3 Marks Question

Question 13 Marks

In a group of students, 225 students know French, 100 know Spanish and 45 know both. Each student knows either French or Spanish. How many students are there in the group.

Answer

Let $F$ and $S$ denote the no. of students who know French and Spanish, respectively.
Given, $n(F)=225, n(S)=100, n(F \cap S)=45$
Using identity,
$\mathrm{n}(\mathrm{F} \cup \mathrm{S})=\mathrm{n}(\mathrm{F})+\mathrm{n}(\mathrm{S})-\mathrm{n}(\mathrm{F} \cap \mathrm{S})$
$=225+100-45$
= 325-45
$=280$

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Question 23 Marks

Find the present value of a regular annuity of ₹ 1000 payable for 3 years at $12 \%$ per annum compounded annually?

Answer

Given $\mathrm{R}=₹ 1000, \mathrm{i}=\frac{12}{100}=0.12$ and $\mathrm{n}=3$
$\mathrm{P}=\mathrm{R}\left[\frac{1-(1+i)^{-n}}{i}\right] \Rightarrow P=1000\left[\frac{1-(1+0.12)^{-3}}{0.12}\right]$
$\Rightarrow P=\frac{1000}{0.12} \mathrm{l}-(1.12)^{-3} \mathrm{~J} \ldots$ (i)
Let $\mathrm{x}=(1.12)^{-3}$
Taking log on both sides, we get
$\log x=-3 \log 1.12=-3 \times 0.0492=-0.1476$
$\Rightarrow \mathrm{x}=\operatorname{antilog}(-0.1476)=\operatorname{antilog}(1.8524)$
$\Rightarrow \mathrm{x}=0.7119$
Substituting the value of $x$ in equation (i), we get
$\mathrm{P}=\frac{1000}{0.12}[1-0.7119]=\frac{1000}{0.12} \times 0.2881=2400.83$

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Question 33 Marks

A family in Agra, U.P consumes 78 SCM of gas in 60 days. The GST 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:

Units 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 78 SCM 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 charges is $10 \%$ of gas consumption charges
$\therefore$ Minimum charges $=10 \%$ 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 43 Marks

Discuss the continuity of the junction $\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=\frac{1}{2}$, where

Answer

We observe that:
(L H.L at $x =\frac{1}{2}$ ) $=\lim _{x \rightarrow 1 / 2^{-}} f ( x )=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}- x \right)\left[\because f(x)=\frac{1}{2}-x\right.$ for $\left.0 \leq x<\frac{1}{2}\right]$
$=\frac{1}{2}-\frac{1}{2}=0$ [Using direct substitution method]

Clearly, $\lim _{x \rightarrow 1 / 2^{-}} f(x) \neq \lim _{x \rightarrow 1 / 2^{+}} f(x)$
Hence, $f ( x )$ is not continuous at $x =\frac{1}{2}$. Clearly, $f ( x )$ has discontinuity of first kind at $x =\frac{1}{2}$.

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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?

Answer

Rohit is the husband of Vanshika

Sumita is the sister of Rohit

Anushka is the sister of Vanshika

So Anushka is Rohit's wife's sister
Anushka is the sister-in-law of Rohit.

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Question 63 Marks

A class consists of 60 boys and 40 girls. In how many ways can a president, vice president, secretary and treasurer can be chosen if the secretary must be a boy, the treasurer must be a girl and a student may not hold more than one office?

Answer

The treasurer can be selected from the 40 girls in ${ }^{40} \mathrm{C}_{1}$ ways, i.e., in 40 ways.
The secretary can be selected from the 60 boys in ${ }^{60} \mathrm{C}_{1}$ ways, i.e., 60 ways.
We now have 100-2 = 98 students left for two positions - president and vice president.
The two can be selected in ${ }^{98} \mathrm{C}_{2}$ ways and can be arranged in 2 ways i.e. the first person selected can be either the president or the vice president.
Hence this can be done $2 \times{ }^{98} \mathrm{C}_{2}$ ways. (OR we can say that the president can be selected in ${ }^{98} \mathrm{C}_{1}$ (i.e. 98 ways) and vice president can be selected in ${ }^{(98-1)} \mathrm{C}_{1}$ ways $=97$ ways).
Total ways of selection $=40 \times 60 \times 98 \times 97=22814400$

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Question 73 Marks

If ${ }^{\mathrm{n}} \mathrm{P}_{4}:{ }^{\mathrm{n}} \mathrm{P}_{2}=12$, find n .

Answer

Given, $\frac{{ }^{n} P_{4}}{{ }^{n} p_{2}}=12$
$\Rightarrow{ }^{n} P_{4}=12^{n} P_{2}$
$\Rightarrow \frac{n!}{n-4!}=12 \frac{n!}{n-2!}$
$\Rightarrow \frac{n-2!}{n-4!}=12$
$\Rightarrow \frac{(n-2)(n-3)(n-4)!}{n-4!}=12$
$\Rightarrow \mathrm{n}^{2}-5 \mathrm{n}+6=12$
$\Rightarrow \mathrm{n}^{2}-5 \mathrm{n}-6=0$
$\Rightarrow \mathrm{n}^{2}-6 \mathrm{n}+\mathrm{n}-6=0$
$\Rightarrow \mathrm{n}(\mathrm{n}-6)+1(\mathrm{n}-6)=0$
$\Rightarrow(\mathrm{n}-6)(\mathrm{n}+1)=0$
$\Rightarrow \mathrm{n}=-1$ or $\mathrm{n}=6$
$\because$ n cannot be negative
$\therefore \mathrm{n}=6$

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Applied Maths 3 Marks Question

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