3 Marks Question

Question 13 Marks

In a class of 50 students, 30 students like Mathematics, 25 like Science, and 16 like both. Find the number of students who like
i. either Mathematics or Science
ii. neither Mathematics nor Science.

Answer

We draw a Venn diagram to solve the problem.
Here $\xi=$ all the students of the class,
$\mathrm{M}=$ students who like Mathematics and
S = students who like Science.

Since 16 students like both Mathematics and Science, we mark 16 in the common region of M and S . Then, as 30 students like Mathematics and of these 16 students have already been marked, therefore, 14 is marked in the remaining portion of M . Also, as 25 students like Science and of these 16 have already been marked, therefore, 9 is marked in the remaining portion of S.
i. The number of students who like either Mathematics or Science
$
=14+16+9=39
$
ii. The number of students who like neither Mathematics nor Science (shown shaded in the diagram)
$
=50-39=11
$

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Question 23 Marks

Mr. Sharma in Delhi is using an MTNL connection with a monthly plan of ₹ 240 , which has no free calls. The call charges are ₹ 1 per minute. Calculate the telephone bill payable for Mr. Sharma if he talks for 470 minutes in a month. Assume the broadband charges to be Nil. A GST of $18 \%$ is levied on the total bill amount.

Answer

Here, the monthly charges are ₹ 240
Since the tariff plan chosen by Mr. Sharma has no free calls and he talks for 470 minutes
$\therefore$ Call charges @ ₹ 1 per minute $=₹(470 \times 1)=₹ 470$
Broband charges = Nil
$\therefore$ Total Bill Amount $=$ Monthly Charges + Call Charges
= ₹ $240+₹ 470=₹ 710$
A GST of $18 \%$ is levied on the total bill amount
GST= $18 \%$ of 710
= ₹ 127.80
$\therefore$ Total Bill Payable $=₹ 710+₹ 127.80$
= ₹ 837.80

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Question 33 Marks

A machinery can be purchased by paying ₹ 600,000 now or ₹ 200,000 initially and two instalments of ₹ 300,000 at the end of second year and ₹ 190,000 at the end of 4th year respectively. To pay cash now, the buyer would have to withdraw the money from an investment earning interest at $8 \%$ compounded annually. Which option is better and by how much in present value terms?

Answer

Let $P_{1}$ be the present value of ₹ 300,000 due 2 years hence at $i=\frac{8}{100}=0.08$ compounded annually. Also, let $P_{2}$ be the present value of ₹ 190,000 due 4 years hence at $i=\frac{8}{100}=0.08$ compounded annually. Then,
$\mathrm{P}_{1}=300,000(1+\mathrm{i})^{-2}$ and $\mathrm{P}_{2}=190,000(1+\mathrm{i})^{-4}$
$\Rightarrow \mathrm{P}_{1}=300,000(1.08)^{-2}$ and $\mathrm{P}_{2}=190,000(1.08)^{-4}$
$\Rightarrow \mathrm{P}_{1}=300,000 \times 0.85733882$ and $\mathrm{P}_{2}=190,000 \times 0.73502985$
$\Rightarrow \mathrm{P}_{1}=257,201.446$ and $\mathrm{P}_{2}=,655.6715$
$\Rightarrow \mathrm{P}_{1}+\mathrm{P}_{2}=396,857.3175 \simeq 396,857.32$
Thus, in the second option the machinery costs ₹ $(200,000+396,857.32)=₹ 596,857.32$
Clearly, in the second option the machinery costs less. Hence, second option is better and it is better by ₹ (600,000-596,857.32) = ₹ 3142.68

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Question 43 Marks

Find the domain of the function f defined by $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$

Answer

Let $\mathrm{f}=\mathrm{g}+\mathrm{h}$, then $\mathrm{g}(\mathrm{x})=\sqrt{4-x}$ and $\mathrm{h}(\mathrm{x})=\frac{1}{\sqrt{x^{2}-1}}$
For $\mathrm{D}_{\mathrm{g}}, \mathrm{g}(\mathrm{x})$ must be a real number
$\Rightarrow 4-\mathrm{x} \geq 0$
$\Rightarrow 4 \geq \mathrm{x} \Rightarrow \mathrm{x} \leq 4 \Rightarrow \mathrm{D}_{\mathrm{g}}=(-\infty, 4]$
For $\mathrm{D}_{\mathrm{h}}, \mathrm{h}(\mathrm{x})$ must be a real number
$\Rightarrow \mathrm{x}^{2}-1>0$
$\Rightarrow(\mathrm{x}+1)(\mathrm{x}-1)>0$
$\Rightarrow \mathrm{x}<-1$ or $\mathrm{x}>1$
$\Rightarrow \mathrm{D}_{\mathrm{h}}=(-\infty,-1) \cup(1, \infty)$
As $f=g+h$, so $D_{f}=D_{g} \cap D_{h}$

From fig. it is clear that $D_{g} \cap D_{h}=(-\infty,-1) \cup(1,4]$
Hence, the domain of the function $\mathrm{f}=(-\infty,-1) \cup(1,4]$

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Question 53 Marks

Rohit is the husband of Vanshika. Sumita is the sister of Rohit. Anushka is the sister of Vanshika. How Anushka is related to Rohit?

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Question 63 Marks

If $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ are two A.M.'s between a and b , prove that
i. $\mathrm{A}_{1}+\mathrm{A}_{2}=\mathrm{a}+\mathrm{b}$
ii. $\left(2 \mathrm{~A}_{1}-\mathrm{A}_{2}\right)\left(2 \mathrm{~A}_{2}-\mathrm{A}_{1}\right)=a b$

Answer

Let $A_{1}, A_{2}$ be two A.M.'s between $a$ and $b$, then $a, A_{1}, A_{2}, b$ are in A.P.
As a, $\mathrm{A}_{1}, \mathrm{~A}_{2}$ are in A.P.
$\therefore 2 \mathrm{~A}_{1}=\mathrm{a}+\mathrm{A}_{2}$
$\Rightarrow 2 A_1- A _2= a \ldots(1)$
Similarly, $\mathrm{A}_{1}, \mathrm{~A}_{2}$, b are in A.P.
$\Rightarrow 2 \mathrm{~A}_{2}-\mathrm{A}_{1}=\mathrm{b} . .$. (2)
i. Adding (1) and (2), we get $\mathrm{A}_{1}+\mathrm{A}_{2}=\mathrm{a}+\mathrm{b}$
ii. Multiplying (1) and (2), we get $\left(2 \mathrm{~A}_{1}-\mathrm{A}_{2}\right)\left(2 \mathrm{~A}_{2}-\mathrm{A}_{1}\right)=a b$

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Question 73 Marks

If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in G.P, show that $\mathrm{a}^{2}+\mathrm{b}^{2}, \mathrm{~b}^{2}+\mathrm{c}^{2}, \mathrm{c}^{2}+\mathrm{d}^{2}$ are in G.P.

Answer

From the question, it is given that $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in G.P
So, bc = ad
$b^{2}=\mathrm{ac}$
$c^{2}=b d$
We have to show that,
$a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P.
Then, $\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$
Consider the LHS $=\left(b^{2}+c^{2}\right)^{2}$
$=b^{4}+c^{4}+2 b^{2} c^{2}$
From the equation (ii) and equation (iii)
$=a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}$
$=c^{2}\left(a^{2}+b^{2}\right)+d^{2}\left(a^{2}+b^{2}\right)$
$=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$
Now consider the
RHS $=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$
By comparing the LHS and RHS
LHS = RHS
Hence it is proved that, $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P.

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Applied Maths 3 Marks Question

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