5 Marks Questions

Question 15 Marks

Mr. Pandey lives in Lucknow, Uttar Pradesh. The reading of the electric meter of his house is found to be 5678 units. If the previous month's reading was 4803 units and the connected load is 4 kW, calculate his electricity bill for that month.
Tariff plan is given below.
Energy charges

Number of units	0-150	151-300	301-500	> 500
Price per unit (in ₹)	₹ 5.50	₹ 6	₹ 6.50	₹ 7

Fixed charges ₹ 110 per kW/month
Energy tax is 5% of tariff rates
Surcharge is ₹ 0.26 per unit

Answer

Given current month's reading = 5678 units
Previous month's reading = 4803 units
Number of units consumed = 5678 - 4803 = 875
Energy charges

Units	Price	Amount
0-150	₹ 5.50	₹ 825
151-300	₹ 6	₹ 900
301-500	₹ 6.50	₹ 1,300
501-875	₹ 7	₹ 2,625
	Total	5650

Fixed charges = ₹(110 × 4) = ₹ 440
Surcharge = ₹(0.26 × 875) = ₹ 227.50
Energy tax = 5% of (₹ 5650 + ₹ 440) = 5% of ₹ 6090 = ₹ 304.50
Electricity bill = ₹ 5650 + ₹ 440 + ₹ 227.50 + ₹ 304.50 = ₹ 6622

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Question 25 Marks

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds.
Further, another set of 15 observations $x _1, x _2, \ldots, x _{15}$, also in seconds, is now available and we have $\sum_{i=1}^{15} x_i=279$ and $\sum_{i=1}^{15} x_i^2=5524$. Calculate the standard derivation based on all 40 observations.

Answer

Given: Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Another set of 15 observations $x_1, x_2, \ldots, x_{15}$ also in seconds, is $\sum_{i=1}^{15} x _i=279$ and $\sum_{i=1}^{15} x _i^2=5524$
To find: the standard derivation based on all 40 observations
As per the given criteria,
In first set,
We have Number of observations, $n _1=25$
Mean, $\overline{ x _1}=18.2$
And standard deviation, $\sigma_1=3.25$
And
In second set,
We have Number of observations, $n_2=15$
$\sum_{i=1}^{15} x _i=279$ and $\sum_{i=1}^{15} x _1^2=5524$
For the first set we have
$\overline{ x _1}=18.2=\frac{\sum x _{ i }}{25}$
$\sum x_i=25 \times 18.2=455$
Therefore the standard deviation becomes,
$\sigma_1^2=\frac{\sum x_i^2}{25}-(18.2)^2$
Substituting the values, we get
$(3.25)^2=\frac{\sum x_i^2}{25}-331.24$
$\begin{array}{l}\Rightarrow 10.5625+331.24=\frac{\sum x_i^2}{25} \\ \Rightarrow \frac{\sum x_i^2}{25}=341.8025 \\ \Rightarrow \sum x_i^2=25 \times 341.8025=8545.06\end{array}$
For the combined standard deviation of the 40 observation, n=40
And
$\begin{array}{l}\Rightarrow \sum x_i^2=8545.06+5524=14069.69 \\ \Rightarrow \sum x_i=455+279=734\end{array}$
Therefore the standard deviation can be written as,
$\sigma=\sqrt{\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2}$
Substituting the values, we get
Therefore the standard deviation can be written as,
$\begin{array}{l}\sigma=\sqrt{\frac{14069.69}{40}-\left(\frac{734}{40}\right)^2} \\ \sigma=\sqrt{351.7265-(18.35)^2} \\ \sigma=\sqrt{351.7265-336.7225} \\ \sigma=\sqrt{15.004} \\ \sigma=3.87\end{array}$

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Question 35 Marks

Using short cut method, find the mean, variance and standard deviation for the data:

Class	25-35	35-45	45-55	55-65	65-75
Frequency	64	132	153	140	51

Answer

From the given information we prepare the following table...

