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Maths (commerce) Solve the Following Question.(2 Marks)

Linear Inequations (p-2) · Maharashtra Board STD 11 Commerce / Arts (MSBSHSE - English Medium). 23 questions.

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Solve the Following Question.(2 Marks)

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23 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
Answer
Let two consecutive positive integers be $2 n-1,2 n+1$ where $n \geq 1 \in Z$,
Given that $2 n-1<10$ and $2 n+1<10$
$
\begin{aligned}
& \therefore 2 n<11 \text { and } 2 n<9 \\
& \therefore 2 n<9 \\
& \therefore n<\frac{9}{2} \ldots . . \text { (i) Also, }(2 n-1)+(2 n+1)>11 \\
& \therefore 4 n>11 \\
& \therefore n>\frac{11}{4} \ldots . . . \text { (ii) }
\end{aligned}
$
From (i) and (ii)
$\frac{11}{4}<n<\frac{9}{2}$ Since, $\mathrm{n}$ is an integer,
$
\therefore \mathrm{n}=3,4
$
$\mathrm{n}=3$ gives $2 \mathrm{n}-1=5,2 \mathrm{n}+1=7$
and $n=4$ gives $2 n-1=7,2 n+1=9$
$\therefore$ The pairs of positive consecutive integers are $(5,7)$ and $(7,9)$.
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Question 22 Marks
To receive Grade ' $A$ ' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in the first four examinations are 87,92 , 94, and 95 , find the minimum marks that Sunita must obtain in the fifth examination to get a grade ' $A$ ' in the course.
Answer
Let $x_1, x_2, x_3, x_4, x_5$ denote the marks in five examinations. Then
$
\begin{aligned}
& \frac{x_1+x_2+x_3+x_4+x_0}{5} \geq 90 \\
& \therefore \frac{87+92+94+95+x_5}{5} \geq 90 \\
& \therefore 368+\mathrm{x}_5 \geq 450
\end{aligned}
$
Subtracting 368 from both sides, we get
$
\therefore \mathrm{x}_5 \geq 82
$
Sunita must obtain a minimum of 82 marks in the 5 th examination to get a grade of A.
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Question 32 Marks
Rajiv obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Answer
Let $x_1, x_2, x_3$ denote the marks in 1st, 2 nd and 3 rd unit test respectively. Then
$
\begin{aligned}
& \frac{x_1+x_2+x_3}{3} \geq 60 \\
& \therefore \frac{70+75+x_3}{3} \geq 60 \\
& \therefore 145+\mathrm{x}_3 \geq 3(60)
\end{aligned}
$
Subtracting 145 from both sides, we get
$
\begin{aligned}
& x_3 \geq 180-145 \\
& \therefore x_3 \geq 35
\end{aligned}
$
Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.
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Question 42 Marks
Solve the inequations: $2|x+3|>1$
Answer
$
2|x+3|>1
$
Dividing by 2 on both sides, we get
$
\begin{aligned}
& |x+3|>\frac{1}{2} \\
& \therefore x+3<-\frac{1}{2} \text { or } x+3>\frac{1}{2} \ldots . .[|x|>\text { a implies } x<-a \text { or } x>a]
\end{aligned}
$
Subtracting 3 from both sides, we get
$
\begin{aligned}
& x<-3-\frac{1}{2} \text { or } x>-3+\frac{1}{2} \\
& \therefore x<\frac{-7}{2} \text { or } x>\frac{-5}{2}
\end{aligned}
$
$\therefore x$ can take all real values less $\frac{-7}{2}$ or it can take values greater than $\frac{-5}{2}$.
$\therefore$ Solution set is $\left(-\infty, \frac{-7}{2}\right) \cup\left(\frac{-7}{2}, \infty\right)$
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Question 52 Marks
Solve the inequations: $|2 x+7| \leq 25$
Answer
$
\begin{aligned}
& |2 x+7|<25 \\
& \therefore-25 \leq 2 x+7 \leq 25 \ldots . .[|x| \leq \text { a implies }-a \leq x \leq a]
\end{aligned}
$
Subtracting 7 from both sides, we get
$-32 \leq 2 x \leq 18$
Dividing by 2 on both sides, we get
$-16 \leq \mathrm{x} \leq 9$
$\therefore x$ can take all real values between -16 and 9 including -16 and 9 .
$\therefore$ the solution set is $[-16,9]$
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Question 62 Marks
Solve the inequations: $2|4-5 x| \geq 9$
Answer
$
\begin{aligned}
& 2|4-5 x| \geq 9 \\
& \therefore|4-5 x| \geq \frac{9}{2} \\
& \therefore 4-5 x \geq \frac{9}{2} \text { or } 4-5 x \leq-\frac{9}{2} \ldots . . .[|x| \geq \text { a implies } x \leq-a \text { or } x \geq a]
\end{aligned}
$
Subtracting 4 from both sides, we get
$
-5 x \geq \frac{1}{2} \text { or }-5 x \leq \frac{-17}{2}
$
Divide by -5 (so inequality sign changes)
$\therefore \mathrm{x} \leq-\frac{1}{10}$ or $\mathrm{x} \geq \frac{17}{10}$
$\therefore x$ takes all real values less than or equal to $-\frac{1}{10}$
or it takes all real values greater or equal to $\frac{17}{10}$.
$\therefore$ the solution set is $\left(-\infty,-\frac{1}{10}\right]$ or $\left[\frac{17}{10}, \infty\right)$
