(Each question 2 marks)

Question 12 Marks

Solve the inequality: $7 \leq \frac{(3 x+11)}{2} \leq 11$

Answer

We have $7 \leqslant \frac{{(3x + 11)}}{2} \leqslant 11$
$ \Rightarrow 14 \leqslant 3x + 11 \leqslant 22 \Rightarrow 3 \leqslant 3x \leqslant 11$
$ \Rightarrow 1 \leqslant x \leqslant \frac{{11}}{3}$

View full question & answer→

Question 22 Marks

Solve the inequality: $-12<4-\frac{3 x}{-5} \leq 2$

Answer

Given inequality is; $-12<4-\frac{3 x}{-5} \leq 2$
$\Rightarrow-12<4-\frac{3 x}{-5} \leq 2$
$\Rightarrow$ - 12 - 4 < 4 - $\frac{3 x}{-5}$ - 4 $\le$ 2 - 4 [subtracting by 4]
$\Rightarrow-16<\frac{-3 x}{-5} \leq-2$
$\Rightarrow-16<\frac{3 x}{5} \leq-2$
Multiplying the inequality by 5.
$\Rightarrow-16 \times 5<\frac{3 x}{5} \times 5 \leq-2 \times 5$
$\Rightarrow$ -80 < 3x $\le$ -10
$\Rightarrow-\frac{80}{3}<x \leq-\frac{10}{3}$
$\therefore$ all real numbers x greater than $-\frac{80}{3}$ but less than or equal to $-\frac{10}{3}$ are a solution of given equality.
x $\in$ ($-\frac{80}{3},-\frac{10}{3}$]

View full question & answer→

Question 32 Marks

Solve the inequality: $-15<\frac{3(x-2)}{5} \leq 0$

Answer

We have $ - 15 < \frac{{3(x - 2)}}{5} \leqslant 0$
$ \Rightarrow - 75 < 3(x - 2) \leqslant 0$$ \Rightarrow - 25 < x - 2 \leqslant 0$
$ \Rightarrow - 23 < x \leqslant 2$

View full question & answer→

Question 42 Marks

Solve the inequality: $-3 \leq 4-\frac{7 x}{2} \leq 18$

Answer

Given inequality $-3 \leq 4-\frac{7 x}{2} \leq 18$
$\Rightarrow-3 \leq 4-\frac{7 x}{2} \leq 18$
$\Rightarrow-3-4 \leq 4-\frac{7 x}{2}-4 \leq 18-4$
$\Rightarrow-7 \leq-\frac{7 x}{2} \leq 14$
Multiplying the inequality by -2.
$\Rightarrow$ (-7) $\times$ (-2) $\geq-\frac{7 x}{2} \times(-2) \geq 14 \times$ (-2)
$\Rightarrow$ 14 $\ge$ 7x $\ge$ -28
$\Rightarrow$ -28 $\le$ 7x $\le$ 14
Dividing the inequality by 7
$\Rightarrow$ -4 $\le$ x $\le$ 2
$\therefore$ all real numbers x greater than or equal to -4 but less than or equal to 2 are the solution of given equality.
x $\in$ [-4, 2] solution set =[-4,2]

View full question & answer→

Question 52 Marks

Solve the inequality: $6 \leq- 3(2x - 4) < 12$

Answer

We have $6 \leqslant - 3(2x - 4) < 12$
$ \Rightarrow 2 \geqslant (2x - 4) > - 4 \Rightarrow 2 \geqslant 2x > 0$
$ \Rightarrow 1 \geqslant x > 0 \Rightarrow 0 < x \leqslant 1$

View full question & answer→

Question 62 Marks

Solve the inequality: 2 $\le$ 3x – 4 $\le$ 5

Answer

Given inequality 2 $\le$ 3x – 4 $\le$ 5
$\Rightarrow$ 2 $\le$ 3x – 4 $\le$ 5
$\Rightarrow$ 2 + 4 $\le$ 3x – 4 + 4 $\le$ 5 + 4
$\Rightarrow$ 6 $\le$ 3x $\le$ 9
$\Rightarrow$ 6/3 $\le$ 3x/3 $\le$ 9/3
$\Rightarrow$ 2 $\le$ x $\le$ 3
$\therefore$ all real numbers x greater than or equal to 2 but less than or equal to 3 are a solution of given equality.
x $\in$ [2, 3] solution set = [2,3]

View full question & answer→

Question 72 Marks

Solve the inequality $x+\frac{x}{2}+\frac{x}{3}<11$ for real x.

Answer

Here $x + \frac{x}{2} + \frac{x}{3} < 11$
$ \Rightarrow \frac{{6x + 3x + 2x}}{6} < 11$
$ \Rightarrow \frac{{11x}}{6} < 11$
Multiplying both sides by 6, we have
11x < 66
Dividing both sides by 11, we have
x < 6
Thus the solution set is $\left( { - \infty ,6} \right)$

View full question & answer→

Question 82 Marks

Solve the inequality 3 (2 – x) $\ge$ 2 (1 – x) for real x.

