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

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[3 marks sum]

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15 questions · timed · auto-graded

Question 13 Marks

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

Answer

Given: $-3+x \leq \frac{8 x}{3}+2 \leq \frac{14}{3}+2 x$, Where $x \in I$
$
\begin{aligned}
& \text { (i) }-3+x \leq \frac{8 x}{3}+2 \\
& -3-2 \leq \frac{8 x}{3}-x \\
& \Rightarrow-5 \leq \frac{5 x}{3} \\
& \Rightarrow-1 \leq \frac{x}{3} \\
& -3 \leq x
\end{aligned}
$
and
$
\begin{aligned}
& \frac{8 x}{3}-2 x \leq \frac{14}{3} \\
& \Rightarrow \frac{2 x}{3} \leq \frac{8}{3} \\
& \Rightarrow x \leq 4
\end{aligned}
$From (i) and (ii),
$
\begin{aligned}
& \Rightarrow-5 \leq \frac{5 x}{3} \text { and } \frac{2 x}{3} \leq \frac{8}{3} \\
& \Rightarrow x \geq-3 \text { and } x \leq 4 \\
& \therefore-3 \leq x \leq 4
\end{aligned}
$
Solution set $=\{-3,-2,-1,0,1,2,3,4\}$
Solution set on number line

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Question 23 Marks

Solve the given inequation and graph the solution on the number line : 2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.

Answer

2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.
(a) 2y – 3 < y + 1
⇒ 2y – y < 1 + 3
⇒ y < 4
⇒ 4 > y ….(i)
(b) y + 1 ≤ 4y + 7
⇒ y – 4y ≤ 7 – 1
⇒ –3y ≤ 6
⇒ 3y ≥ 6
$\Rightarrow y \geq \frac{6}{-3}$
⇒ y ≥ –2 ....(ii)
From (i) and (ii),
4 > y ≥ –2 or –2 ≤ y < 4
Now representing it on a number given below

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Question 33 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
$
Hence, solution set is {x : -4 < x < 5, x ∈ R}
The solution set is represented on the number line as below.
$
\begin{aligned}
& \Rightarrow 4 x-19<\frac{3 x}{5}-2 \text { and } \frac{3 x}{5}-2 \leq \frac{-2}{5}+x, x \in R \\
& \Rightarrow 4 x-\frac{3 x}{5}<17 \text { and } 2+\frac{2}{5} \leq x-\frac{3 x}{5}, x \in R \\
& \Rightarrow \frac{17 x}{5}<17 \text { and } \frac{-8}{5}<\frac{2 x}{5}, x \in R \\
& \Rightarrow x <5 \text { and }-4 \leq x , x \in R \\
& \Rightarrow-4 \leq x <5, x \in R
\end{aligned}
$
Hence, solution set is $\{x:-4 \leq x<5, x \in R\}$
The solution set is represented on the number line as below.

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Question 43 Marks

A = {x : 11x – 5 > 7x + 3, x ∈R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}. Find the range of set A ∩ B and represent it on a number line

Answer

$
\begin{aligned}
& A=\{x: 11 x-5>7 x+3, x \in R\} \\
& B=\{x: 18 x-9 \geq 15+12 x, x \in R\}
\end{aligned}
$
Now, $A=11 x-5>7 x+3$
$
\begin{aligned}
& \Rightarrow 11 x-7 x>3+5 \\
& \Rightarrow 4 x>8 \\
& \Rightarrow x>2, x \in R \\
& B=18 x-9 \geq 15+12 x \\
& \Rightarrow 18 x-12 x \geq 15+9 \\
& \Rightarrow 6 x \geq 24 \\
& \Rightarrow x \geq 4 \\
& \therefore A \cap B=x \geq 4, x \in R
\end{aligned}
$
Hence Range of $A \cap B=\{x: x \geq 4, x \in R\}$ and its graph will be.

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Question 53 Marks

Solve the inequation : $-2 \frac{1}{2}+2 x \leq \frac{4 x}{3} \leq \frac{4}{3}+2 x, x \in W$. Graph the solution set on the number line.

Answer

$
\begin{aligned}
& -2 \frac{1}{2}+2 x \leq \frac{4 x}{3} \leq \frac{4}{3}+2 x, x \in W \\
& -\frac{5}{2}+2 x \leq \frac{4 x}{3} \leq \frac{4}{3}+2 x \\
& -\frac{5}{2}+2 x \leq \frac{4 x}{3} \text { and } \frac{4 x}{3} \leq \frac{4}{3}+2 x \\
& 2 x-\frac{4 x}{3} \leq \frac{5}{2} \text { and } \frac{4 x}{3}-2 x \leq \frac{4}{3} \\
& 12 x-8 x \leq 15 \text { and } 4 x -6 x \leq 4 \\
& 4 x \leq 15 \text { and }-2 x \leq 4 \\
& x \leq \frac{15}{4} \text { and }-x \leq 4 \\
& x \leq \frac{15}{4} \text { and } x \geq 4 \\
& x \leq \frac{15}{4} \text { and }-4 \leq x \\
& \therefore-2 \leq x \leq \frac{15}{4} \\
& \therefore x =0,1,2,3
\end{aligned}
$
Solution set $\{x: x \in W, x \leq 3\}$
Solution set on number line

