Case study (4 Marks)

Question 14 Marks

A manufacturing company produces certain goods. The company manager used to make a data record on daily basis about the cost and revenue of these goods separately. The cost and revenue function of a product are given by $C(x)=20 x+4000$ and $R(x)=60 x+2000$, respectively, where $x$ is the number of goods produced and sold.

Based on above information, answer the following questions.

(i) How many goods must be sold to realise some profit?
(a) $\mathrm{x}<\mathbf{5 0}$
(b) $x>50$
(c) $x \geq 50$
(d) $\mathbf{x} \leq \mathbf{5 0}$

(ii) If the cost and revenue functions of a product are given by $C(x)=3 x+400$ and $R(x)=$ $5 x+20$ respectively, where $x$ is the number of items produced by the manufacturer, then how many items must be sold to realise some profit?
(a) $x \leq 190$
(b) $x \geq 190$
(c) $x<190$
(d) $x>190$

(iii) Let $\mathbf{x}$ and $\mathbf{b}$ are real numbers. If $\mathbf{b} > \mathbf{0}$ and $\mathbf{x}< \mathbf{b}$, then
(a) $x$ is always positive
(b) $\mathbf{X}$ is always negative
(c) $\mathrm{x}$ is real number
(d) None of these

(iv) The solution set of $\mathbf{3}-\mathbf{5}<\mathrm{x}+\mathbf{7}$, when $\mathrm{x}$ is a whole number is given by
(a) $\{0,1,2,3,4,5\}$
(b) $(-\infty, 6)$
(c) $[0,5]$
(d) None of these

(v) Graph of inequality $x>2$ on the number line is represented by
(a)

(b)

(c)

(d) None of the above

Answer

(i) (b) We know that, Profit = Revenue - Cost
$\therefore$ In order to realise some profit, revenue should be greater than the cost.
Thus, we should have $\mathbf{R}(\mathbf{x})>\mathbf{C}(\mathbf{x})$
$
\begin{aligned}
& \Rightarrow 60 \mathrm{x}+2000 \quad>20 \mathrm{x}+4000 \\
& \Rightarrow \quad 60 \mathrm{x}+2000-20 \mathrm{x}>20 \mathrm{x}+4000-20 \mathrm{x} \\
& \text { [ subtracting } 20 \mathrm{x} \text { from both sides ] } \\
& \Rightarrow \quad 40 \mathrm{x}+2000>4000 \\
& \Rightarrow \quad 40 x+2000-2000>4000-2000 \\
& \Rightarrow \quad \text { [ subtracting } 2000 \text { from both sides ] } \\
& \Rightarrow40 \mathrm{x}>2000 \\
& \Rightarrow \frac{40 \mathrm{x}}{40}>\frac{2000}{40} \\
& \Rightarrow x>50 \\
\end{aligned}
$
[subtracting 2000 from both sides]
[dividing both sides by 40 ]
$
\Rightarrow x>50
$
Hence, the manufacturer must sell more than 50 items to realise some profit.

(iii) (d) We have, $\mathbf{b}>\mathbf{0}$
and $\mathbf{x}<\mathbf{b}$
Its mean $\mathbf{x}$ is always less than some positive quantity.
$\therefore \mathrm{x}$ may be a real number.

(iv) (a) We have, $3 \mathbf{x}-\mathbf{5}<\mathbf{x}+\mathbf{7}$
$
\begin{aligned}
& \Rightarrow 3 \mathrm{x}-5+5<\mathrm{x}+7+5 \quad \text { [adding } 5 \text { both sides] } \\
& \Rightarrow \quad 3 \mathrm{x}<\mathrm{x}+12 \\
& \Rightarrow \quad 3 \mathrm{x}-\mathrm{x}<\mathrm{x}+12-\mathrm{x} \\
&
\end{aligned}
$
[adding 5 both sides]
[subtracting $\mathbf{x}$ from both sides]
$
\begin{aligned}
& \Rightarrow 2 x<12 \\
& \Rightarrow \frac{2 x}{2}<\frac{12}{2}
\end{aligned}
$
[dividing both sides by 2]
$
\Rightarrow \mathrm{x}<6
$
Now, if $\mathbf{x}$ is a whole number, then the solution set $\{0,1,2,3,4,5\}$.

(v) (b)

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Question 24 Marks

Shweta was teaching "method to solve a linear inequality in one variable" to her daughter.
Step I Collect all terms involving the variable (x) on one side and constant terms on other side with the help of above rules and then reduce it in the form $\mathbf{a x}<\mathbf{b}$ or $\mathbf{a x} \leq \mathbf{b}$ or $\mathbf{a x}>\mathbf{b}$ or $\mathbf{a x} \geq \mathbf{b}$.
Step II Divide this inequality by the coefficient of variable (x). This gives the solution set of given inequality.
Step III Write the solution set.