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Question 72 Marks
Solve the inequations: $-1<3-\frac{x}{5} \leq 1$
Answer
$-1<3-\frac{x}{5} \leq 1$
Subtracting 3 from both sides, we get
$-4<-\frac{x}{5}<-2$ Multiplying by -1 throughout (so inequality sign changes) $\therefore 4>\frac{x}{5}>2$ i.e., $2<\frac{x}{5}<4$
Multiplying by 5 on both sides, we get
$
10<x<20
$
i.e., $x$ takes all real values between 10 and 20 .
$\therefore$ the solution set is $(10,20)$
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Question 82 Marks
Solve the inequations: $-8 \leq-(3 x-5)<13$
Answer
$-8<-(3 x-5)<13$ Multiplying by -1 throughout (so inequality sign changes) $8 \geq 3 x-5>-13$ i.e., $-13<3 x-5 \leq 8$
Adding 5 on both the sides, we get
$
-8<3 \mathrm{x} \leq 13
$
Dividing, by 3 on both sides, we get
$
\therefore-\frac{8}{3}<x \leq \frac{13}{3}
$
i.e., $x$ takes all real values between $-\frac{8}{3}$ and $\frac{13}{3}$ including $\frac{13}{3}$.
$\therefore$ the solution set is $\left(-\frac{8}{3}, \frac{13}{3}\right]$
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Question 92 Marks
Solve the inequations: $\frac{3}{4} x-6 \leq x-7$
Answer
$
\frac{3}{4} x-6 \leq x-7
$
Multiplying by 4 on both sides, we get
$
3 x-24 \leq 4 x-28
$
Subtracting $3 x$ from both sides, we get
$
-24 \leq x-28
$
Adding 28 on both the sides, we get
$
\therefore 4 \leq \mathrm{x} \text { i.e., } \mathrm{x} \geq 4
$
i.e., $\mathrm{x}$ takes all real values greater or equal to 4 .
$\therefore$ the solution set is $[4, \infty)$
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Question 102 Marks
Solve the inequations: $4-2 x<3(3-x)$
Answer
$
\begin{aligned}
& 4-2 x<3(3-x) \\
& \therefore 4-2 x<9-3 x
\end{aligned}
$
Adding $3 x$ on both sides, we get
$
4+x<9
$
Subtracting 4 from both sides, we get
$
x<5
$
i.e., $\mathrm{x}$ takes all real values less than 5
$\therefore$ the solution set is $(-\infty, 5)$
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Question 112 Marks
Solve the inequations: $3 \mathrm{x}+1 \geq 6 \mathrm{x}-4$
Answer
$
3 x+1 \geq 6 x-4
$
Subtracting $3 x$ from both sides, we get
$
1 \geq 3 \mathrm{x}-4
$
Adding 4 on both sides, we get
$
5 \geq 3 x
$
Dividing by 3 on both sides, we get
$
\frac{5}{3} \geq x
$
i.e., $x \leq \frac{5}{3}$
i.e, $x$ takes all real values less than or equal to $\frac{5}{3}$. $\therefore$ the solution set is $\left(-\infty, \frac{5}{3}\right]$
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Question 122 Marks
Solve the inequations: $\frac{x-2}{x+5}>0$
Answer
$
\frac{x-2}{x+5}>0
$
Since $\frac{a}{b}>0$,
when $a>0$ and $b>0$ or $a<0$ and $b<0 b$
$\therefore$ either $x-2>0$ and $x+5>0$
or $x-2<0$ and $x+5<0$ Case $1: x-2>0$ and $x+5>0$
$\therefore x>2$ and $x>-5$
$\therefore \mathrm{x}>2$
$\therefore$ solution set $=(2, \infty)$
Case II:
$
\begin{aligned}
& x-2<0 \text { and } x+5<0 \\
& \therefore x<2 \text { and } x<-5 \\
& \therefore x<-5 \\
& \therefore \text { solution set }=(-\infty,-5)
\end{aligned}
$
$\therefore$ solution set $=(-\infty,-5)$
$\therefore$ the solution set of the given inequation is $(-\infty,-5) \cup(2, \infty)$
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Question 132 Marks
Solve the inequations: $\frac{x+5}{x-3}<0$
Answer
$
\frac{x+5}{x-3}<0
$
Since $\frac{a}{b}<0$, when $\mathrm{a}>0$ and $\mathrm{b}<0$ or $\mathrm{a}<0$ and $\mathrm{b}>0$
$\therefore$ either $x+5>0$ and $x-3<0$
or $x+5<0$ and $x-3>0$
Case l:
$
\begin{aligned}
& x+5>0 \text { and } x-3<0 \therefore x>-5 \text { and } x<3 \\
& \therefore-5<x<3 \\
& \therefore \text { solution set }=(-5,3) \\
& \text { Case II: } \\
& x+5<0 \text { and } x-3>0 \\
& \therefore x<-5 \text { and } x>3
\end{aligned}
$
Case II:
$
\begin{aligned}
& x+5<0 \text { and } x-3>0 \\
& \therefore x<-5 \text { and } x>3
\end{aligned}
$
which is not possible
$\therefore$ solution set $=\Phi$
$\therefore$ solution set of the given inequation is $(-5,3)$
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Question 142 Marks
Solve the inequations: $5 x+7>4-2 x$
Answer
$
5 x+7>4-2 x
$
Adding $2 x$ on both sides, we get
$
7 x+7>4
$
Subtracting 7 from both sides, we get
$
7 x>-3
$
Dividing by 7 on both sides, we get
$
\therefore x>-\frac{3}{7}
$
i.e., $x$ takes all real values greater than $-\frac{3}{7}$ $\therefore$ the solution set is $\left(-\frac{3}{7}, \infty\right)$
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