[3 marks sum] - Mathematics STD 10 Questions - Vidyadip

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[3 marks sum]

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18 questions · timed · auto-graded

Question 13 Marks

Solve the equation and represent the solution set on the number line.
$-3+x \leq \frac{8 x}{3}+2 \leq \frac{14}{3}+2 x$, where $x \in I$

Answer

$-3+x \leq \frac{8 x}{3}+2 \leq \frac{14}{3}+2 x$, where $x \in I$,
$-3+x \leq \frac{8 x}{3}+2, \quad \frac{8 x}{3}+2 \leq \frac{14}{3}+2 x$
or $\frac{8 x}{3}-x \geq-3-2, \frac{8 x}{3}-2 x \leq \frac{14}{3}-2$
or $\frac{5 x}{3} \geq-5, \quad \frac{2 x}{3} \leq \frac{8}{3}$
5x ≥ - 15, 2x ≤ 8
x ≥ -3, x ≤ 4
Solution set (-3, -2, -1, 0, 1, 2, 3, 4}
Number line

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Question 23 Marks

For each inequality, determine which of the given numbers are in the solution set:
16 - 5 x ≤ - 4; 4, -3, 10.

Answer

If x = 4, then 16 - 5x = 16 - 5 x 4 = -4
Since, -4 ≤ - 4 is true.
So, x = 4 is in the solution of 16 - 5x - 4
If x = -3, then 16 - 5x = 16 - 5 x -3 = 31
Since, 31 ≤ - 4 is false.
So, x = -3 is not in the solution of 16 - 5x ≤ - 4
If x = 10, then 16 - 5x = 16 - 5 x 10 = -34
Since, -34 ≤ -4 is true.
So, x = 10 is in the solution of 16 - 5x ≤ -4.

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Question 33 Marks

For each inequality, determine which of the given numbers are in the solution set:
2x + 3 >11; -3, 4, 5, 7

Answer

If x = -3
Then 2x + 3 = 2 x (-3) + 3 = -3
Since, -3 > 11 is false.
So -3 is not in the solution of 2x + 3 > 11
If, x = 4, en 2x + 3 = 2 x 4 + 3 = 11
since 11 > 11 is false.
So 4 is not in the solution of 2x + 3 > 11
If x = 5, en 2x + 3 = 2 x 5 + 3 = 13
Since,13 > 11 is true ;
So, 5 is in the solution of 2x + 3 >11
Similarly, x = 7 is in the solution of 2x 3 > 11.

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Question 43 Marks

Given that x ∈ R, solve the following inequality and graph the solution on the number line:
-1 ≤ 3 + 4x < 23

Answer

Given : - 1 ≤ 3 + 4x < 23; x ∈ R
⇒ - 1 ≤ 3 + 4x and 3 + 4x < 23
⇒ - 4 ≤ 4x and 4x < 20
$\Rightarrow-\frac{4}{4} \leq x$ and $x<\frac{20}{4}$
⇒ - 1 ≤ x < 5
Solution Set = {x : - 1 ≤ x < 5; x ∈ R}
Solution on the number line is -

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Question 53 Marks

Solve the following inequalities and graph their solution set :
$\frac{x+8}{x+1}>1$.

Answer

The inequality $\frac{x+8}{x+1}$ is equivalent to
$\frac{x+8}{x+1}-1>0 \Leftrightarrow \frac{x+8-x-1}{x+1}>0 $
$ \Rightarrow \frac{7}{x+1}$
$= 0$
$\text { but } \frac{a}{b}>0, a>0 \Leftrightarrow b>$
Thus, $\frac{7}{x+1}>0,7>0$
$\Rightarrow x + 1 > 0$ or $x > -1$
The graph of this set is

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Question 63 Marks

Solve the following inequalities and graph their solution set :
$\frac{2 x-5}{x+2}<2$

Answer

The inequality $\frac{2 x-5}{x+2}<2$ is
eqivalent to $\frac{2 x-5}{x+2}-2<0 \Leftrightarrow \frac{2 x-5-2 x-4}{x+2}<0$
$\Rightarrow \frac{-9}{x+2}<0 $
$ \text { But } \frac{a}{b}<0, a<0$
$\Rightarrow b > 0$
Thus, $\frac{-9}{x+2}<0,-9<0$
$\Rightarrow x > -2 > 0$
$\Rightarrow x > -2$
The graph of this solution is $x > -2.$

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Question 73 Marks

Find the solution set of the following inequalities and draw the graph of their solutions sets :
$\left|\frac{x-5}{3}\right|<6$

Answer

We have $\left|\frac{x-5}{3}\right|<6$
Using prop. $| x | < a \Rightarrow - a < x < a$
$\left|\frac{x-5}{3}\right|<6$
$\Rightarrow-6<\frac{x-5}{3}<6$
$\Rightarrow -18< x -5 <18$
$\Rightarrow 5 - 18< x <18 + 5$
$\Rightarrow -13< x <23$
So $\left|\frac{x-5}{3}\right|<6$
$\Rightarrow \{x : - 13< x <23\}$
The graph of this set is

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Question 83 Marks

Find the solution set of the following inequalities and draw the graph of their solutions sets :
| x - 1 | > 3

Answer

We have
| x - 1 | > 3
Using prop. | x | ≥ a ⇔ ≥ a or x ≤ - a
Then | x - 1 | > 3 ⇔ x - 1 > 3 or x - 1 < (-3)
⇒ x > 4 or x < - 2
so | x - 1 | > 3 {x : x < -2 or x > 4}
The graph of this set is

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Question 93 Marks

Find the solution set of the following inequalities and draw the graph of their solutions sets :
$|x+5|<8$.

