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Mathematics [2 Mark Question Answer]

Linear Inequations · ICSE STD 10 (ICSE - English Medium). 25 questions.

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[2 Mark Question Answer]

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25 questions · timed · auto-graded

Question 12 Marks
Solve x – 3 (2 + x) > 2 (3x – 1), x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}. Also represent its solution on the number line.
Answer
$
\begin{aligned}
& x-3(2+x)>2(3 x-1) \\
& \Rightarrow x-6-3 x>6 x-2 \\
& \Rightarrow x-3 x-6 x>-2+6 \\
& \Rightarrow-8 x>4 \\
& \Rightarrow x<\frac{-4}{8} \\
& \Rightarrow x<-\frac{1}{2} \\
& x \in\{-3,-2,-1,0,1,2\}
\end{aligned}
$
$\therefore$ Solution set $=\{-3,-2,-1\}$
Solution set on Number Line :
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Question 22 Marks
Solve: $\frac{2 x-3}{4} \geq \frac{1}{2}, x \in\{0,1,2, \ldots, 8\}$
Answer
$\begin{aligned} & \frac{2 x-3}{4} \geq \frac{1}{2} \\ & \Rightarrow 2 x-3 \geq \frac{4}{2} \\ & \Rightarrow 2 x-3 \geq 2 \\ & \Rightarrow 2 x \geq 2+3 \\ & \Rightarrow 2 x \geq 5 \\ & \Rightarrow x \geq \frac{5}{2} \\ & \because x \in\{0,1,2, \ldots \ldots . . . .8\} \\ & \therefore \text { Solution set }=\{3,4,5,6,7,8\} .\end{aligned}$
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Question 32 Marks
If $x$ is a negative integer, find the solution set of $\frac{2}{3}+\frac{1}{3}(x+1)>0$.
Answer
$
\begin{aligned}
& \frac{2}{3}+\frac{1}{3} x+\frac{1}{3}>0 \\
& \Rightarrow \frac{1}{3} x+1>0 \\
& \Rightarrow \frac{1}{3} x>-1 \\
& \Rightarrow x>-1 \times \frac{3}{1} \\
& \Rightarrow x>-3
\end{aligned}
$
$x$ is a negative integer
Solution set $=\{-2,-1\}$.
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Question 42 Marks
One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.
Answer
Let the length of the shortest pole $=x$ metre
Length of pole which is buried in mud $=\frac{x}{3}$
Length of pole which is in the water $=\frac{x}{6}$
According to this problem,
$
\begin{aligned}
& x-\left[\frac{x}{3}+\frac{x}{6}\right] \geq 3 \\
& \Rightarrow x-\left(\frac{2 x+x}{6}\right) \geq 3 \\
& \Rightarrow x-\frac{x}{2} \geq 3 \\
& \Rightarrow \frac{x}{2} \geq 3 \\
& \Rightarrow x \geq 6
\end{aligned}
$
$\therefore$ Length of pole (shortest in length) $=6$ meters.
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Question 52 Marks
If P is the solution set of – 3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find
(i) P ∩ Q
(ii) Q – P.
Answer
$
\begin{aligned}
& \text { (i) }-3 x+4<2 x-3 \\
& -3 x-2 x<-3-4 \Rightarrow-5 x<-7 \\
& -x<-\frac{7}{5} \Rightarrow x>\frac{7}{5}
\end{aligned}
$
$\therefore$ Solution set $P =\{2,3,4,5, \ldots .$.
$
\begin{aligned}
& \text { (ii) } 4 x-5<12 \\
& 4 x<12+5 \Rightarrow 4 x<17 \\
& x<\frac{17}{4}
\end{aligned}
$
$\therefore$ Solution set $Q=\{4,3,2,1,0\}$
(i) $P \cap Q=\{2,3,4\}$
(ii) $Q-P=\{1,0\}$.
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Question 62 Marks
If x ∈ I, A is the solution set of 2 (x – 1) < 3 x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩ B.
Answer
$ \begin{aligned} & 2(x-1)<3 x-1 \\ & 2 x-2<3 x-1 \\ & 2 x-3 x<-1+2 \\ & \Rightarrow-x<1 x>-1 \end{aligned} $
Solution set $A=\{0,1,2,3, \ldots$.
