1 Marks Question

Question 11 Mark

Solve: 3x + 8 >2, when x is a real number.

Answer

It is given in the question that,
3x + 8 >2
$\Rightarrow$ 3x + 8 – 8 >2 – 8
$\Rightarrow$ 3x >- 6
$\Rightarrow$ $\frac{3 x}{3}>\frac{-6}{3}$
$\Rightarrow$ x > -2, where x is a real number.
It can be clearly observed that the solutions of 3x + 8 > 2 will be given by x > -2 which states that all the real numbers that are greater than -2.
$\therefore$ solution set = (-2, $\infty$)

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Question 21 Mark

Solve: 3x + 8 >2, when x is an integer.

Answer

It is given in the question that,
3x + 8 > 2
$\Rightarrow$ 3x + 8 – 8 >2 – 8
$\Rightarrow$ 3x >- 6
$\Rightarrow$ $\frac{3 x}{3}>\frac{-6}{3}$
$\Rightarrow$ x > -2, where x is an integer
It can be clearly observed that the integer number greater than -2 are -1, 0, 1, 2,...
Thus, solution of 3x + 8 > 2 is -1, 0, 1, 2,… when x is an integer.
$\therefore$ solution set = {-1, 0, 1, 2,…}

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Question 31 Mark

Solve: 5x – 3 < 7, when x is a real number.

Answer

It is given in the question that,
5x – 3 < 7
Adding 3 both side we get,
$\Rightarrow$ 5x – 3 + 3 < 7 + 3
$\Rightarrow$ 5x < 10
Dividing both sides by 5 we get,
$\Rightarrow$ $\frac{5 x}{5}<\frac{10}{5}$
$\Rightarrow$ x < 2
When x is a real number,
Solution set = (-$\infty$, 2)

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Question 41 Mark

Solve: 5x – 3 < 7, when x is an integer.

Answer

It is given in the question that,
5x – 3 < 7
$\Rightarrow$ 5x – 3 + 3 < 7 + 3
$\Rightarrow$ 5x < 10
$\Rightarrow$ $\frac{5 x}{5}<\frac{10}{5}$
$\Rightarrow$ x < 2, When x is an integer
It can be clearly observed that the integer number less than 2 are…, -2, -1, 0, 1.
Thus, solution of 5x – 3 < 7 is …,-2, -1, 0, 1, when x is an integer.
$\therefore$ solution set = {…, -2, -1, 0, 1}

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Question 51 Mark

Solve: –12x > 30, when x is an integer.

Answer

It is given in the question that,
– 12x > 30
Dividing the inequality by -12 on both sides we get,
$x<\frac{-5}{2}$
When x is an integer
It can be clearly observed that the integer number less than$\frac{-5}{2}$are…, -5, -4, -3.
Thus, solution of –12x > 30 is …, -5, -4, -3, when x is an integer.
$\therefore$ solution set = {…, -5, -4, -3}

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Question 61 Mark

Solve – 12x > 30, when x is a natural number.

Answer

It is given in the question that,
– 12x > 30
Dividing the inequality by -12 into both sides we get,
$x<\frac{-5}{2}$
When x is a natural integer.
It can be clearly observed that there is no natural number less than$\frac{-5}{2}$because $\frac{5}{-2}$ is a negative number and natural numbers are positive numbers.
$\therefore$ There would be no solution to the given inequality when x is a natural number.

solution set = $\phi$

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Question 71 Mark

Solve: 24x < 100, when x is an integer.

Answer

We have, 24x < 100
$\Rightarrow \quad \frac{24 x}{24}<\frac{100}{24}$ [dividing both sides by 24]
$\Rightarrow \quad x<\frac{25}{6}$
When x is an integer.
Hence the solution set of inequality is {...., -4, -3, -2, -1, 0, 1, 2, 3, 4}

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Question 81 Mark

Solve: 24x < 100, when x is a natural number.

Answer

We have 24x < 100
$\Rightarrow \quad \frac{24 x}{24}<\frac{100}{24}$ [dividing both sides by 24]
$\Rightarrow \quad x<\frac{25}{6}$
When x is a natural number, then solutions of the inequality are given by $x<\frac{25}{6}$ i.e., all natural numbers x which are less than $\frac{25}{6}.$
Hence, the solution set is {1, 2, 3, 4}

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Question 91 Mark

Solve 5x – 3 < 3x +1 when x is a real number.

Answer

We have, 5x –3 < 3x + 1
or 5x –3 + 3 < 3x +1 +3
$\Rightarrow$ 5x < 3x +4
$\Rightarrow$ 5x – 3x < 3x + 4 – 3x
$\Rightarrow$ 2x < 4
$\Rightarrow$ x < 2
Therefore, the solution set of the inequality is x $\in$ (– $\infty$, 2).

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Question 101 Mark

Solve 5x – 3 < 3x +1 when x is an integer.

Answer

We have, 5x –3 < 3x + 1
or 5x –3 + 3 < 3x +1 +3
or 5x < 3x +4
or 5x – 3x < 3x + 4 – 3x
or 2x < 4
or x < 2
x < 2 is a solution for x$\in$R
When x is an integer, the solutions of the given inequality are
{............... – 4, – 3, – 2, – 1, 0, 1}

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Question 111 Mark

Solve the following system of inequalities graphically

$5x + 4y \le$ 40
$x \ge 2$
$y \ge 3$

Answer

Given,
$5x + 4y \leq 40 ..... (1)$
$x \geq 2 ..... (2)$
$y \geq 3 .... (3)$
We first draw the graph of the line
$5x + 4y = 40, x = 2$ and $y = 3$
Then we note that the inequality $(1)$ represents shaded region below the line $5x + 4y = 40$ and inequality $(2)$ represents the shaded region right of line $x = 2$ but inequality $(3)$ represents the shaded region above the line $y = 3$.
Hence, the shaded region in the figure

including all the points on the lines are also the solution of the given system of the linear inequalities.
In many practical situations involving the system of inequalities the variable $x$ and $y$ often represent quantities that cannot have negative values, for example, the number of units produced, the number of articles purchased, the number of hours worked, etc.
Clearly, in such cases, $x \ge 0, y \ge$ 0 and the solution region lies only in the first quadrant.

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Question 121 Mark

Solve 30 x < 200 when x is an integer.

Answer

We are given 30 x < 200
$\Leftrightarrow$ $\frac{30 x}{30}<\frac{200}{30}$
i.e., x < 20 / 3.
When x is an integer, the solutions of the given inequality are
..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}

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Question 131 Mark

Solve 30 x < 200 when x is a natural number.

Answer

We are given 30 x < 200
or $\frac{30 x}{30}<\frac{200}{30}$
i.e., x < 20 / 3.
x < 6.66 for x $\in$R
When x $\in$ Natural number,
In this case, the following values of x make the statement true.
1, 2, 3, 4, 5, 6.
The solution set of the given linear inequality is {1, 2, 3, 4, 5, 6}.

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