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Applied Maths 3 Marks Question

Model Paper 5 · CBSE STD 12 Science (CBSE - English Medium). 8 questions.

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Question 13 Marks
A random sample of 10 boys had the following I.Q's: $70,120,110,101,88,83,95,98,107,100$. Do these data support the assumption of a population mean I.Q. of 100 ? Find a reasonable range in which most of the mean I.Q. values of samples of 10 boys lie. (Given to $(0.05)=2.262)$
Answer
We have,
$\mu=$ Population mean $=100, n =$ Sample size $=10$
We define
Null Hypothesis $H _0$ : The data are consistent with the assumption of a mean I.Q. of 100 in the population
Alternate hypothesis $H _1$ : The mean I.Q. of population $\neq 100$
Let the sample statistic $t$ be given by
$
t=\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}, \text { where } S^2=\frac{1}{n-1} \sum_{i=1}^n\left(x_i-\bar{X}\right)^2
$
Let us now compute $\bar{X}$ and $S ^2$.
Computation of $\bar{X}$ and S
$x _{ i }$$d_1=x_i-90$$d _{ i }{ }^2$
70-20400
12030900
11020400
10111121
88-24
83-749
95525
98864
10717289
10010100
 $\sum d_i=72$$\sum d _{ i }^2=2352$
Here, $d_i=x_i-90$
$
\begin{array}{l}
\therefore \bar{X}=90+\frac{1}{10} \sum d_{i}=90+\frac{72}{10}=972\left[Using: \bar{X}=A+\frac{1}{n} \sum d_{i}\right] \\
S^2=\frac{1}{n-1}\left\{\sum d_i^2-\frac{1}{n}\left(\sum d_i\right)^2\right\}=\frac{1}{9}\left\{2352-\frac{(72)^2}{10}\right\}=\frac{1833.6}{9}=203.73 \\
\therefore t=\frac{\bar{X}-\mu}{S / \sqrt{n}} \Rightarrow t=\frac{97.2-100}{\sqrt{\frac{203.73}{10}}}=\frac{-2.8}{\sqrt{20.37}}=\frac{-2.8}{4514}=-0.62 \\
\Rightarrow|t|=0.62
\end{array}
$
The sample statistict follows student's $t$-distribution with $v=(10-1)=9$ degrees of freedom. It is given that $t_9(0.05)=2.262$
$\therefore$ Calculated $| t |<$ tabulated $t _9(0.05)$
So, the null hypothesis may be accepted at $5 \%$ level of significance.
Hence, the assumption of a population mean I.Q. of 100 is valid.
The $95 \%$ confidence limits within which the mean I.Q. values of samples of 10 boys will lie are $\bar{X}-\frac{S}{\sqrt{n}} t _9(0.05)$ and $\bar{X}+\frac{S}{\sqrt{n}} t _9(0.05)$
or $97.2-\sqrt{\frac{203.73}{10}} \times 2.262$ and $97.2+\sqrt{\frac{203.73}{10}} \times 2.262$
or, $97.2-4514 \times 2.262$ and $97.2+4.514 \times 2.262$
or, $97.2-10.21$ and $97.2+10.21$
or, 86.99 and 107.41
Hence, the required $95 \%$ confidence interval is $[86.99,107.41]$
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Question 23 Marks
i. Obtain the three year moving averages for the following series of observations:
Year19951996199719981999200020012002
Annual Sales (in 000 ₹)3.64.34.33.44.45.43.42.4
ii. Obtain the five year moving average.
iii. Construct also the 4-year centred moving average.
