[1 Mark Question Answer]

Question 11 Mark

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.

Answer

x – 3 < 2x – 1
x – 2x < – 1 + 3
⇒ – x < 2 x > – 2
But x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4, 5, 6, 7, 9}

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Question 21 Mark

Solve : 2 (x – 2) < 3x – 2, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3} .

Answer

2(x – 2) < 3x – 2
⇒ 2x – 4 < 3x – 2
⇒ 2x – 3x < – 2 + 4
⇒ – x < 2
⇒ x > – 2
Solution set = { – 1, 0, 1, 2, 3}.

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Question 31 Mark

Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.

Answer

Let the greatest integer = x
According to the condition,
2x + 7 > 3x
⇒ 2x – 3x > – 7
⇒ – x > – 7
⇒ x < 7
Value of x which is greatest = 6.

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Question 41 Mark

List the solution set of 30 – 4 (2.x – 1) < 30, given that x is a positive integer.

Answer

$
\begin{aligned}
& 30-4(2 x-1)<30 \\
& 30-8 x+4<30 \\
& -8 x<30-30-4 \\
& -8 x<-4 x>\frac{-4}{-8} \\
& \Rightarrow x>\frac{1}{2}
\end{aligned}
$
$x$ is a positive integer
$
x=\{1,2,3,4 \ldots . . .\} \text {. }
$

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Question 51 Mark

Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when
(i) x ∈ I
(ii) x ∈ W
(iii) x ∈ N.

Answer

20 – 5 x < 5 (x + 8)
⇒ 20 – 5x < 5x + 40
⇒ – 5x – 5x < 40 – 20
⇒ – 10x < 20
⇒ – x < 2
⇒ x > – 2
(i) When x ∈ I, then smallest value = – 1.
(ii) When x ∈ W, then smallest value = 0.
(iii) When x ∈ N, then smallest value = 1.

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Question 61 Mark

If $x \in I$, find the smallest value of $x$ which satisfies the inequation $2 x+\frac{5}{2}>\frac{5 x}{3}+2$

Answer

$
\begin{aligned}
& 2 x+\frac{5}{2}>\frac{5 x}{3}+2 \\
& \Rightarrow 2 x-\frac{5 x}{3}>2-\frac{5}{2} \\
& \Rightarrow 12 x-10 x>12-15 \\
& \Rightarrow 2 x>-3 \\
& \Rightarrow x>-\frac{3}{2}
\end{aligned}
$
Smallest value of $x=-1$.

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Question 71 Mark

If $x \in W$, find the solution set of $\frac{3}{5} x-\frac{2 x-1}{1}>1$ Also graph the solution set on the number line, if possible.

Answer

$
\begin{aligned}
& \frac{3}{5} x-\frac{2 x-1}{1}>1 \\
& 9 x-(10 x-5)>15 \ldots(\text { L.C.M. of } 5,3=15) \\
& \Rightarrow 9 x-10 x+5>15 \\
& \Rightarrow-x>15-5 \\
& \Rightarrow-x>10 \\
& \Rightarrow x<-10
\end{aligned}
$
But $x \in W$
Solution set $=\Phi$
Hence it can't be represented on number line.

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Question 81 Mark

List the solution set of the inequation $\frac{1}{2}+8 x>5 x-\frac{3}{2}, x \in Z$

Answer

$\begin{aligned} & \frac{1}{2}+8 x>5 x-\frac{3}{2} \\ & \Rightarrow 8 x-5 x>-\frac{3}{2}-\frac{1}{2} \\ & \Rightarrow 3 x>-2 \\ & \Rightarrow x>-\frac{2}{3} \\ & \because x \in Z \\ & \therefore \text { Solution set }=\{0,1,2,3,4, \ldots \ldots .\} .\end{aligned}$

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Question 91 Mark

Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.

Answer

Let the integer be x.
According to the given conditions,
⇒ 5x - 6 < 4x
⇒ 5x - 4x < 6
⇒ x < 6
Since x is positive, x can only have the values 1, 2, 3, 4, 5.
Solution set = {x : x < 6} = { 1, 2, 3, 4, 5}.

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Question 101 Mark

Solve the inequations : 6x – 5 < 3x + 4, x ∈ I.

Answer

6x – 5 < 3x + 4
6x – 3x < 4 + 5
⇒ 3x <9
⇒ x < 3
x ∈ I
Solution Set = { -1, -2, 2, 1, 0….. }.

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