Answer

Here $3(2 - x) \geqslant 2(1 - x)$
$ \Rightarrow 6 - 3x \geqslant 2 - 2x$
$ \Rightarrow - 3x + 2x \geqslant 2 - 6$
$ \Rightarrow - x \leqslant - 4$
Dividing both sides by -1, we have
$\frac{{ - x}}{{ - 1}} < \frac{{ - 4}}{{ - 1}} \Rightarrow x \leqslant 4$
Thus the solution set is $\left( { - \infty ,4} \right]$.

View full question & answer→

Question 92 Marks

Solve the inequality 3(x – 1) $\leq$ 2 (x – 3) for real x.

Answer

It is given in the question that,
3(x – 1) $\leq$ 2 (x – 3)
$\Rightarrow$ 3x – 3 $\leq$ 2x – 6
$\Rightarrow$ 3x – 3+ 3 $\leq$ 2x – 6+ 3
$\Rightarrow$ 3x $\leq$ 2x – 3
$\Rightarrow$ 3x – 2x $\leq$ 2x – 3 – 2x
$\Rightarrow$ x $\leq$ -3
$\therefore$ The solutions of the given inequality are defined by all the real numbers less than or equal to -3.
Thus, the solutions set to the given inequality are (-$\infty $, - 3]

View full question & answer→

Question 102 Marks

Solve the inequality 3x – 7 > 5x – 1 for real x.

View full question & answer→

Question 112 Marks

Solve the inequality 4x + 3 < 5x + 7 for real x.

Answer

Here 4x + 3 < 5x +7, 4x - 5x < 7 - 3
$ \Rightarrow - x < 4$
Dividing both sides by -1, we have
$\frac{{ - x}}{{ - 1}} > \frac{4}{{ - 1}}$$ \Rightarrow x > - 4$
Thus the solution set is $( - 4,\infty )$

View full question & answer→

Question 122 Marks

Solve the inequality $\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$ for real x.

Answer

Here $\frac{{(2x - 1)}}{3} \geqslant \frac{{(3x - 2)}}{4} - \frac{{(2 - x)}}{5}$
$ \Rightarrow \frac{{2x}}{3} - \frac{1}{3} \geqslant \frac{{3x}}{4} - \frac{2}{4} - \frac{2}{5} + \frac{x}{5}$
$ \Rightarrow \frac{{2x}}{3} - \frac{3x}{4}-\frac{x}{5}\geqslant \ \frac{-2}{4} - \frac{2}{5} + \frac{1}{3}$
$ \Rightarrow \frac{{40x - 45x - 12x}}{{60}} \geqslant \frac{{ - 30 - 24 + 20}}{{60}}$
$ \Rightarrow \frac{{ - 17x}}{{60}} \geqslant \frac{{ - 34}}{{60}}$
Multiplying both sides by 60, we have
$ - 17x \geqslant - 34$
Dividing both sides by -17, we have
$\frac{{ - 17x}}{{ - 17}} \leqslant \frac{{ - 34}}{{ - 17}}$
$ \Rightarrow x \leqslant 2$
Thus the solution set is $\left( { - \infty ,2} \right]$

View full question & answer→

Question 132 Marks

Solve the inequality $\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$ for real x.

Answer

Here $\frac{x}{4} < \frac{{(5x - 2)}}{3} - \frac{{(7x - 3)}}{5}$
$ \Rightarrow \frac{x}{4} < \frac{{5x}}{3} - \frac{2}{3} - \frac{{7x}}{5} + \frac{3}{5}$ $ \Rightarrow \frac{{15x - 100x + 84x}}{{60}} < \frac{{ - 10 + 9}}{{15}}$
$ \Rightarrow \frac{{ - x}}{{60}} < \frac{{ - 1}}{{15}}$
Multiplying both sides by 60, we have
-x < -4
Dividing both sides by -1, we have
x > 4
Thus the solution set is $(4,\infty )$

View full question & answer→

Question 142 Marks

Solve the inequality 37 – (3x + 5) $\ge$ 9x – 8 (x – 3) for real x.

Answer

Here $37 - (3x + 5) \geqslant 9x - 8(x - 3)$
$ \Rightarrow 37 - 3x - 5 \geqslant 9x - 8x + 24$
$ \Rightarrow 32 - 3x \geqslant x + 24$
$ \Rightarrow - 3x - x \geqslant 24 - 32$
$ \Rightarrow - 4x \geqslant - 8$
Dividing both sides by -4, we have
$\frac{{ - 4x}}{{ - 4}} \leqslant \frac{{ - 8}}{{ - 4}}$
$ \Rightarrow x \leqslant 2$
Thus the solution set is $\left( { - \infty ,2} \right]$

View full question & answer→

Question 152 Marks

Solve the inequality 2 (2x + 3) – 10 < 6 (x – 2) for real x.

