹ 5253	₹ 5253
Variance of the distribution of wages	100	121

i. Which firm A or B shows greater variability in individual wages? (1)
ii. Find the standard deviation of the distribution of wages for frim B. (1)
iii. Find the coefficient of variation of the distribution of wages for firm A. (2)
OR
Find the amount paid by firm A. (2)

Answer

i. coefficient of variation of wages, of firm $A=0.19$ coefficient of variation of wages, of firm $B=\frac{121}{5253} \times 100=0.21$
$\therefore$ Firm B shows greater variability in individual wages.
ii. Standard deviation, $\sigma=\sqrt{\sigma^2}=\sqrt{121}=11$
iii. Variance of distribution of wages, $\sigma^2=100$
Standard deviation, $\sigma=\sqrt{\sigma^2}=\sqrt{100}=10$
coefficient of Variation $=\frac{\sigma}{\bar{x}} \times 100$
$\begin{array}{l}=\frac{10}{5,253} \times 100 \\ =0.19\end{array}$
OR
No. of wage earners $=586$
Mean of monthly wages, $\bar{x}$=₹ 5253
Amount paid by firm A = ₹ $(586 \times 5253$)=₹ 3078258

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Question 24 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 words, with or without meaning can be made from the letters of the word, MONDAY, assuming that no letter is repeated if
(i) 4 letters are used at a time
(ii) all letters are used at a time

Answer

Total number of letters in word MONDAY = 6
Number of vowels in word MONDAY = 2
(i) Number of letters used = 4
$\begin{array}{l}\therefore \text { Number of permutations }={ }^6 P_4=\frac{6!}{(6-4)!} \\ =\frac{6!}{2!}=\frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}=360\end{array}$
(ii) Number of letters used = 6
$\begin{array}{l}\therefore \text { Number of permutations }={ }^6 P_6 \\
=\frac{6!}{0!}=6 \times 5 \times 4 \times 3 \times 2 \times 1=720\end{array}$

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Question 34 Marks

The girder of a railway bridge is a parabola with its vertex at the highest point, 10 m above the ends. Its span is 100 m.

i. Find the coordinates of the focus of the parabola. (1)
ii. Find the equation of girder of bridge and find the length of latus rectum of girder of bridge. (1)
iii. Find the height of the bridge at 20m from the mid-point. (2)
OR
Find the radius of circle with centre at focus of the parabola and passes through the vertex of parabola. (2)

Answer

i. From the diagram equation of parabola is $x^2=-4 a y$
Vertex is 10 m hight and spam is 100 m
parabola passes through ( $50,-10)$
Hence, $50^2=-4 a(-10)$
$\begin{array}{l}\Rightarrow 2500=40 a \\ \Rightarrow a =\frac{2500}{40}=62.5\end{array}$
Hence coordinates of focus = (-a, 0) = (-62.5, 0)
ii. Equation of parabola is $x^2=-4 a y$ and $a=\frac{2500}{40}=62.5$
Equation is $x^2=-4\left(\frac{2500}{40}\right) y$
$\Rightarrow x^2=-250 y$
Length of latus rectum is $4 a =4 \times 62.5=250 m$
iii. Equation parabola $x^2=-250 y$
Coordinates of the point at 20 m from mid point $=(20, y)$
Substituting in the equation of parabola
$\begin{array}{l}\Rightarrow 400=-250 y \\ \Rightarrow y=\frac{-400}{250}=-1.6\end{array}$
height of the bridge = 10 - 1.6 = 8.4m
OR
vertex of parabola is (0, 0) and focus is (0, -62.5)
$\Rightarrow(0,-62.5)$ is center and $(0,0)$ is on the circle
$\Rightarrow r =0-(-62.5)=62.5 m$

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MATHS Case study (4 Marks)

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