Answer

We have
$|x+5|<8$
Using prop. | x | < a ⇔ - a < x < a
⇒ -8 < x + 5 < 8
⇒ -5 -8 < x < 8 - 5
⇒ -13 < x < 3
The graph of this set is

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Question 103 Marks

Solve the following inequalities in the given universal set :
5x - 3 < 6x - 2; x ∈ N

Answer

We have
5x - 3 < 6x - 2; x ∈ N
⇒ 5x - 6x < -2 + -3
⇒ -x < 1
⇒ x > -1
As x ∈ N, so x be the set of all natural numbers.
The given set can be represent on number line as x = N.

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Question 113 Marks

Solve the following inequalities in the given universal set :
4x + 2 ≤ 2x - 7; x ∈ I

Answer

We have
4x + 2 ≤ 2x - 7; x ∈ I
⇒ 4x - 2x ≤ -7 -2
⇒ 2x ≤ -9
⇒ x ≤ -9/2
As x ∈ I, x can take values -5, -6, -7, ......,
so x = {-5, -6, -7, -8, .......,}
This set can be drawn on number line as x ≤ -9/2.

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Question 123 Marks

Solve the following inequalities in the given universal set :
3x - 5 > x + 7: x ∈ N

Answer

We have
3x - 5 > x + 7
⇒ 3x - x > 7 + 5
⇒ 2x > 12
⇒ x > 6
As x ∈ N, x can take values 7, 8, 9, 10, ......
This set is drawn on the number line as follow.

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Question 133 Marks

Solve the following inequalities and represent the solution set on a number line:
$3>\frac{2(3-4 x)}{7} \geq-2$.

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Question 143 Marks

Solve the following inequalities and represent the solution set on a number line:
$0<\frac{3 x-2}{4} \leq 2$

Answer

The given inequality is$0<\frac{3 x-2}{4} \leq 2$
Which is equivalent to
0 < 3x - 2 ≤ 8
⇒ 2 < 3x ≤ 8 + 2
⇒ 2 < 3x ≤ 10
⇒ 2/3 < x ≤ 10/3
The graph of this set is 2/3 < x ≤ 10/3.

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Question 153 Marks

Solve the following inequalities and represent the solution on a number line:
$\frac{3 x}{2}+\frac{1}{4}>\frac{5 x}{8}-\frac{1}{2}$

Answer

The given inequality is
$\frac{3 x}{2}+\frac{1}{4}>\frac{5 x}{8}-\frac{1}{2} $
$ \Rightarrow \frac{3 x}{2}-\frac{5 x}{8}>-\frac{1}{2}-\frac{1}{4}$
$\Rightarrow \frac{12 x-5 x}{8}>\frac{-2-1}{4} $
$ \Rightarrow \frac{7 x}{8}>\frac{-3}{4}$
$\Rightarrow 4(7x) > -3 x 8$
$\Rightarrow 28x > -24$
$\Rightarrow x>\frac{-24}{28}$
$\Rightarrow x > -6/7$
The graph of the solution set is $x > -6/7.$

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Question 163 Marks

Solve the following inequalities and represent the solution on a number line:
$\frac{2 x+5}{4}>\frac{4-3 x}{6}$

Answer

The given inequality is

$\frac{2 x+5}{4}>\frac{4-3 x}{6}$
⇒ 6(2x + 5) > 4(4 - 3x)
⇒ 12x + 30 > 16 - 12x
⇒ 12x + 12x > 16 - 30
⇒ 24x > -14
$\Rightarrow x>-\frac{14}{24}$
x > -7/12.
The graph of solution is x > -7/12

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Question 173 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

The given inequation has two parts:
$-2 \frac{2}{3} \leq x+\frac{1}{3} \text { and } x+\frac{1}{3}<3 \frac{1}{3} $
$ -\frac{8}{3} \leq x+\frac{1}{3} \text { and } x+\frac{1}{3}<\frac{10}{3}$
$-\frac{8}{3}-\frac{1}{3} \leq x \text { and } x<\frac{10}{3}-\frac{1}{3} $
$ -\frac{9}{3} \leq x \text { and } x \leq \frac{9}{3}$
$-3 \leq x$ and $x < 3$
$-3 \leq x < 3$
$\therefore$ The required graph line is:

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Question 183 Marks

Solve the following inequation and represent the solution set on the number line :
$4 x-19<\frac{3 x}{5}-2 \leq \frac{-2}{5}+x, x \in R$

Answer

$4 x-19<\frac{3 x}{5}-2 \leq \frac{-2}{5}+x, x \in R $
$ \therefore 4 x-19<\frac{3 x}{5}-2$
$4 x-\frac{3 x}{5}<-2+19$
$ \frac{17 x}{5}<17$
$x < 5,$
and $\frac{3 x}{5}-2 \leq \frac{-2}{5}+x$
$\frac{3 x}{5}-x \leq \frac{-2}{5}+2$
$-2x \leq 8$
$x \geq - 4$
$\Rightarrow - 4 ≤ x ≤ 5$

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[3 marks sum] - Mathematics STD 10 Questions - Vidyadip