5 Marks Questions

Question 15 Marks

Mr. Rishabh lives in Bengaluru, Karnataka. He consumed 37 kL of water in one month. Calculate his water bill for that month. Water tariff plan is given below.

Units of consumption (in kL)upto	upto 8	8 to 25	25-50	>50
Price per unit (in ₹)	7	11	25	45

Maintenance charge = ₹ 56 per month
Sewerage charges flat ₹ 14 for those whose consumption is upto 8 kL and $25 \%$ of water charges whose consumption is more than 8 kL .

Answer

Here, the consumption of water is given to be 37 kL ,
According to the given tariff plan
Water consumption charge $=₹(8 \times 7+17 \times 11+12 \times 25)$
$=₹(56+187+300)$
= ₹ (543)
Sewerage charge for consumption above 8 KL is $25 \%$ of the consumption charges.
$\therefore$ Sewerage charge $=25 \%$ of 543
= 135.75
Maintenance charges $=₹ 56$ per month
Total water bill = consumption charge + Sewerage charge + Maintenance charge
= ₹ 543 + ₹ $135.75+₹ 56$
= ₹ 734.75

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Question 25 Marks

Find the mean deviation about the median for the following data:

x_i	3	6	9	12	13	15	21	22
$f_i$	3	4	5	2	4	5	4	3

Answer

Table for mean deviation about median

$x_i$	$I _{ i }$	cf	$\left\| d _{ i }\right\|=\left\| x _{ i }-13\right\|$	$f _{ i }\left\| d _{ i }\right\|$
3	3	3	10	30
6	4	7	7	28
9	5	12	4	20
12	2	14	1	2
13	4	18	0	0
15	5	23	2	10
21	4	27	8	32
22	3	30	9	27
	$\sum_{i=1}^n f_i=30$			$\sum_{i=1}^n J_i\left\|d_i\right\|=149$

$\mathrm{N}=30, \mathrm{~N}$ is even, $\frac{N}{2}=15$ or $\frac{2(30+1)}{4}=15.5$ lies between 15 and 16.
Median $=\left(\frac{\text { Value of } 15 \text { th term }+ \text { Value of } 16 \text { th term }}{2}\right)=\frac{13+13}{2}=13$
Mean deviation $=\frac{\sum_{i=1}^{8} f_{i}\left|d_{i}\right|}{\sum_{i=1}^{8} f_{i}}=\frac{149}{30}=4.97$

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Question 35 Marks

Find $\operatorname{Cov}(X, Y)$ between $X$ and $Y$, if:

X:	1	2	3	4	5
Y:	2	4	6	8	10

Answer

$x_1$	$y_j$	$x _1-\bar{X}$	$y_i-\bar{Y}$	$\left(x_i-\bar{X}\right)\left(y_i-\bar{Y}\right)$
1	2	-2	-4	8
2	4	-1	-2	2
3	6	0	0	0
4	8	1	2	2
5	10	2	4	8
$\Sigma x_i=15$	$2 y_i=30$		$\Sigma\left(x_i-\bar{X}\right)\left(y_i-\bar{Y}\right)=20$

We have, $\mathrm{n}=5, \Sigma \mathrm{x}_{\mathrm{i}}=15$ and $\Sigma \mathrm{y}_{\mathrm{i}}=30$
$\therefore \bar{X}=\frac{1}{n} \sum \mathrm{x}_{\mathrm{i}}=\frac{15}{5}=3, \bar{Y}=\frac{1}{n} \sum \mathrm{y}_{\mathrm{i}}=\frac{30}{5}=6, \Sigma\left(\mathrm{x}_{\mathrm{i}}-\bar{X}\right)\left(\mathrm{y}_{\mathrm{i}}-\bar{Y}\right)=20$ and, $\mathrm{n}=5$
$\therefore \operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n} \sum\left(\mathrm{x}_{\mathrm{i}}-\bar{X}\right)\left(\mathrm{y}_{\mathrm{i}}-\bar{Y}\right)=\frac{20}{5}=4$
It is clear from the above illustration that if $\bar{X}$ and $\bar{Y}$ are not integers, then the calculations for the covariance by using formula (i) will be cumbersome and time consuming. We therefore develop an alternate formula as discussed below.
We have,
$\operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n} \sum_{i=1}^{n}\left(\mathrm{x}_{\mathrm{i}}-\bar{X}\right)\left(\mathrm{y}_{\mathrm{i}}-\bar{Y}\right)$
$\Rightarrow \operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n} \sum_{i=1}^{n}\left\{\mathrm{x}_{\mathrm{i}} \mathrm{y}_{\mathrm{i}}-\mathrm{x}_{\mathrm{i}} \bar{Y}-\mathrm{y}_{\mathrm{i}} \bar{X}+\bar{X} \bar{Y}\right)$
$\Rightarrow \operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n}\left\{\sum_{i=1}^{n} x_{i} y_{i}-\bar{Y} \sum_{i=1}^{n} x_{i}-\bar{X} \sum_{i=1}^{n} y_{i}+n \bar{X} \bar{Y}\right\}$
$\Rightarrow \operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n}\left\{\sum_{i=1}^{n} x_{i} y_{i}-\bar{Y}(n \bar{X})-\bar{X}(n \bar{Y})+n \bar{X} \bar{Y}\right\}\left[\because \sum_{i=1}^{n} x_{i}=n \bar{X}, \sum_{i=1}^{n} \mathrm{y}=\mathrm{n} \bar{Y}\right]$
$\Rightarrow \operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n}\left\{\sum_{i=1}^{n} x_{i} y_{i}-n \bar{X} \bar{Y}\right\}$
$=\frac{1}{n} \sum_{i=1}^n x_i y_i-\bar{X} \bar{Y}=\frac{1}{n} \sum_{i=1}^n x_i y_i-\left\{\frac{1}{n} \sum_{i=1}^n x_i\right\}\left\{\frac{1}{n} \sum_{i=1}^n y_i\right\}$
Hence, $\operatorname{Cov}(\mathrm{X}, \mathrm{Y})=\frac{1}{n} \sum_{i=1}^{n} x_{i} y_{i}-\left\{\frac{1}{n} \sum_{i=1}^{n} x_{i}\right\}\left\{\frac{1}{n} \sum_{i=1}^{n} y_{i}\right\}\quad \ldots (ii)$
i.e., $\operatorname{Cov}(\mathrm{X}, \mathrm{Y})=($ Mean of the product of values of $(\mathrm{X}$ and Y$)$ - (Product of means of X and Y$)$.