Answer
i. First 3-year moving average is $\frac{3.6+4.3+4.3}{3}=\frac{12.2}{3}=4.067$, and is placed against 2nd year i.e. 1996; second 3-year moving average is $\frac{4.3+4.3+3.4}{3}=\frac{12.0}{3}=4.0$, and is placed against 3rd year i.e. 1997 , and so on. Thus, we have:
Image
ii. First 5 -yearly moving average is $\frac{3.6+4.3+4.3+3.4+4.4}{5}=\frac{20.0}{5}=4.00$, and is placed against 3rd year i.e. 1997. Second 5 -yearly moving average is $\frac{4.3+4.3+3.4+4.4+5.4}{5}=\frac{21.8}{5}=4.36$, and is placed against 4 th year i.e. 1998 , and so on. Thus, we have:
Image
iii. In the 4-year moving averages, the first step of averaging of 4 values each results in placing these in between years - so we take averages of each two successive moving averages to synchronise them with given time frame. Thus, we have the following table:
Image
Note that values of 4th column are not synchronised with first column, but values of 5th column are synchronised.
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Question 33 Marks
Answer
We know that sum of the probabilities in a probability distribution is always unity.
$
\begin{array}{l}
\therefore 0.1+k+0.2+2 k+0.3+k=1 \\
\Rightarrow 0.6+4 k=1 \Rightarrow 4 k=0.4 \Rightarrow k=0.1
\end{array}
$
Calculation of mean and variance:
xipiPixi$p_i x_i^2$
-20.1-0.20.4
-10.1-0.10.1
00.200
10.20.20.2
20.30.61.2
30.10.30.9
  $\Sigma p_i x_i=0.8$$\Sigma p_i x_i^2=2.8$
Therefore, $\Sigma p_i x_i=0.8$ and $\Sigma p_i x_i^2=2.8$
$
\begin{array}{l}
\therefore \text { Mean }=\Sigma p_i x_i=0.8 \\
\text { and Variance }=\Sigma p_i x_i^2-\left(\Sigma p_i x_i\right)^2=2.8-(0.8)^2=2.8-0.64=2.16
\end{array}
$
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Question 43 Marks
Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.
Answer
Let X be a random variable denoting the number of tails in three tosses of a coin. Then, X can take the values $0,1,2$ and 3 Now, we have,
$
\begin{array}{l}
P(X=0)=P(HHH)=\frac{1}{8} \\
P(X=1)=P(THH \text { or HHT or HTH })=\frac{3}{8} \\
P(X=2)=P(TTH \text { or THT or HTT })=\frac{3}{8} \\
P(X=3)=P(TTT)=\frac{1}{8}
\end{array}
$
Thus, the probability distribution of X is as follows:
X0123
P(X)$\frac{1}{8}$$\frac{3}{8}$$\frac{3}{8}$$\frac{1}{8}$
Computation of mean and variance:
xipi$p _{ i } x _{ i }$$p_i x_i^2$
0$\frac{1}{8}$00
1$\frac{3}{8}$$\frac{3}{8}$$\frac{3}{8}$
2$\frac{6}{8}$$\frac{6}{8}$$\frac{12}{8}$
3$\frac{3}{8}$$\frac{3}{8}$$\frac{9}{8}$
  $\sum p_i x_i=\frac{3}{2}$$\sum p_i x_i^2=3$
Therefore, mean $=\sum p_i x_i=\frac{3}{2}$
$
\begin{array}{l}
\text { Variance }=\sum p_i x_i^2-(\text { Mean })^2 \\
=3-\left(\frac{3}{2}\right)^2 \\
=3-\frac{9}{4} \\
=\frac{3}{4}
\end{array}
$
$
\text { Standard deviation }=\sqrt{\text { Variance }}
$
$
\begin{array}{l}
=\sqrt{\frac{3}{4}} \\
=0.87
\end{array}
$
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Question 53 Marks
The demand and supply functions for a commodity are $p = x ^2-6 x +16$ and $p =\frac{1}{3} x^2+\frac{4}{3} x+4$ respectively. Find each of the following assuming $x \leq 5$ :
i. The equilibrium point.
ii. The consumer's surplus at the equilibrium point.
iii. The producer's surplus at the equilibrium point.
