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Applied Maths 3 Marks Question

Model Paper 3 · CBSE STD 12 Science (CBSE - English Medium). 8 questions.

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3 Marks Question

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Question 13 Marks
Ten cartons are taken at random from an automatic filling machine. The mean net weight of the cartons is 11.8 kg and the standard deviation 0.15 kg. Does the sample mean differ significantly from the intended weight of 12 kg? [Given that for d.f. $=9, \mathrm{t}_{0.05}=2.26$]
Answer
$\mu=$ Population mean $=12 \mathrm{Kg}$
$\bar{X}=$ Sample mean $=11.8 \mathrm{Kg}$
$\mathrm{n}=10$
Sample standard deviation $=\mathrm{s}=0.15$
Null Hypothesis $\mathrm{H}_{0}=$ There is no significance between the sample mean
$\bar{X}$ and the population mean $\mu$.
Alternate Hypothesis $\mathrm{H}_{1}=$ There is significance between the sample mean $\bar{X}$ and the population mean $\mu$
Let t be the test statistic given by
$t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n-1}}}$
$t=\left(\frac{11.8-12}{0.15}\right) \times 3$
$=-4$
The test statistic t follows student t-distribution with (10-1)=9 degrees of freedom
It is given that $\mathrm{t}_{0.05}=2.26$
We observe that,
$|t|=4>2.26$
$\Longrightarrow$ Calculate $|\mathrm{t}|>$ tabulated $\mathrm{t}_{9}(0.05)$
So, the null hypothesis is rejected at a $5 \%$ level of significance.
Hence there is a significance between the sample mean $\bar{X}$ and the population mean $\mu$.

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Question 23 Marks
From the following time series obtain trent value by 3 yearly moving averages.
YearSales ( in ₹ 000)YearSales ( in ₹ 000)
20088201312
200912201416
201010201517
201113201614
201215201717
Answer
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Question 33 Marks
Let X denote the no of hours you study during a randomly selected school day. The probability that X can take the values x, has the following form where K is some unknown constant
$P(\chi=x)= \begin{cases}0.1, & \text { if } x=0 \\ k x, & \text { if } x=1, \text { or } 2 \\ K(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { otherwise }\end{cases}$
i. Find the value of $K$.
ii. What is the probability that you study at least two hours? Exactly two hours. At most two hours.
Answer
The probability distribution of x is
X01234
P(X)0.1K2 K2 KK
i. $\sum_{i=1}^{n} p i=1$
$
0.1+\mathrm{K}+2 \mathrm{~K}+2 \mathrm{~K}+\mathrm{K}=1
$
$
K=0.15
$
ii. $p$ (study atleast two hours) $=p(x \geq 2)$
$
\begin{aligned}
& =2 \mathrm{~K}+2 \mathrm{~K}+\mathrm{K} \\
& =5 \mathrm{~K} \\
& =5 \times 0.15 \\
& =0.75 \\
& \mathrm{p} \text { (Study exactly two hours) }=\mathrm{p}(\mathrm{x}=2) \\
& =2 \mathrm{~K} \\
& =2 \times 0.15 \\
& =0.3 \\
& \mathrm{p}(\text { Study at most two hours })=\mathrm{p}(\mathrm{x} \leq 2) \\
& =\mathrm{p}(\mathrm{x}=0)+\mathrm{p}(\mathrm{x}=1)+\mathrm{p}(\mathrm{x}=2) \\
& =0.1+\mathrm{k}+2 \mathrm{k} \\
& =0.1+3 \mathrm{k}=0.1+3(0.15) \\
& =0.1+0.45=0.55
\end{aligned}
$
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Question 43 Marks
A bag contains 8 red and 5 white balls. Two successive draws of all 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and second draw yields 3 red balls.
Answer
Let E : Event that 3 balls in the first draw are all white.
F : Event that 3 balls in the second draw are all red.