$ \begin{aligned} & 4 x-3 \leq 8+x \\ & 4 x-x \leq 8+3 \\ & \Rightarrow 3 x \leq 11 \\ & \Rightarrow x \leq \frac{11}{3} \end{aligned} $
Solution set $B=\{3,2,1,0,-1 \ldots\}$
$ A \cap B=\{0,1,2,3\} . $
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Question 72 Marks
Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line.
Answer
$\begin{aligned} & 2 x-5 \leq 5 x+4<112 x-5 \leq 5 x+4 \\ & \Rightarrow 2 x-5-4 \leq 5 x \text { and } 5 x+4<11 \\ & \Rightarrow 2 x-9 \leq 5 x \text { and } 5 x<11-4 \\ & \text { and } 5 x<7 \\ & \Rightarrow 2 x-5 x \leq 9 \text { and } x<\frac{7}{5} \\ & \Rightarrow 3 x>-9 \text { and } x<1.4 \\ & \Rightarrow x>-3\end{aligned}$
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Question 82 Marks
Solve : 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.
Answer
$
\begin{aligned}
& 5-4 x>2-3 x \\
& -4 x+3 x>2-5 \\
& \Rightarrow-x>-3 \\
& \Rightarrow x<3 \\
& x \in w
\end{aligned}
$
solution set $\{0,1,2\}$
Solution set on Number Line:
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Question 92 Marks
Solve the following inequation and graph the solution on the number line. $-2 \frac{2}{3} \leq x+\frac{1}{3}<3+\frac{1}{3}$ $x \in R$
Answer
$
\begin{aligned}
& \text { Given }-2 \frac{2}{3} \leq x+\frac{1}{3}<3+\frac{1}{3} x \in R \\
& -\frac{8}{3} \leq x+\frac{1}{3}<\frac{10}{3}
\end{aligned}
$
Multiplying by 3, L.C.M. of fractions, we get
$
\begin{aligned}
& -8 \leq 3 x+1<10 \\
& -8-1 \leq 3 x+1-1<10-1 \ldots \text {... [Add - 1] } \\
& -9 \leq 3 x<9 \\
& -3 \leq x<3 \quad \ldots[\text {Dividing by } 3]
\end{aligned}
$
Hence the solution set is $\{x: x \in R,-3 \leq x<3\}$

The graph of the solution set is shown by the thick portion of the number line. The solid circle at -3 indicates that the number -3 is included among the solutions where as the open circle at 3 indicates that 3 is not included among the solutions.
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Question 102 Marks
Given that x ∈ R, solve the following inequation and graph the solution on the number line: – 1 ≤ 3 + 4x < 23.
Answer
We have
$
\begin{aligned}
& -1 \leq 3+4 x<23 \\
& \Rightarrow-1-3 \leq 4 x<23-3 \\
& \Rightarrow-4 \leq 4 x<20 \\
& \Rightarrow-1 \leq x<5, x \in R
\end{aligned}
$
Solution Set $=\{-1 \leq x<5 ; x \in R\}$
The graph of the solution set is shown below:
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Question 112 Marks
Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line.
Answer
-3 ≤ 3 – 2x < 9
– 3 ≤ 3 – 2x and 3 – 2x < 9
2x ≤ 3 + 3 and – 2x < 9 – 3
2x ≤ 6 and – 2x < 6
⇒ x ≤ 3 and – x < 3
⇒ x ≤ – 3 and – 3 < x
– 3 < x ≤ 3.
Solution set= {x : x ∈ R, – 3 < x ≤ 3)
Solution on number line
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Question 122 Marks
Solve $\frac{3 x}{5}-\frac{2 x-1}{3}>1, x \in R$ and represent the solution set on the number line.
Answer
$
\begin{aligned}
& \frac{3 x}{5}-\frac{2 x-1}{3}>1 \\
& \Rightarrow 9 x-(10 x-5)>15 \\
& \Rightarrow 9 x-10 x+5>15 \\
& \Rightarrow-x>15-5 \\
& \Rightarrow-x>10 \\
& \Rightarrow x<-10 \\
& x \in R .
