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Applied Maths 3 Marks Question

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

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3 Marks Question

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8 questions · timed · auto-graded

Question 13 Marks
The I.Q.'s (intelligence quotients) of 16 students from one area of a city showed a mean of 107 with a standard deviation of 10 while the I.Q.'s of 14 students from another area of the city showed a mean of 112 with a standard deviation of 8. Is there a significant difference between the I.Q.’s of the two groups at
i. $1\%$
ii. $5\%$ level of significance?
Answer
Performing independent samples t-test,not assuming equal variances.
Assumptions: both populations must be normal.
The null hypothesis: the mean IQs are equal.
The alternative hypothesis: the mean IQs are different.
Degrees of freedom: df= min(N1, N2) - 1 = 13
The standard error:
$SE =\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{\frac{10^2}{16}+\frac{8^2}{14}}=3.2896$
The test statistics:
$t =\frac{\left(\bar{X}_1-\bar{X}_2\right)-0}{S E}=\frac{107-112}{3.2896}=-1.52$
The two-tailed cumulative probability value associated with the given t-statistic can be determined from the Student’s t-distribution table or calculated using the technology (function T.DIST.2T() of MS Excel).
For df = 13 and t = -1.52, p = 0.152
Since the p-value is greater than both $\alpha$ values, fail to reject the null hypothesis at both significance levels.
The samples do not provide sufficient evidence to conclude the difference between the mean IQs.
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Question 23 Marks
The average number, in lakhs, of working days lost in strikes during each year of the period 2001-2010 was
2001200220032004200520062007200820092010
1.51.81.92.22.63.72.26.43.65.4
Calculate the three-yearly moving averages and draw the moving averages graph.
Answer
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Question 33 Marks
If X is a normal variate with mean 30 and S.D. 5. Find:
i. $P (26 \leq X \leq 40)$
ii. $P(X \geq 45)$
iii. $P (| X -30|>5)$
Answer
We have, $\mu=30$ and $\sigma=5$
Let Z be the standard normal variate. Then,
$Z =\frac{X-\mu}{\sigma} \Rightarrow Z=\frac{X-30}{5}$
i. When $X=26$, we obtain: $Z=\frac{26-30}{5}=-0.8$
When $X =40$, we obtain : $Z =\frac{40-30}{5}=2$
$\begin{array}{l}\therefore P (26 \leq X \leq 40) \\ = P (-0.8 \leq Z \leq 2) \\ = P (-0.8 \leq Z \leq 0)+ P (0 \leq Z \leq 2) \\ = P (0 \leq Z \leq 0.8)+ P (0 \leq 2 \leq 2) \\ =0.2881+0.4772=0.7563 \text { [See table] }\end{array}$
Image
ii. When X = 45, we obtain
$\begin{array}{l} Z =\frac{45-30}{5}=3 \\ \therefore P ( X \geq 45) \\ = P ( Z \geq 3) \\ =0.5- P (0 \leq Z \leq 3) \\ =0.5-0.49865 \text { [See Table] } \\ =0.00135\end{array}$
Image
iii. $P(|X - 30| > 5)$
$\begin{array}{l}=1- P (| X -30| \leq 5) \\ =1- P (30-5 \leq X \leq 30+5) \\ =1- P (25 \leq X \leq 35)\end{array}$
Now, $X=25 \Rightarrow Z=\frac{25-30}{5}=-1$ and, $X=35 \Rightarrow Z=\frac{35-30}{5}=1$.
$\begin{array}{l}\therefore P (| X -30|>5) \\ =1- P (-1 \leq Z \leq 1) \\ =1-2 . P (0 \leq Z \leq 1) \\ =1-2 \times 0.3413 \\ =1-0.6826 \\ =0.3174\end{array}$
Image
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Question 43 Marks
In a normal distribution $31\%$ of the articles are under 45 and $8\%$ are over $64.$ Calculate the mean and standard deviation of the distribution.