Based on above information, answer the following questions.

(i) The solution set of $\mathbf{2 4 x}<\mathbf{1 0 0}$, when $\mathrm{x}$ is a natural number is
(a) $\{1,2,3,4\}$ (b) $(1,4)$ (c) $[1,4]$ (d) None of these

(ii) The solution set of $24100 \mathrm{x}<$, when $\mathrm{x}$ is an integer is
(a) $\{\ldots \ldots-4,-3,-2,-1,0,1,2,3,4\}$ (b) $(-\infty, 4]$ (c) $[4, \infty]$ (d) None of the above

(iii) The solution set of $-\mathbf{5 x}+\mathbf{2 5}>0$ is
(a) $[5, \infty)$ (b) $(-\infty, 5]$ (c) $(5, \infty)$ (d) $(-\infty, 5)$

(iv) The solution set of $\mathbf{3 x}-\mathbf{5}<\mathbf{x + 7}$ is
(a) $(6, \infty)$ (b) $[6, \infty)$ (c) $(-\infty, 6)$ (d) $(-\infty, 6]$

(v) The solution set of $x+\frac{x}{2}+\frac{x}{3}<11$ is
(a) $(-\infty, 6]$ (b) $(-\infty, 6)$ (c) $[6, \infty)$ (d) None of these

Answer

(i) (a) We have, $\mathbf{2 4 x}<\mathbf{1 0 0}$
On dividing both sides by 24 , we get
$
=\frac{24 x}{24}<\frac{100}{24} \Rightarrow x<\frac{25}{6}
$
When $\mathbf{x}$ is a natural number, then solutions of the inequality $x<\frac{25}{6}$, are all natural numbers, which are less than $\frac{25}{6}$. In this case, the following values of $x$ make the statement true.
$
\mathrm{x}=1,2,3,4
$
Hence, the solution set of inequality is $\{\mathbf{1 , 2 , 3 , 4}\}$.

(ii) (a) We have, $24 \mathrm{x}<\mathbf{1 0 0}$
On dividing both sides by 24 , we get
$
=\frac{24 x}{24}<\frac{100}{24} \Rightarrow x<\frac{25}{6}
$
When $\mathbf{x}$ is an integer.
In this case, solutions of given inequality are
$
\ldots,-4,-3,-2,-1,0,1,2,3,4
$
Hence, the solution set of inequality is
$
\{\ldots,-4,-3,-2,-1,0,1,2,3,4\}
$

(iii) (d) We have, $-5 x+25>0$
On adding $5 \mathbf{x}$ both sides, we get
$
-5 x+25+5 x>0+5 x \Rightarrow 25>5 x \Rightarrow 5 x<25
$
On dividing both sides by 5 , we get
$
\Rightarrow \frac{5 x}{5}<\frac{25}{5} \Rightarrow x<5
$
Hence, the required solution set is $(-\infty, 5)$.

(iv)
(c) We have, $3 \mathbf{x}-\mathbf{5} < \mathbf{x}+\mathbf{7}$
$\Rightarrow 3 x-5+5 < x+7+5$
$\Rightarrow 3 \mathrm{x}<\mathrm{x}+12$
$\Rightarrow 3 \mathrm{x}-\mathrm{x}<\mathrm{x}+12-\mathrm{x}$
[adding 5 both sides]
[subtracting $\mathbf{x}$ from both sides]
$
\begin{aligned}
& \Rightarrow \quad 2 x<12 \\
& \Rightarrow \quad \frac{2 x}{2}<\frac{12}{2} \\
& \Rightarrow \quad x<6
\end{aligned}
$
[dividing both sides by 2]
The solution set is $\{\mathbf{x}: \mathbf{x} \in \mathbf{R}$ and $\mathbf{x}<6\}$ i.e., any real number less than 6 . This can also be written as $(-\infty, 6)$.

(v) (b) We have, $x+\frac{x}{2}+\frac{x}{3}<11$
$
\Rightarrow \frac{6 x+3 x+2 x}{6}<11 \Rightarrow \frac{11 x}{6}<11
$

On multiplying both sides by $\frac{6}{11}$, we get
$
\begin{aligned}
& \frac{11 x}{6} \times \frac{6}{11}<11 \times \frac{6}{11} \Rightarrow x<6 \\
& \therefore \quad x \in(-\infty, 6) \\
&
\end{aligned}
$

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MATHS Case study (4 Marks)

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