Class	Frequency $\left( f _{ i }\right)$	$x _{ i }$	$d _{ i }= x _{ i }-50$	$f _{ i } d _{ i }$	$f_i d_i^2$
25-350	64	30	-20	-1280	25600
35-45	132	40	-10	-1320	13200
45-55	153	50	0	0	0
55-65	140	60	1400	14000	65-75
65-75	51	70	20	1020	20400
$\sum f_i=540$				$\sum f_i d_i=-180$	73200

Mean $=A+\frac{\sum f_i d_i}{\sum f_i} \times h$
$=50+\frac{(-180)}{540} \times 5=40$
Variance $\left(\sigma^2\right)$
Standard deviation $(\sigma)=\sqrt{\frac{\sum f_i d_i^2}{\sum f_i}-\left(\frac{\sum f_i d_i}{\sum f_i}\right)^2}=\sqrt{\frac{73200}{540}-\left(\frac{-180}{540}\right)^2}$
$=\sqrt{135.55-0.111}=\sqrt{135.44}=11.64$
Standard deviation $(\sigma)=\sqrt{\text { Variance }}=\sqrt{11.64}$

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Question 45 Marks

For what integers m and n does $\lim _{x \rightarrow 0} f(x)$ and $\lim _{ x \rightarrow 1} f( x )$ exist, if $f(x)=\left\{\begin{array}{lc}m x^2+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^2+m, & x>1\end{array}\right.$

Answer

Given,
$f(x)=\left\{\begin{array}{cc}m x^2+n, & x<0 \\ n x+m, & 0 \leqslant x \leqslant 1 \\ n x^2+m, & x>1\end{array}\right.$
Limit at x = 0,
$LHL =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} mx ^2+ n$
$=\lim _{h \rightarrow 0} m(0- h )^2+ n$
[putting $x =0- h$ as $x \rightarrow 0$, then $h \rightarrow 0$ ]
= n
RHL $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} nx + m$
$=\lim _{h \rightarrow 0} n(0+ h )+ m = m$
[putting $x =0+ h$ as $x \rightarrow 0$, then $h \rightarrow 0$ ]
As $\lim _{x \rightarrow 0} f(x)$ exists, then
$LHL = RHL \Rightarrow n = m$
$\because$ Limit at $x =1$
Now, LHL $=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}( nx + m )$
$=\lim _{h \rightarrow 0}[ n (1- h )+ m ]$
[putting $x =1- h$ as $x \rightarrow 0$, then $h \rightarrow 0$
= n + m
Here, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)$ exists for all $m , n \in Z$.

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Question 55 Marks

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?

Answer

Total letters of the word MISSISSIPPI = 11
Here M =1, I = 4, S = 4 and P = 2
$\begin{array}{l}\therefore \text { Number of permutations }=\frac{11!}{4!4!2!} \\ =\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!\times 4 \times 3 \times 2 \times 1 \times 2 \times 1}=34650\end{array}$
When the four I's come together then it becomes one letter so total number of letters in the word when all I's come together = 8
$\therefore$ Number of Permutations $=\frac{8!}{4!2!}=\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!\times 2 \times 1}=840$
Number of permutations when four I's do not come together = 34650 - 840 = 33810

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Question 65 Marks

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are. in the ratio $2: 1$, find the value of $n$.

Answer

We have $\frac{n!}{2!(n-2)!}: \frac{n!}{4!(n-4)!}=2: 1$
$\begin{array}{l}\Rightarrow \frac{n!}{2!(n-2)!} \div \frac{n!}{4!(n-4)!}=\frac{2}{1} \\ \Rightarrow \frac{4!(n-4)!}{2!(n-2)!}=2 \\ \Rightarrow \frac{4 \times 3 \times 2!}{2!} \times \frac{(n-4)!}{(n-2)(n-3)(n-4)!}=2 \\ \Rightarrow n ^2-5 n +6=6 \\ \Rightarrow n ^2-5 n =0, \text { we have } n =5 \text { or } n =0\end{array}$
Rejecting n = 0, we have n = 5.

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Applied Maths 5 Marks Questions

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