Now, 3 balls can be drawn out of 13 in ${ }^{13} \mathrm{C}_{3}$ ways and 3 white balls can be drawn out of 5 in ${ }^{5} \mathrm{C}_{3}$ ways
$P(E)=\frac{{ }^{5} C_{3}}{{ }^{13} C_{3}}=\frac{5!}{3!\times 2!} \times \frac{3!\times 10!}{13!}=\frac{5}{143}$
Since, 3 balls are not replaced before the second draw, we are left with 8 red and 2 white balls.
Now, 3 balls can be drawn in ${ }^{10} \mathrm{C}_{3}$ ways and 3 red balls can be drawn in ${ }^{8} \mathrm{C}_{3}$ ways.
$P\left(\frac{F}{E}\right)=\frac{{ }^{8} C_{3}}{{ }^{10} C_{3}}$
$=\frac{8!}{3!\times 5!} \times \frac{3!\times 7!}{10!}=\frac{7}{15}$
$\therefore P(E \cap F)=P(E) . P\left(\frac{F}{E}\right)=\frac{5}{143} \times \frac{7}{15}$
$=\frac{7}{429}$.
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Question 53 Marks
If the marginal revenue function for output x is given by $\mathrm{MR}=\frac{6}{(x+2)^{2}}+5$, find the total revenue function and the demand function.
Answer
$M R=\frac{6}{(x+2)^{2}}+5$
$\mathrm{R}=\int(M R) d x+C$
$=\int\left(\frac{6}{(x+2)^{2}}+5\right) d x+\mathrm{C}$
$=6 \int \frac{1}{(x+2)^{2}} d x+5 \int d x+C$
$=6 \frac{-1}{(x+2)}+5 x+c$
$=\frac{-6}{(x+2)}+5 x+c$
When $\mathrm{R}=0$ and $\mathrm{x}=0$
$0=\frac{-6}{(0+2)}+5(0)+C$
$\mathrm{C}=3$
$\therefore \mathrm{R}=\frac{-6}{(x+2)}+5 x+3$
and demand function is given by,
$\mathrm{p}=\frac{R}{x}$
$p =\frac{\frac{-6}{(x+2)}+5 x+3}{x}$
$\mathrm{p}=\frac{-6}{x(x+2)}+5+\frac{3}{x}$
Where p is the price, when number of units sold x.
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Question 63 Marks
Madhu exchanged her old car valued at ₹ $1,50,000$ with a new one priced at ₹$ 6,50,000$. She paid ₹ $x$ as down payment and the balance in 20 monthly equal instalments of ₹ 21,000 each. The rate of interest offered to her is $9 \%$ p.a. Find the value of $x$. [Given that: $(1.0075)^{-20}=0.86118985$]
Answer
Madhu paid the balance in 20 monthly installments of ₹ 21000 each
Let Principle $=\mathrm{P}, \mathrm{i}=\frac{9}{1200}=0.0075, \mathrm{n}=20$ and $\mathrm{E}=21000$
$E=\frac{P i}{1-(1+i)^{-n}}$
$\Rightarrow 21000=\frac{P \times(0.0075)}{1-(1.0075)^{-20}}$
$\Rightarrow 21000=\frac{P \times(0.0075)}{1-0.8611}$
$\Rightarrow 21000=\frac{P \times(0.0075)}{0.1389}$
$\Rightarrow 21000 \times 0.1389=P \times(0.0075)$
$\Rightarrow \mathrm{P}=388920$
Thus, the balance is ₹ 388920
Madhu exchanged her old car valued at ₹ 150000 and a new one priced at ₹ 650000 .
So, Madhu had ₹ 500000 after the exchange.
She paid approximately ₹ 388920in the form of monthly installments.
Therefore, the down payment $\mathrm{x}=500000-388920=111080$.
Hence, the value of x is 111080.