Answer
The demand and supply functions are $p=D(x)$ and $p=S(x)$, where $D(x)=x^2=6 x+16$ and $S(x)=\frac{1}{3} x^2+\frac{4}{3} x+4$
i. The equilibrium point $\left(x_0, P_0\right)$ is the point at which the demand-supply curves intersect. Therefore, the equilibrium point is obtained by setting $D(x)=S(x)$.
Now, $D ( x )= S ( x )$
$
\begin{array}{l}
\Rightarrow x^2-6 x+16=\frac{1}{3} x^2+\frac{4}{3} x+4 \\
\Rightarrow \frac{2}{3} x^2-\frac{22}{3} x+12=0 \Rightarrow x^2-11 x+18=0 \Rightarrow(x-2)(x-9) \Rightarrow x=2[\because x \leq 5]
\end{array}
$
Putting $x=2$ either in $p=D(x)$ or in $p=S(x)$, we obtain $p=8$. Thus, $x_0=2$ and $p_0=8$. Hence, $(2,8)$ is the equilibrium point.
ii. The consumer's surplus (CS) at the equilibrium point $(2,8)$ is given by
$
\begin{array}{l}
CS=\int_0^{x_0} D(x) d x-p_0 x_0 \\
\Rightarrow CS=\int_0^2\left(x^2-6 x+16\right) d x-8 \times 2 \\
\Rightarrow CS=\left[\frac{x^3}{3}-3 x^2+16 x\right]_0^2-16=\left(\frac{8}{3}-12+32\right)-16=\frac{20}{3}
\end{array}
$
iii. The producer's surplus (PS) at the equilibrium point $(2,8)$ is given by
$
\begin{array}{l}
PS=p_0 x_0-\int_0^{x_0} S(x) d x \\
\Rightarrow PS=8 \times 2-\int_0^2\left(\frac{1}{3} x^2+\frac{4}{3} x+4\right) d x \\
\Rightarrow PS=16-\left[\frac{x^3}{9}+\frac{2}{3} x^2+4 x\right]_0^2=16-\left(\frac{8}{9}+\frac{8}{3}+8\right)=\frac{40}{9}
\end{array}
$
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Question 63 Marks
The cost of a washing machine depreciates by ₹720 during the second year and by ₹648 during the third year.
Calculate:
i. the rate of depreciation per annum.
ii. the original cost of the machine.
iii. the value of the machine at the end of third year.
Answer
i. Let the original cost of the washing machine be ₹P and the rate of depreciation be r \% p.a. Then the value of machine (in ₹) after one year, two years and 3 years are $P (1- i ), P (1- i )^2$ and $P (1- i )^3$ respectively, where $i =\frac{r}{100}$.
According to given,
$
\begin{array}{l}
P(1-i)-P(1-i)^2=720 \text { and } P(1-i)^2-P(1-i)^3=648 \\
\Rightarrow P(1-i)[1-(1-i)]=720 \text { and } P(1-i)^2[1-(1-i)]=648 . \\
\Rightarrow P(1-i) i=720 \ldots(i) \text { and } P(1-i)^2 \cdot i=648
\end{array}
$
Dividing (ii) by (i), we get
$
\begin{array}{l}
1-i=\frac{648}{720} \Rightarrow 1-i=\frac{9}{10} \\
\Rightarrow i=1-\frac{9}{10} \Rightarrow i=\frac{1}{10} \Rightarrow \frac{r}{100}=\frac{1}{10} \\
\Rightarrow r=10
\end{array}
$
Hence, the rate of depreciation $=10 \%$ p.a.
ii. Putting $i =\frac{1}{10}$ in equation (i), we get
$
P\left(1-\frac{1}{10}\right) \times \frac{1}{10}=720 \Rightarrow P \times \frac{9}{100}=720 \Rightarrow P=8000
$
Hence, the original cost of the machine $=₹ 8000$
iii. The value of machine at the end of third year $=P(1-i)^3$
$
\begin{array}{l}
=8000\left(1-\frac{1}{10}\right)^3=8000(0.9)^3 \\
=8000 \times 0.729=5832
\end{array}
$
Hence, the value of the machine at the end of the third year $=₹ 5832$
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Question 73 Marks
Obtain the differential equation of all circles of radius r
Answer
The equation of the family of circles of radius $r$ is
$
(x-a)^2+(y-b)^2=2 \ldots \text { (i) }
$
where a and bare a parameters.