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Question 45 Marks

Discuss the continuity of the function $f ( x )$ at $x =1$, defined by $f(x)-\left\{\begin{array}{ll}\frac{3}{2}-x, & \text { if } \frac{1}{2} \leq x < 1 \\ \frac{3}{2}, & \text { if } x-1 \\ \frac{3}{2}+x, & \text { if } 1 < x \leq 2\end{array}\right.$

Answer

$\begin{array}{c}
L H L=\lim _{x \rightarrow 1}\left(\frac{3}{2}-x\right)=\frac{3}{2}-1=\frac{1}{2} \\
R H L=\lim _{x \rightarrow 1^{+}}\left(\frac{3}{2}+x\right)=\frac{3}{2}+1=\frac{5}{2}
\end{array}
$
$ \underset{x=1}{L H L} \neq R \underset{x=1}{R H}$. Hence, the function is discontinuous at $x =1$.
$
\begin{array}{l}
x=\lim _{h \rightarrow 0} \frac{\sqrt{1+k(-h)}-\sqrt{1-k(-h)}}{-h} \\
=\lim _{h \rightarrow 0} \frac{(\sqrt{1-k h}-\sqrt{1+k h})(\sqrt{1-k h}+\sqrt{1+k h})}{-h[\sqrt{1-k h}+\sqrt{1+k h}]} \\
=\lim _{h \rightarrow 0} \frac{1-k h-1-k h}{-h[\sqrt{1-k h}+\sqrt{1+k h}]} \\
=\lim _{h \rightarrow 0} \frac{-2 k h}{-h[\sqrt{1-k h}+\sqrt{1+k h}]} \\
=\lim _{h \rightarrow 0} \frac{2 k}{\sqrt{1-k h}+\sqrt{1+k h}}=\frac{2 k}{1+1}=k \quad \ldots(\text { iii }) \\
R H L=\lim _{x \rightarrow 0} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\
=\lim _{h \rightarrow 0} \frac{2(h)+1}{h-1}=\frac{0+1}{0-1}=-1 \quad \ldots \text { (iv) }
\end{array}
$
From (i), (ii), (iii) and (iv), we get
$
k=-1 \text {, for function to be continuous at } x=0 \text {. }
$

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Question 55 Marks

In how many ways can 9 examination papers be arranged so that the best and the worst papers are never together?

Answer

The number of arrangements in which the best and the worst papers never come together can be obtained by subtracting from the total number of arrangements, the number of arrangements in which the best and worst come together.
The total number of arrangements of 9 papers $={ }^{9} \mathrm{P}_{9}=9$ !
Considering the best and the worst papers as one paper, we have 8 papers which can be arranged in ${ }^{8} \mathrm{P}_{8}=8$ ! ways. But, the best and worst papers can be put together in 2 ! ways. So, the number of permutations in which the best and the worst papers can be put together $=(2!\times 8!)$.
Hence, the number of ways in which the best and the worst papers never come together $=9!-2!\times 8!=9 \times 8!-2 \times 8!=7 \times 8!$ $=282240$

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Question 65 Marks

How many 3-digit numbers are there which have exactly one of their digits as 6 ?

Answer

i. When six is at unit's place then possible numbers $=8 \times 9 \times 1=72$.
ii. When six is at ten's place then possible numbers $=8 \times 1 \times 9=72$.
iii. When six is at hundred's place then possible numbers $=1 \times 9 \times 9=81$.
Required number of ways $=72+72+81=225$.
(in case I and II hundred place cannot have zero and six).

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Applied Maths 5 Marks Questions

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