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Question 63 Marks

Solving the following inequation, write the solution set and represent it on the number line. $-3(x-$ 7) $\geq 15-7 x >\frac{x+1}{3}, n \in R$

Answer

$
\begin{aligned}
& -3(x-7) \geq 15-7 x>\frac{x+1}{3}, n \in R \\
& -3(x-7) \geq 15-7 x \Rightarrow 3 x+21 \geq 15-7 x \\
& -3 x+7 x \geq 15-21 \Rightarrow 4 x \geq-6 \\
& \Rightarrow x \geq \frac{-6}{4} \\
& \Rightarrow x \geq \frac{-3}{2} \\
& \Rightarrow \frac{-3}{2} \leq x
\end{aligned}
$
and
$
\begin{aligned}
& 15-7 x>\frac{x+1}{3} \\
& \Rightarrow 45-21 x>x+1 \\
& \Rightarrow 45-1>x+21 x \\
& \Rightarrow 44>22 x \\
& 2>x \Rightarrow x=2 \\
& \therefore \frac{-3}{2} \leq x<2, x \in R
\end{aligned}
$

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Question 73 Marks

Solve $\frac{2 x+1}{2}+2(3-x) \geq 7, x \in R$. Also graph the solution set on the number line

Answer

$
\begin{aligned}
& \frac{2 x+1}{2}+2(3-x) \geq 7, x \in R \\
& \Rightarrow \frac{2 x+1}{2}+6-2 x \geq 7 \\
& \Rightarrow \frac{2 x+1}{2}-2 x \geq 7-6 \\
& \Rightarrow \frac{2 x+1-4 x}{2} \geq 1 \\
& \Rightarrow 2 x+1-4 x \geq 2 \\
& \Rightarrow-2 x \geq 2-1 \\
& \Rightarrow-2 x \geq 1 \\
& \Rightarrow-x \geq \frac{1}{2} \\
& \Rightarrow x \leq-\frac{1}{2} \\
& \therefore \text { Solution set }\left\{x: x \in R , x \leq-\frac{1}{2}\right\}
\end{aligned}
$
Solution on number line:

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Question 83 Marks

Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line.

Answer

$
\begin{aligned}
& 2 \leq 2 x-3 \leq 5 \\
& 2 \leq 2 x-3 \text { and } 2 x-3 \leq 5 \\
& 2+3 \leq 2 x \text { and } 2 x \leq 5+3 \\
& 5 \leq 2 x \text { and } 2 x \leq 8 \\
& \therefore \frac{5}{2} \leq x \text { and } x \leq 4 \\
& \therefore \frac{5}{2} \leq x \leq 4 \\
& \therefore \text { Solution set }=\left\{x: x \in R, \frac{5}{2} \leq x \leq 4\right\}
\end{aligned}
$
Solution set on number line

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Question 93 Marks

Solve the inequation $=12+1 \frac{5}{6} \leq 5+3 x, x \in R$. Represent the solution on a number line.

Answer

$
\begin{aligned}
& 12+1 \frac{5}{6} x \leq 5+3 x \\
& \Rightarrow 12+\frac{11}{6} x \leq 5+3 x \\
& \Rightarrow 72+11 x \leq 30+18 x \quad \ldots(\text { Multiplying by } 6 \text { ) } \\
& \Rightarrow 11 x-18 x \leq 30-72 \\
& \Rightarrow-7 x \leq-42 \\
& \Rightarrow-x \leq-\frac{42}{7} \\
& \Rightarrow-x \leq-6 \\
& \Rightarrow x \geq 6 \\
& \therefore x \in R
\end{aligned}
$
$\therefore$ Solution set $=\{x: x \in R, x \geq 6\}$
Solution set on Number line

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Question 103 Marks

Given that $x \in I$, solve the inequation and graph the solution on the number line:
$
3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2
$

Answer

$
\begin{aligned}
& 3 \geq \frac{x-4}{2}+\frac{x}{3} \text { and } 3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2 \\
& \text { (i) } 3 \geq \frac{3 x-12+2 x}{6} \\
& \Rightarrow 3 \geq \frac{5 x-12}{6} \\
& \Rightarrow 18 \geq 5 x-12 \\
& \Rightarrow 5 x-12 \leq 18 \\
& \Rightarrow 5 x \leq 18+12 \\
& \Rightarrow 5 x \leq 30 \\
& \Rightarrow x \leq 6
\end{aligned}
$
(ii) $\frac{x-4}{2}+\frac{x}{3} \geq 2$
$
\begin{aligned}
& \frac{3 x-12+2 x}{6} \geq 2 \\
& \Rightarrow \frac{5 x-12}{6} \geq 2 \\
& \Rightarrow 5 x-12 \geq 12 \\
& \Rightarrow 5 x \geq 12+12, x \geq \frac{24}{5} \\
& \Rightarrow x \geq 4 \frac{4}{5} \\
& \therefore x=\{5,6\}
\end{aligned}
$
Number line:

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Question 113 Marks

Find the values of $x$, which satisfy the inequation : $-2 \leq \frac{1}{2}-\frac{2 x}{3} \leq 1 \frac{5}{6}, x \in N$. Graph the solution set on the number line.

