3 Marks Question

Question 13 Marks

In a survey of 450 people, it was found that 110 play cricket, 160 play tennis and 70 play both cricket as well tennis. How many play neither cricket nor tennis?

Answer

Let C and T denotes the students who play cricket and tennis, respectively.
Given, $\mathrm{n}(\mathrm{C})=110, \mathrm{n}(\mathrm{T})=160, \mathrm{n}(\mathrm{C} \cap \mathrm{T})=70, \mathrm{n}(\mathrm{U})=450$.
Using identity,
$\mathrm{n}(\mathrm{C} \cup \mathrm{T})=\mathrm{n}(\mathrm{C})+\mathrm{n}(\mathrm{T})-\mathrm{n}(\mathrm{C} \cap \mathrm{T})$
$=110+160-70$
= 200
$\therefore$ No. of students play neither cricket nor tennis
$=n(U)-n(C \cap T)$
$=450-200$
$=250$

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Question 23 Marks

A family in Agra, U.P. consumes 78 SCM of gas in 60 days. The VAT is included in the gas charges and the minimum charge is $10 \%$ of the gas consumption charges. The PNG rate in Agra are as follows:

Unit of Consumption(in SCM)	Price Per Unit
up to 45 SCM / 60 days	₹ 29.50
above 45 SCM / 60 days	₹ 42.61

Calculate the bimonthly PNG bill of the family.

Answer

Here, the consumption of gas is given to be 78SCM for 60 days.
According to the given tariff plan:
Gas consumption charges $=₹[(45 \times 29.50)+(33 \times 42.61)]$
$=₹(1327.50+1406.13)$
= ₹ 2733.63
The minimum charge is $10 \%$ of the gas consumption charge.
$\therefore$ Minimum charge $=10 \%$ of ₹ 2733.63
=₹ 273.36
$\therefore$ Bimonthly PNG bill of the family $=$ Gas consumption charges + Minimum charges
= ₹ 2733.63 + ₹ 273.36
= ₹ 3006.99

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Question 33 Marks

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at $4 \%$ per annum is ₹ 1. Find the sum.

Answer

Let the sum be ₹ x. Then,
Compound Interest $=\mathrm{A}_{\mathrm{n}}-\mathrm{P} \Rightarrow \mathrm{I}=(1+\mathrm{i})^{\mathrm{n}}-1$
Her, $\mathrm{P}=\mathrm{x}, \mathrm{i}=4 \%=\frac{4}{100}=0.04, \mathrm{n}=2$
Compound Interest $=x(1+0.04)^{2}-\mathrm{x}$
$=\frac{676}{625} \mathrm{x}-\mathrm{x}$
$=\frac{51}{625} \mathrm{x}$
Now, simple interest,
$\mathrm{I}=\mathrm{P}$ it
$=\mathrm{x} \times 0.04 \times 2$
$=\frac{2 x}{25}$
Since, given difference $=$ ₹ 1
i.e., $\frac{51}{625} x-\frac{2 x}{25}=1$
$\Rightarrow \mathrm{x}=625$.

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Question 43 Marks

Find the domain and the range of the given function: $\mathrm{f}(\mathrm{x})=\sqrt{x^{2}-4}$

Answer

Given $\mathrm{f}(\mathrm{x})=\sqrt{x^{2}-4}$
For $\mathrm{D}_{\mathrm{f}}, \mathrm{f}(\mathrm{x})$ must be a real number $\Rightarrow \sqrt{x^{2}-4}$ must be a real number
$x^{2}-4 \geq 0 \Rightarrow(x+2)(x-2) \geq 0$
either $\mathrm{x} \leq-2$ or $\mathrm{x} \geq 2$
$\Rightarrow \mathrm{D}_{\mathrm{f}}=(-\infty,-2] \cup[2, \infty)$
For $\mathrm{R}_{\mathrm{f}}$, let $\mathrm{y}=\sqrt{x^{2}-4} \ldots$ (i)
As square root of a real number is always non-negative, $\mathrm{y} \geq 0$
On squaring (i), we get $y^{2}=x^{2}-4$
$\Rightarrow x^{2}=y^{2}+4$ but $x^{2} \geq 0$ for all $x \in D_{f}$
$\Rightarrow \mathrm{y}^{2}+4 \geq 0 \Rightarrow \mathrm{y}^{2} \geq-4$, which is true for all $\mathrm{y} \in \mathbf{R}$. Also $\mathrm{y} \geq 0$
$\Rightarrow \mathrm{R}_{\mathrm{f}}=[0, \infty)$

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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 $5, \mathrm{x}, \mathrm{y}, \mathrm{z}, 405$ are the first five terms of a GP, find the values of $\mathrm{x}, \mathrm{y}$ and z .

Answer

5, $\mathrm{x}, \mathrm{y}, \mathrm{z}, 405$ are in GP.
Let the ratio be $r$, also $a_{5}=405$
$\therefore 5(\mathrm{r})^{4}=405$
$\Rightarrow(\mathrm{r})^{4}=81=(3)^{4} \Rightarrow \mathrm{r}=3$
$\Rightarrow \mathrm{x}=5 \times 3=15$,
$y=15 \times 3=45$,
$z=45 \times 3=135$

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Question 73 Marks

The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is $\frac{1}{3}$. Calculate the first and the thirteenth term.

Answer

Let 'a' be the first term and 'd' be a common difference.
$\mathrm{S}_{6}=42$
$\left.\frac{6}{2}\right\rvert\,(2 a+5 d)=14$
$2 \mathrm{a}=14-5 \mathrm{~d} . . .(\mathrm{i})$
$\frac{a_{10}}{a_{30}}=\frac{1}{3}$
$\frac{a+9 d}{a+29 d}=\frac{1}{3}$
$\frac{2 a+18 d}{2 a+58 d}=\frac{1}{3}$
$\frac{14-5 d+18 d}{14-5 d+58 d}=\frac{1}{3}$ From (i)
$\frac{14+13 d}{14+53 d}=\frac{1}{3}$
$42+39 \mathrm{~d}=14+53 \mathrm{~d}$
$28=14 \mathrm{~d}$
$\Rightarrow \mathrm{d}=2$
From (i)
$2 \mathrm{a}+5 \mathrm{~d}=14$
$\Rightarrow \mathrm{a}=2$
Now, $\mathrm{a}_{13}=2+12 \times 2=26$

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Applied Maths 3 Marks Question

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