Answer
Let X be a normal distribution random variable, then
$P(X < 45) = 31\%$ i.e. 0.31 and $P(X > 64) = 8\%$ i.e. 0.08
Let the mean and the standard deviation of the distribution be $\mu$ and $\sigma$ respectively, then
$\begin{array}{l}\text { for } X =45, Z =\frac{45-\mu}{\sigma} \text { and for } X =64, Z =\frac{64-\mu}{\sigma} \\ \therefore P ( X <45)= P \left( Z <\frac{45-\mu}{\sigma}\right)=0.31\end{array}$
$\begin{array}{l}\Rightarrow P \left( Z <\frac{45-\mu}{\sigma}\right)= P ( Z <-0.5) \text { (using table) } \\ \Rightarrow \frac{45-\mu}{\sigma}=-0.5 \ldots \text { (i) }\end{array}$
$\begin{array}{l}\text { and } P ( X >64)= P \left( Z >\frac{64-\mu}{\sigma}\right)=0.08=1-0.92 \\ \Rightarrow P \left( Z >\frac{64-\mu}{\sigma}\right)=1- P ( Z \leq 1.4)= P ( Z >1.4) \text { (using table) } \\ \Rightarrow \frac{64-\mu}{\sigma}=1.4 \ldots\text{(ii)}\end{array}$
Dividing equation (i) by (ii), we get
$\begin{array}{l}\frac{45-\mu}{64-\mu}=\frac{-0.5}{1.4} \Rightarrow 63-1.4 \mu=-32+0.5 \mu \\ \Rightarrow 95=1.9 \mu \Rightarrow \mu=50.\end{array}$
Substituting $\mu=50$ in equation (i), we get
$\frac{45-50}{\sigma}=-0.5 \Rightarrow \sigma=10$
Hence, mean 50 and standard deviation = $\sigma=10$
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Question 53 Marks
Suppose that the demand function for a certain commodity is $p=20-4 x^2$ and the marginal cost is $\text{MC} = 2x + 6,$ where x is the number of units produced. Find the consumer’s surplus at the sales level $x _0$ where profit is maximized.
Answer
Let P be the profit function and R be the revenue function. Then,
$R = px =20 x -4 x ^3$ and $MR =\frac{d R}{d x}=20-12 x ^2$
Now, P = R - C
$\begin{array}{l}\Rightarrow \frac{d P}{d x}=\frac{d R}{d x}-\frac{d C}{d x} \\ \Rightarrow \frac{d P}{d x}= MR - MC \\ \Rightarrow \frac{d P}{d x}=\left(20-12 x ^2\right)-(2 x +6) \\ \Rightarrow \frac{d P}{d x}=14-2 x -12 x ^2 \text { and } \frac{d^2 P}{d x^2}=-2-24 x \end{array}$
For maximum value of P, we must have
$\frac{d P}{d x}=0$
$\begin{array}{l}\Rightarrow 14-2 x -12 x ^2=0 \\ \Rightarrow 6 x ^2+ x -7=0 \\ \Rightarrow 6 x ^2+7 x -6 x -7=0 \Rightarrow(6 x -7)( x -1)=0 \Rightarrow x -1=0 \Rightarrow x =1\end{array}$
Clearly, $\left(\frac{d^2 p}{d x^2}\right)_{x=1}=-2-24=-26<0$
Hence, the profit is maximum when x = 1. Therefore, $x_0=1$. Putting $x_0=1$ in $p=20-4 x^2$ , we obtain $p_0=16$.