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Question 73 Marks
It is known that, if the interest is compounded continuously, the principal changes at the rate equal to the product of the rate of bank interest per annum and the principal.
i. If the interest is compounded continuously at $5 \%$ per annum, in how many years will ₹ 100 double itself?
ii. At what interest rate will ₹ 100 double itself in 10 years ?( $\left.\log _{e} 2=0.6931\right)$
iii. How much will ₹ 1000 be worth at $5 \%$ interest after 10 years? ( $e^{0.5}=1.648$ )
Answer
If $P$ denotes the principal at any time $t$ and the rate of interest be $r \%$ per annum compounded continuously, then according to the law given in the problem, we get
$
\frac{d P}{d t}=\frac{P r}{100}
$
$\Rightarrow \frac{d P}{P}=\frac{r}{100} d t$
$\Rightarrow \int \frac{1}{P} d P=\frac{r}{100} \int d t$
$\Rightarrow \log \mathrm{P}=\frac{r t}{100}+\mathrm{C} \ldots$ (i)
Let $\mathrm{P}_{0}$ be the initial principal i.e. at $\mathrm{t}=0, \mathrm{P}=\mathrm{P}_{0}$
Putting $\mathrm{P}=\mathrm{P}_{0}$ in (i), we get
$\log \mathrm{P}_{0}=\mathrm{C}$
Putting $\mathrm{C}=\log \mathrm{P}_{0}$ in (i), we get
$\log \mathrm{P}=\frac{r t}{100}+\log \mathrm{P}_{0}$
$\Rightarrow \log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100}$ ...(ii)
i. In this case, we have
$
\mathrm{r}=5, \mathrm{P}_{0}=$ ₹ $100 \text { and } \mathrm{P}=$ ₹ $200=2 \mathrm{P}_{0}
$
Substituting these values in (ii), we have
$\log 2=\frac{5}{100} t \Rightarrow \mathrm{t}=20 \log _{\mathrm{e}} 2=20 \times 0.6931$ years $=13.862$ years.
ii. In this case, we have
$
P_{0}=$ ₹ $100, P=$ ₹ $ 200 = 2 P_{0} \text { and } t=10 \text { years. }
$
Substituting these values in (ii), we get
$\log 2=\frac{10 r}{100} t \Rightarrow r=10 \log 2=10 \times 0.6931=6.931$
Hence, $r=6.931 \%$ per annum.
iii. In this case, we have
$
P_{0}=$ ₹ $1000, r=5 \text { and } t = 10
$
Substituting these values in (ii), we get
$
\begin{aligned}
& \log \left(\frac{P}{1000}\right)=\frac{5 \times 10}{100}=\frac{1}{2}=0.5 \Rightarrow \frac{P}{1000}=\mathrm{e}^{0.5} \Rightarrow \mathrm{P}=1000 \times 1.648=1648 \\
& \mathrm{P}=\text { ₹ } 1648
\end{aligned}
$
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Question 83 Marks
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
Answer
Let A be the quantity of bacteria present in culture at any time t and initial quantity of bacteria is $\mathrm{A}_{0}$
$\frac{d A}{d A} \propto A$
$\frac{d A}{d t}=\lambda A$
$\frac{d A}{A}=\lambda d t$
$\int \frac{d A}{A}=\int \lambda d t$
$\log \mathrm{A}=\lambda \mathrm{t}+\mathrm{c} \ldots$ (i)
Initially, $\mathrm{A}=\mathrm{A}_{0}, \mathrm{t}=0$
$\log \mathrm{A}_{0}=0+\mathrm{c}$
$\log A_{0}=c$
Now equation (i) becomes,
$\log \mathrm{A}=\lambda \mathrm{t}+\log \mathrm{A}_{0}$
$\log \left(\frac{A}{A_{0}}\right)=\lambda t$ ...(ii)
Given $\mathrm{A}=2 \mathrm{~A}_{0}$ when $\mathrm{t}=6$ hours
$\log \left(\frac{A}{A_{0}}\right)=6 \lambda$
$\frac{\log 2}{6}=\lambda$
Now equation (ii) becomes,
$\log \left(\frac{A}{A_{0}}\right)=\frac{\log 2}{6} t$
Now, $A=8 \mathrm{~A}_{0}$
so, $\log \left(\frac{8 A_{0}}{A_{0}}\right)=\frac{\log 2}{6} t$
$\log 2^{3}=\frac{\log 2}{6} t$
$3 \log 2=\frac{\log 2}{6} t$
$18=\mathrm{t}$
Hence, Bacteria becomes 8 times in 18 hours.
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3 Marks Question - Applied Maths STD 12 Science Questions - Vidyadip