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 one is white and one is black.

Answer

Consider the following events:
$\mathrm{W}_{1}=$ Drawing a white ball from first bag, $\mathrm{W}_{2}=$ Drawing a white ball from second bag.
$\mathrm{B}_{1}=$ Drawing a black ball from first bag, $\mathrm{B}_{2}=$ 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 (one white ball and one black ball)
$=\mathrm{P}[($ black from 1st and white from 2nd) or (white from 1st and black from 2nd) $]$
$=P\left[\left(B_{1} \cap W_{2}\right) \cup\left(W_{1} \cap B_{2}\right)\right]$
$=P\left(B_{1} \cap W_{2}\right)+P\left(W_{1} \cap B_{2}\right)$ [By addition theorem for mutually exclusive events]
$=\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{W}_{2}\right)+\mathrm{P}\left(\mathrm{W}_{1}\right) \mathrm{P}\left(\mathrm{B}_{2}\right)\left[\because \mathrm{B}_{1}\right.$ and $\mathrm{W}_{2} ; \mathrm{B}_{2} \& \mathrm{~W}_{1}$ are pairs of independent events $]$
$=\frac{2}{6} \times \frac{3}{8}+\frac{4}{6} \times \frac{5}{8}=\frac{13}{24}$

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Question 22 Marks

Differentiate the function $\sqrt{x}\left(2 x^{2}+3\right)$ with respect to x .

Answer

Let $\mathrm{y}=\sqrt{x}\left(2 x^{2}+3\right)$
$=2 x^{\frac{5}{2}}+3 x^{\frac{1}{2}}$
$\frac{d y}{d x}=2 \cdot \frac{5}{2} x^{\frac{3}{2}}+3 \times \frac{1}{2} \cdot x^{\frac{-1}{2}}$
$\Rightarrow \frac{d y}{d x}=5 x^{\frac{3}{2}}+\frac{3}{2 \sqrt{x}}$

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Question 32 Marks

Find the derivative of the given function from the first principle: $\frac{a x+b}{c x+d}$

Answer

Let $\mathrm{f}(\mathrm{x})=\frac{a x+b}{c x+d}$, note that f is not defined at $\mathrm{x}=\frac{d}{c}$
By def., $\mathrm{f}^{\prime}(\mathrm{x})=\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\operatorname{Lt}_{h \rightarrow 0} \frac{\frac{a(x+h)+b}{c(x+h)+d}-\frac{a x+b}{c x+d}}{h}$
$=\operatorname{Lt}_{h \rightarrow 0} \frac{(a x+b+a h)(c x+d)-(c x+d+c h)(a x+b)}{h(c x+d+c h)(c x+d)}$
$=\operatorname{Lt}_{h \rightarrow 0} \frac{a h(c x+d)-c h(a x+b)}{h(c x+d+c h)(c x+d)}=\operatorname{Lt}_{h \rightarrow 0} \frac{a d-b c}{(c x+d+c h)(c x+d)}$
$=\frac{a d-b c}{(c x+d+0)(c x+d)}=\frac{a d-b c}{(c x+d)^{2}}, x \neq-\frac{d}{c}$

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Question 42 Marks

At what time between 5:00 and 6:00 will the hands of clock be at right angles?

Answer

At 5 O'clock, the hour hand is at 5 and the minute hand is at 12 . It means the angle between the two hands of the clock is $150^{\circ}$. The hands of the clock will be at right angles twice between 5:00 and 6:00
For the first time: The minute hand had to cover a relative distance of $60^{\circ}$.
So the time required $=\frac{60}{5.5}$ minutes $=\frac{60 \times 2}{11}$ minutes $=10 \frac{10}{11}$ minutes
$=10 \mathrm{~min} 55 \mathrm{sec}$
Hence, the hands of the clock are at right angles at 5:10:55
For the second time: The minute hand had to cover a relative distance of $240^{\circ}$.
So the time required $=\frac{240}{5.5}$ minutes $=\frac{240 \times 2}{11}$ minutes $=43 \frac{7}{11}$ minutes
$=43 \mathrm{~min} 38 \mathrm{sec}$
Hence, the hands of the clock are at right angles at 5:43:38.

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Question 52 Marks

In a certain code language 327 means Truth is eternal, 7983 means 'Enmity is not eternal' and 9426 means Truth does not perish. What does Enmity mean in this language?

Answer

In the first and second statements, common codes are 3 and 7, and common words are 'is' and 'eternal'. In the first and third statements, the common code is 2 and the common word is 'truth'.
So code 2 means 'truth'
In the second and third statements, the common code is 9 and the common word is 'not'.
So code 8 stands for the word 'Enmity'.

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Question 62 Marks

Pointing to a girl, Rishabh said Her mother's brother is the only son of my mother's father. How is the girl's mother related to Rishabh?

Answer

Her mother's brother means a girl's maternal uncle.
Only son of my mother's father means Rishabh's maternal uncle.
So girl's maternal uncle is Rishabh's maternal uncle
$\Rightarrow$ girl's mother's brother is Rishabh's mother's brother
$\Rightarrow$ girl's mother and Rishabh's mother are sisters
girl's mother is the maternal Aunt of Rishabh.

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Question 72 Marks

Rohit takes twice as much time as Mohit and thrice as much time as Sonit to complete a piece of work. If working together, they can complete the work in 4 days, find the time taken by each of them separately to complete the work.

Answer

Suppose Rohit can complete the given piece of work in n days. Then, Mohit and Sonit alone take $\frac{n}{2}$ and $\frac{n}{3}$ days respectively to complete the same piece of work. It is given that all three working together can complete the work in 4 days.
$\therefore \frac{1}{4}=\frac{1}{n}+\frac{1}{\frac{n}{2}}+\frac{1}{\frac{n}{3}} \Rightarrow \frac{1}{4}=\frac{1}{n}+\frac{2}{n}+\frac{3}{n} \Rightarrow \frac{1}{4}=\frac{6}{n} \Rightarrow \mathrm{n}=24$
Hence, Rohit, Mohit and Sonit alone can complete the work in 24 days, 12 days and 8 days respectively

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