\end{aligned}
$
$\therefore$ Solution set $=\{x: x \in R, x<-10\}$
Solution set on the number line
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Question 132 Marks
Solve : $\frac{4 x-10}{3} \leq \frac{5 x-7}{2} x \in R$ and represent the solution set on the number line.
Answer
$
\begin{aligned}
& \frac{4 x-10}{3} \leq \frac{5 x-7}{2} \\
& \Rightarrow 8 x-20 \leq 15 x-21 \quad \ldots(\text { L.C.M. of } 3,2=6) \\
& \Rightarrow 8 x-15 x \leq-21+20 \\
& \Rightarrow-7 x \leq-1 \\
& \Rightarrow-x \leq-\frac{1}{7} \\
& \Rightarrow x>\frac{1}{7} \\
& \because x \in R \\
& \therefore \text { Solution set }=\left\{x: x \in R, x>\frac{1}{7}\right\}
\end{aligned}
$
$\therefore$ Solution set on the number line
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Question 142 Marks
Solve 2(x – 3)< 1, x ∈ {1, 2, 3, …. 10}
Answer
$
\begin{aligned}
& 2(x-3)<1 \\
& \Rightarrow x-3<\frac{1}{2} \\
& \Rightarrow x<\frac{1}{2}+3 \\
& \Rightarrow x<3 \frac{1}{2}
\end{aligned}
$
But $x \in\{1,2,3 . . . .10\}$
Solution set $=\{1,2,3\}$
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Question 152 Marks
If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.
Answer
$ \begin{aligned} & 2+4 x<2 x-5 \leq 3 x \\ & 2+4 x<2 x-5 \text { and } 2 x-5 \leq 3 x=>4 x-2 x<-5-2 \text {, and } 2 x-3 x \leq 5 \\ & \Rightarrow 2 x<-7 \text { and }-x \leq 5 \\ & \Rightarrow x<-\frac{7}{2} \text { and } x \geq-5 \text { and }-5 \leq x \\ & \therefore-5 \leq x<-\frac{7}{2} \\ & \because x \in Z \end{aligned} $
$\therefore$ Solution set $=\{-5,-4\}$
Solution set on Number line
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Question 162 Marks
Solve : 1 ≥ 15 – 7x > 2x – 27, x ∈ N
Answer
$
\begin{aligned}
& 1 \geq 15-7 x>2 x-27 \\
& 1 \geq 15-7 x \text { and } 15-7 x>2 x-27 \\
& \Rightarrow 7 x \geq 15-1 \text { and }-7 x-2 x>-27-15 \\
& \Rightarrow 7 x \geq 14 \text { and }-9 x>-42 \\
& \Rightarrow x \geq 2 \text { and }-x>-\frac{42}{9} \\
& \Rightarrow 2 \leq x \text { and }-x>-\frac{14}{3} \text { and } x<\frac{14}{3} \\
& 2 \leq x<\frac{14}{3}
\end{aligned}
$
But $x \in N$
$\therefore$ Solution set $=\{2,3,4\}$.
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Question 172 Marks
Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1 < x + 4.
Answer
3 < 2x – 1 < x + 4.
⇒ – 3 < 2x – 1 and 2x – 1 < x + 4
⇒ – 2x < – 1 + 3 and 2x – x < 4 + 1
⇒ – 2x < 2 and x < 5
⇒ – x < 1
⇒ x > – 1
– 1 < x < 5
x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4}.
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Question 182 Marks
$\frac{2 x+3}{3} \geq \frac{3 x-1}{4}$ where $x$ is positive even integer.
Answer
$
\begin{aligned}
& \frac{2 x+3}{3} \geq \frac{3 x-1}{4} \\
& \Rightarrow \frac{2 x}{3}+\frac{3}{3} \geq \frac{3 x}{4}-\frac{1}{4} \\
& \Rightarrow \frac{2 x}{3}-\frac{3 x}{4} \geq \frac{-1}{4}-1 \\
& \Rightarrow \frac{8 x-9 x}{12} \geq-\frac{5}{4} \\
& \Rightarrow \frac{-x}{12} \geq \frac{-5}{4} \\
& \Rightarrow \frac{x}{12} \leq \frac{5}{4} \\
& \Rightarrow x \leq \frac{5}{4} \times 12 \\
& \Rightarrow x \leq 15
\end{aligned}
$
$\because x$ is positive even integer
$
\therefore x=\{2,4,6,8,10,12,14\} .