The consumer's surplus at $x_0=1$ is given by
$\begin{array}{l} CS =\int_0^1 p d x-p_0 x_0 \\ \Rightarrow CS =\int_0^1\left(20-4 x^2\right) d x-16 \times 1 \\ \Rightarrow CS =\left[20 x-\frac{4}{3} x^3\right]_0^1-16=20-\frac{4}{3}-16=\frac{8}{3}\end{array}$
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Question 63 Marks
Mr. M borrowed ₹10,00,000 from a bank to purchase a house and decided to repay by monthly equal instalments in 10 years. The bank charges interest at 9% compounded annually. The bank calculated his EMI as ₹ 12,668. Find the principal and interest paid in first year. [Given $a_{108] 0.0075}=73.83916$ ]
Answer
Principal left unpaid after one year (12 payments)
= present value of remaining 108 payments = $R a_{n\rceil i}$
Where $R =12,668, n =108$ and $i =\frac{0.09}{12}=0.0075$
$\begin{array}{l}=12,668 \times a_{108\rceil0.0075} \\ =12,668 \times 73.83916=\text { ₹ } 9,35,395\end{array}$
Principal paid during first year $= 10,00,000 - 9,35,395 =$ ₹ $64,605$
Interest paid during first year
$=(12,668\times12)-64,605$
$=$ ₹ $87,651$
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Question 73 Marks
Find the differential equation of all the circles in the first quadrant which touch the coordinate axes.
Answer
The equation of the family of circles in the first quadrant which touch the coordinate axes is
$(x-a)^2+(y-a)^2=a^2 \ldots\text{(i)}$
where a is a parameter.
This equation contains one arbitrary constant, so we shall differentiate it once only to get a differential equation of first order.
Differentiating (i) with respect to x, we get
$\begin{array}{l}2( x - a )+2( y - a ) \frac{d y}{d x}=0 \\ \Rightarrow x - a +( y - a ) \frac{d y}{d x}=0 \\ \Rightarrow a =\frac{x+y \frac{d y}{d x}}{1+\frac{d y}{d x}}\end{array}$
$\Rightarrow a =\frac{x+p y}{1+p}$, where $p =\frac{d y}{d x}$
Substituting the value of a in (i), we get
Image
$\left(x-\frac{x+p y}{1+p}\right)^2+\left(y-\frac{x+p y}{1+p}\right)^2=\left(\frac{x+p y}{1+p}\right)^2$
$\begin{array}{l}\Rightarrow( xp - py )^2+( y - x )^2=( x + py )^2 \\ \Rightarrow( x - y )^2 p ^2+( x - y )^2=( x + py )^2 \\ \Rightarrow( x - y )^2\left( p ^2+1\right)=( x + py )^2 \\ \Rightarrow( x - y )^2\left\{1+\left(\frac{d y}{d x}\right)^2\right\}=\left(x+y \frac{d y}{d x}\right)^2, \text { which is the required differential equation. }\end{array}$
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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, $x \frac{d y}{d x}+ y = x \log x$
$\Rightarrow \frac{d y}{d x}+\frac{1}{x} y=\log x \ldots\text{(i)}$
This is linear differential equation of the form $\frac{d y}{d x}+ Py = Q$ with $P =\frac{1}{x}$ and $Q =\log x$
$\therefore$ $I.F.=e^{\int \frac{1}{x} d x}= e ^{\log x }= x [\because x >0]$
Multiplying both sides of (i) by I.F. = x, we get
$x\frac{dy}{dx}+y=x\text{ log x}$
Integrating with respect to $x,$ we get,
$yx=$ $\int \underset{I I}{x}$ $\log x ~d x$ [Using: y (I.F.) $=\int Q$ (I.F.) $\left.dx + C \right]$
$\Rightarrow yx =\frac{x^2}{2}(\log x ) \frac{1}{2} \int xdx$
$\Rightarrow yx =\frac{x^2}{2}(\log x )-\frac{x^2}{4}+ C \ldots\text{(ii)}$
It is given that $\text{y}(1)=\frac{1}{4}$ i.e. $\text{y}=\frac{1}{4}$ where $\text{x}=1.$ Putting $\text{x}=1$ and $\text{y}=\frac{1}{4}$ in $\text{(ii)},$ we get
$\frac{1}{4}=0-\frac{1}{4}+ C \Rightarrow C =\frac{1}{2}$
Putting $\text{C}=\frac{1}{2}$ in $\text{(ii)},$ 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, $\text{y}=\frac{1}{2}$ $x$ log $x-\frac{x}{4}+\frac{1}{2x}$ is the solution of the given differential equation.
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3 Marks Question - Applied Maths STD 12 Science Questions - Vidyadip