Case study (4 Marks)

Question 14 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, $\text{MONDAY,}$ 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 $\text{MONDAY} = 6$
Number of vowels in word $\text{MONDAY} = 2$
$(i)$ Number of letters used $= 4$
$\therefore$ Number of permutations $={ }^6 P_4=\frac{6!}{(6-4)!}$
$=\frac{6!}{2!}=\frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}=360$
$(ii)$ Number of letters used $= 6$
$\therefore$ Number of permutations $={ }^6 P_6$
$=\frac{6!}{0!}$
$=6 \times 5 \times 4 \times 3 \times 2 \times 1$
$=720$

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Question 24 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 $=\frac{\sigma}{\bar{x}} \times 100$
$=\frac{10}{5,253} \times 100$
$=0.19$
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 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 $20\ m$ 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, $50^2=-4 a(-10)$
$\Rightarrow 2500=40 a$
$\Rightarrow a =\frac{2500}{40}=62.5$
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
$\Rightarrow 400=-250 y$
$\Rightarrow y=\frac{-400}{250}=-1.6$
height of the bridge $= 10 - 1.6 = 8.4\ m$
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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