Clearly equation (i) contains two arbitrary constants. So, let us differentiate it two times with respect to $x$.
Differentiating (i) with respect to $x$, we get
$
\begin{array}{l}
2(x-a)+2(y-b) \frac{d y}{d x}=0 \\
\Rightarrow(x-a)+(y-b) \frac{d y}{d x}=0\ldots \text { (ii) }
\end{array}
$
Differentiating (ii) with respect to $x$, we get
$
\begin{array}{l}
1+(y-b) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 \ldots \text { (iii) }\\
\Rightarrow y-b=-\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}} \ldots \text { (iv) }
\end{array}
$
Putting this value of ( $y - b$ ) in (ii), we obtain
$
x-a=\frac{\left.1+\left(\frac{d y}{d z}\right)^2\right\} \frac{d y}{d z}}{\frac{d^2 y}{d^2}} \ldots(v)
$
Substituting the values of $(x-a)$ and $(y-b)$ in (i), we get
$
\frac{\left\{1+\left.\left(\frac{d y}{d x}\right)^2\right|^2\left(\frac{d y}{d x}\right)^2\right.}{\left(\frac{d^2 y}{d x^2}\right)^2}+\frac{\left.\left\lvert\, 1+\left(\frac{d y}{d x}\right)^2\right.\right\}^2}{\left(\frac{d^2 y}{d x^2}\right)^2}=r^2 \Rightarrow\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^3=r^2\left(\frac{d^2 y}{d x^2}\right)^2
$
This is the required differential equation.
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Question 83 Marks
Solve the initial value problem: $x \frac{d y}{d x}+ y = x \log x , y (1)=\frac{1}{4}$
Answer
We have, $x \frac{d y}{d x}+ y = x \log x$
$
\Rightarrow \frac{d y}{d x}+\frac{1}{x} y=\log x
$
This is linear differential equation of the form $\frac{d y}{d x}+ Py = Q$ with $P =\frac{1}{x}$ and $Q =\log x$
$
\therefore \text { I.F. }=e^{\int \frac{1}{x} d x}=e^{\log x}=x[\because x>0]
$
Multiplying both sides of (i) by I.F. $=x$, we get
$
x \frac{d y}{d x}+y=x \log x
$
Integrating with respect to $x$, we ge
$
\begin{array}{l}
yx=\int \underset{I I}{x} \log x d x\left[\text { Using: } y(\text { L.F. })=\int Q(\text { I.F. }) dx+C\right] \\
\Rightarrow yx=\frac{x^2}{2}(\log x) \frac{1}{2} \int xdx \\
\Rightarrow yx=\frac{x^2}{2}(\log x)-\frac{x^2}{4}+C \ldots \text { (ii) }
\end{array}
$
It is given that $y(1)=\frac{1}{4}$ i.e. $y=\frac{1}{4}$ where $x=1$. Putting $x=1$ and $y=\frac{1}{4}$ in (ii), we get
$
\frac{1}{4}=0-\frac{1}{4}+C \Rightarrow C=\frac{1}{2}
$
Putting $C =\frac{1}{2}$ in (ii), we get
$
x y=\frac{x^2}{2}(\log x)-\frac{x^2}{2}+\frac{1}{2} \Rightarrow y=\frac{1}{2} x \log x=\frac{x}{4}+\frac{1}{2 x}
$
Hence, $y =\frac{1}{2} x \log x -\frac{\pi}{4}+\frac{1}{2 x}$ is the solution of the given differential equation.
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