$
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Question 192 Marks
$\frac{x}{2}+5 \leq \frac{x}{3}+6$ where $x$ is a positive odd integer.
Answer
$
\begin{aligned}
& \frac{x}{2}+5 \leq \frac{x}{3}+6 \\
& \Rightarrow \frac{x}{2}-\frac{x}{3} \leq 6-5 \\
& \Rightarrow \frac{3 x-2 x}{6} \leq 1 \\
& \Rightarrow \frac{x}{6} \leq 1 \\
& \Rightarrow x \leq 6
\end{aligned}
$
$\because x$ is a positive odd integer
$
\therefore x=\{1,3,5\} .
$
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Question 202 Marks
List the solution set of $\frac{11-2 x}{5} \geq \frac{9-3 x}{8}+\frac{3}{4}, x \in N$
Answer
$
\begin{aligned}
& \frac{11-2 x}{5} \geq \frac{9-3 x}{8}+\frac{3}{4} \\
& \Rightarrow 88-16 x \geq 45-15 x+30 \quad \ldots(\text { L.C.M. of } 8,5,4=40\} \\
& \Rightarrow-16 x+15 x \geq 45+30-88 \\
& \Rightarrow-x \geq-13 \\
& \Rightarrow x \leq 13 \\
& x \leq N
\end{aligned}
$
Solution set $=\{1,2,3,4,5, \ldots, 13\}$.
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Question 212 Marks
Given A = {x : x ∈ I, – 4 ≤ x ≤ 4}, solve 2x – 3 < 3 where x has the domain A Graph the solution set on the number line.
Answer
$
\begin{aligned}
& 2 x-3<3 \\
& \Rightarrow 2 x<3+3 \\
& \Rightarrow 2 x<6 \\
& \Rightarrow x<3
\end{aligned}
$
But $x$ has the domain $A=\{x: x \in I-4 \leq x \leq 4\}$
Solution set $=\{-4,-3,-2,-1,0,1,2\}$
Solution set on Number line:
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Question 222 Marks
Solve the inequation 3x -11 < 3 where x ∈ {1, 2, 3,……, 10}. Also represent its solution on a number line
Answer
$
\begin{aligned}
& 3 x-11<3 \\
& \Rightarrow 3 x<3+11 \\
& \Rightarrow 3 x<14 x<\frac{14}{3}
\end{aligned}
$
But $x \in 6\{1,2,3, \ldots . . ., 10\}$
Solution set is $(1,2,3,4\}$
Ans. Solution set on number line
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Question 232 Marks
If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.
Answer
$
\begin{aligned}
& -1<3-2 x \leq 7 \\
& -1<3-2 x \text { and } 3-2 x \leq 7 \\
& \Rightarrow 2 x<3+1 \text { and }-2 x \leq 7-3 \\
& \Rightarrow 2 x<4 \text { and }-2 x \leq 4 \\
& \Rightarrow x<2 \text { and }-x \leq 2 \\
& \text { and } x \geq-2 \text { or }-2 \leq x \\
& x \in R
\end{aligned}
$
Solution set $-2 \leq x<2$
Solution set on number line
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Question 242 Marks
Find the solution set of the inequation x + 5 < 2 x + 3 ; x ∈ R Graph the solution set on the number line.
Answer
$
\begin{aligned}
& x+5 \leq 2 x+3 \\
& x-2 x \leq 3-5 \\
& \Rightarrow-x \leq-2 \\
& \Rightarrow x \geq 2 \\
& \because x \in R
\end{aligned}
$
$\therefore$ Solution set $=\{2,3,4,5, \ldots \ldots\}$
Solution set on number line
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Question 252 Marks
Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.
Answer
$
\begin{aligned}
& 5 x -2<3(3- x ) \\
& \Rightarrow 5 x -2 \leq 9-3 x \\
& \Rightarrow 5 x +3 x \leq 9+2 \\
& \Rightarrow 8 x \leq 11 \\
& \Rightarrow x \leq \frac{11}{8} \\
& \because x \in\{-2,-1,0,1,2,3,4\} \\
& \therefore \text { Solution set }=\{-2,-1,0,1\}
\end{aligned}
$
Solution set on umber line
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