Answer

$
\begin{aligned}
& -2 \leq \frac{1}{2}-\frac{2 x}{3} \leq 1 \frac{5}{6}, x \in N \\
& \Rightarrow-2-\frac{1}{2} \leq \frac{1}{2}-\frac{2 x}{3}-\frac{1}{2} \leq \frac{11}{6}-\frac{1}{2}
\end{aligned}
$
[By subtracting $\frac{1}{2}$ on both sides of inequality]
$
\begin{aligned}
& \Rightarrow-\frac{5}{2} \leq \frac{2 x}{3} \leq \frac{8}{6} \\
& \Rightarrow-15 \leq-4 x \leq 8 \\
& \Rightarrow 15 \geq 4 x \geq-8 \\
& \Rightarrow \frac{15}{4} \geq x \geq-2 \\
& 3 \frac{3}{4} \geq x \geq-2
\end{aligned}
$
But $x \in N$, hence only possible solution for $x=\{1,2,3\}$

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Question 123 Marks

Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3

Answer

Let first least natural number $=x$
then second number $=x+1$
and third number $=x+2$
According to the condition $\frac{1}{3}(x+2)-\frac{1}{5}(x) \geq 3$
$5 x+10-3 x \geq 45$...(Multiplying by 15 the L.C.M. of 3 and 5 )
$2 x \geq 45-10 \Rightarrow 2 x \geq 35$
$
x \geq \frac{35}{12} \Rightarrow x \geq 17 \frac{1}{2}
$
$\because x$ is a natural least number
$
\therefore x =18
$
$\therefore$ First least natural number $=18$
Second number $=18+1=19$
and third number $h=18+2=20$
Hence least natural numbers are $18,19,20$.

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Question 133 Marks

If $x \in R$, solve $2 x-3 \geq x+\frac{1-x}{3}>\frac{2}{5} x$

Answer

$
\begin{aligned}
& 2 x-3 \geq x+\frac{1-x}{3}>\frac{2}{5} x \\
& 2 x-3 \geq x+\frac{1-x}{3} \text { and } x+\frac{1-x}{3}>\frac{2}{5} x \\
& 2 x-3 \geq \frac{3 x+1-x}{3} \text { and } \frac{3 x+1-x}{3}>\frac{2}{5} x \\
& 6 x -9 \geq 3 x +1- x \text { and } 15 x +5-5 x >6 x \\
& 6 x -3 x + x \geq 1+9 \text { and } 15 x -6 x -5 x >-5 \\
& 4 x \geq 10 \text { and } 4 x >-5 \\
& x \geq \frac{10}{4} \text { and } x>-\frac{5}{4} \\
& x \geq \frac{5}{2} \\
& \therefore x \geq \frac{5}{2} \\
& \because x \in R \\
& \therefore \text { Solution set }=\left\{x: x \in R , x \geq \frac{5}{2}\right\}
\end{aligned}
$
Solution set on number line

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Question 143 Marks

Find the range of values of a, which satisfy $7 \leq – 4x + 2 < 12, x \in R.$ Graph these values of a on the real number line.

Answer

$7 \leq – 4x + 2 < 12$
$7 < – 4x + 2$ and $– 4x + 2 < 12$
$4x \leq 2 - 7$ and $-4x < 12 - 2$
$4x \leq -5$ and $-4x < 10$
$x \leq \frac{-5}{4} \text { and }-x<\frac{10}{4}$
$x \leq \frac{-5}{4} \text { and }-x<\frac{5}{4}$
or
$x>-\frac{5}{2}$
$\because x \in R$
$\therefore \text { Solution set }-\frac{5}{2}$
Solution set on number line

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Question 153 Marks

Solve the inequation : $\frac{5 x+1}{7}-4\left(\frac{x}{7}+\frac{2}{5}\right) \leq 1 \frac{3}{5}+\frac{3 x-1}{7}, x \in R$

Answer

$
\begin{aligned}
& \frac{5 x+1}{7}-4\left(\frac{x}{7}+\frac{2}{5}\right) \leq 1 \frac{3}{5}+\frac{3 x-1}{7} \\
& \frac{5 x+1}{7}-4\left(\frac{x}{7}+\frac{2}{5}\right) \leq \frac{8}{5}+\frac{3 x-1}{7}
\end{aligned}
$
Multiplying by L.C.M. of 7 and 5 i.e. 35
$
\begin{aligned}
& 25 x+5-4(5 x+14) \leq 56+15 x-5 \\
& 25 x+5-20 x-56 \leq 56-5-5+56 \\
& 25 x-20 x-15 x \leq 56-5-5+56 \\
& -10 x \leq 102 \\
& -x \leq \frac{102}{10} \\
& \Rightarrow-x \leq \frac{51}{5} \\
& \Rightarrow x \geq-\frac{51}{5} \\
& \because x \in R
\end{aligned}
$
$\therefore$ Solution set $=\left\{x: x \in R , x \geq-\frac{51}{5}\right\}$

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