2 Marks Questions

Question 12 Marks

A bag contains 4 white and 2 black balls. Another contains 3 white and 5 black balls. If one ball is drawn from each bag, find the probability that both are black.

Answer

Consider the following events:
$\mathrm{W}_{1}=$ Drawing a white ball from the first bag, $\mathrm{W}_{2}=$ Drawing a white ball from second bag.
$B_{1}=$ Drawing a black ball from first bag, $B_{2}=$ Drawing a black ball from second bag.
Clearly, $\mathrm{P}\left(\mathrm{W}_{1}\right)=\frac{4}{6}, \mathrm{P}\left(\mathrm{B}_{1}\right)=\frac{2}{6}, \mathrm{P}\left(\mathrm{W}_{2}\right)=\frac{3}{8}$ and $\mathrm{P}\left(\mathrm{B}_{2}\right)=\frac{5}{8}$
Therefore, required probability is given by,
P (both balls are black) $=\mathrm{P}[$ (black ball from 1st bag) and (black ball from 2nd bag)]
$=P\left(B_{1} \cap B_{2}\right)$
$=\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{B}_{2}\right)\left[\because \mathrm{B}_{1}\right.$ and $\mathrm{B}_{2}$ are independent events $\left.)\right]$
$=\frac{2}{6} \times \frac{5}{8}=\frac{5}{24}$

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Question 22 Marks

Differentiate $3^{x}+x^{3}+4 x-5$ with respect to $x$.

Answer

Let, $\mathrm{f}(\mathrm{x})=3^{\mathrm{x}}+\mathrm{x}^{3}+4 \mathrm{x}-5$
$\therefore \frac{d}{d x} f(x)=\mathrm{f}^{\prime}(\mathrm{x})$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{d}{d x}\left(3^{x}+x^{3}+4 x-5\right)$
$=3^{\mathrm{x}} \log _{\mathrm{e}} 3+3 \mathrm{x}^{2}+4 \times 1-0$
$=3^{x} \log _{e} 3+3 x^{2}+4$.

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Question 32 Marks

Differentiate the function with respect to $\mathrm{x}: 3 \mathrm{e}^{-3 \mathrm{x}} \log (1+2)$

Answer

Let $y=3 e^{-3 x} \log (1+x)$
Differentiate it with respect to x we get,
$\frac{d y}{d x}=3 \frac{d}{d x}\left[\mathrm{e}^{-3 \mathrm{x}} \log (1+\mathrm{x})\right]$
$=3\left\{\mathrm{e}^{-3 \mathrm{x}} \frac{1}{1+x}+\log (1+\mathrm{x})\left(-3 \mathrm{e}^{-3 \mathrm{x}}\right)\right\}$ [Using product rule and chain rule
$=3 \frac{e^{-3 x}}{1+x}-3 e^{-3 \mathrm{x}} \log (1+\mathrm{x})$
$=3 \mathrm{e}^{-3 \mathrm{x}}\left\{\frac{1}{1+x}-3 \log (1+\mathrm{x})\right\}$
So, $\frac{d}{d x}\left[3 \mathrm{e}^{-3 \mathrm{x}} \log (1+\mathrm{x})\right]=3 \mathrm{e}^{-3 \mathrm{x}}\left\{\frac{1}{1+x}-3 \log (1+\mathrm{x})\right\}$

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Question 42 Marks

The marks obtained by 15 students in a monthly test are: $11,09,07,03,18,21,13,15,18,04,06,17,22,13,15$
i. Find the average marks of 15 students.
ii. Find the average of their marks when the marks of each student are increased by 2.

Answer

i. The sum of marks of all 15 students
$=11+9+7+3+18+21+13+15+18+4+6+17+22+13+15=192$
$\therefore$Average marks $=\frac{192}{15}=12.8$
ii. When the marks of each student are increased by 2 , then the sum of their marks increases by $15 \times 2$ i.e. by 30
$\therefore$ The new sum of marks of all students $=192+30=222$
$\therefore$ The new average marks $=\frac{222}{15}=14.8$

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Question 52 Marks

Find the values of the letter and give a reason for the steps involved.

Answer

We have to find the value of A And B
Since, $B+1$ we get 8 , i.e., the number whose ones digit is 8
$\therefore$ The only possible value of B is 7 .
So, the question has been decoded as

Now, $\mathrm{A}+7$ we get 1 , i.e. the number whose ones digit is 1
Clearly, $4+7=11$
$\therefore$ the value of $\mathrm{A}=4$.
So, the question has been decoded as

Hence $\mathrm{A}=4$ and $\mathrm{B}=7$

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Question 62 Marks

Which of the two conclusions is/are true on the basis of given statements:
Statements I: Some rats are cats
II: Some cats are not dogs
Conclusions I: Some cats are rats
II: Some dogs are rats

Answer

For the given statements; two possible Venn diagrams are shown below:

Since conclusion, I can be deduced from both the Venn diagrams, so the conclusion I is true. But conclusion II cannot be deduced from Venn diagram (i), so conclusion II is false.

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Question 72 Marks

Which day of the week was on 21st October 1948 ?

Answer

Date is 21st October 1948.

Last two digits of the year	48
Quotient on dividing 48 by 4	12
Date (21st)	21
Manth (October)	07
Year (1948)	00
Total	88

Remainder on dividing 88 by 7 is 4 .
4 is associated with Thursday.
$\therefore$ Day was Thursday.

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