Answer

Here 2(2x + 3) - 10 < 6(x - 2)
$ \Rightarrow 4x + 6 - 10 < 6x - 12$
$ \Rightarrow 4x - 4 < 6x - 12$
$ \Rightarrow 4x - 6x < - 12 + 4$
$ \Rightarrow - 2x < - 8$
Dividing both sides by -2, we have
$\frac{{ - 2x}}{{ - 2}} > \frac{{ - 8}}{{ - 2}}$
$ \Rightarrow x > 4$
Thus the solution set is $(4,\infty )$

View full question & answer→

Question 162 Marks

Solve the inequality $\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$ for real x.

Answer

Here $\frac{1}{2}\left( {\frac{{3x}}{5} + 4} \right) \geqslant \frac{1}{3}(x - 6)$
$ \Rightarrow \frac{{3x}}{{10}} + 2 \geqslant \frac{x}{3} - 2$
$ \Rightarrow \frac{{3x}}{{10}} - \frac{x}{3} \geqslant - 2 - 2$
$ \Rightarrow \frac{{9x - 10x}}{{30}} \geqslant - 4$
$ \Rightarrow \frac{{ - x}}{{30}} \geqslant - 4$
Multiplying both sides by 30, we have
$ - x \geqslant - 120$
Dividing both sides by -1, we have
$x \leqslant 120$
Thus the solution set is $\left[ { - \infty ,120} \right]$

View full question & answer→

Question 172 Marks

Solve the inequality $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$ for real x.

Answer

Here $\frac{{3(x - 2)}}{5} \leqslant \frac{{5(2 - x)}}{3}$
$ \Rightarrow \frac{{3x - 6}}{5} \leqslant \frac{{10 - 5x}}{3}$
$ \Rightarrow \frac{{3x}}{5} - \frac{6}{5} \leqslant \frac{{10}}{3} - \frac{{5x}}{3}$
$ \Rightarrow \frac{{3x}}{5} + \frac{{5x}}{3} \leqslant \frac{{10}}{3} + \frac{6}{5}$
$ \Rightarrow \frac{{9x + 25x}}{{15}} \leqslant \frac{{50 + 18}}{{15}}$
$ \Rightarrow \frac{{34x}}{{15}} \leqslant \frac{{68}}{{15}}$
Multiplying both sides by 15, we have
$34x \leqslant 68$
Dividing both sides by 34, we have
$x \leqslant 2$
Thus the solution set is $\left( { - \infty ,2} \right]$

View full question & answer→

Question 182 Marks

Solve the inequality $\frac{x}{3}>\frac{x}{2}+1$ for real x.

Answer

Here $\frac{x}{3} > \frac{x}{2} + 1$
$ \Rightarrow \frac{x}{3} - \frac{x}{2} > 1$
$ \Rightarrow \frac{{2x - 3x}}{6} > 1$
$ \Rightarrow \frac{{ - x}}{6} > 1$
Multiplying both sides by 6, we have
-x > 6
Dividing both sides by -1, we have x < -6
Thus the solution set is $( - \infty , - 6)$

View full question & answer→

Question 192 Marks

Solve 3x + 2y > 6 graphically.

Answer

Graph of 3x + 2y = 6 is given as a dotted line in the figure.

This line divides the XY-plane into two half-planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half-planes and determine if this point satisfies the given inequality or not. we note that
3 (0) + 2 (0) > 6
$\Rightarrow$ 0 > 6, which is not true.
Hence, half-plane I is not the region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality. In other words, the shaded half-plane II excluding the points on the line is the solution region of the inequality.

View full question & answer→

Question 202 Marks

Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.

Answer

Let x be the smaller of the two consecutive odd natural numbers. Then the other odd integer is x+2.
It is given that both the natural number are greater than 10 and their sum is less than 40.
$\therefore$ x > 10 and, x + x + 2 < 40
$\Rightarrow$ x > 10 and 2x < 38
$\Rightarrow$ x > 10 and x <19
$\Rightarrow$ 10 < x < 19
$\Rightarrow$ x = 11,13,15,17 [$\because$ x is an odd number]
Hence, the required pairs of odd natural number are (11, 13), (13, 15), (15, 17) and (17, 19).

View full question & answer→

Question 212 Marks

The marks obtained by a student of Class XI in first and second terminal examinations are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.

Answer

Let x be the marks obtained by student in the annual examination.
Average = $\frac{62+48+x}{3} \geq 60$
$\Rightarrow$ $\frac{62+48+x}{3} \geq 60$
$\Rightarrow$ 110 + x $\ge$ 60
$\Rightarrow$ x $\ge$ 70
Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

View full question & answer→

Question 222 Marks

Solve $\frac{3 x-4}{2} \geq \frac{x+1}{4}-1$. Show the graph of the solution on number line.

Answer

We have
$\frac{3 x-4}{2} \geq \frac{x+1}{4}-1$
$\Rightarrow$ $\frac{3 x-4}{2} \geq \frac{x-3}{4}$
$\Rightarrow$ 2 (3x – 4) $\ge$ (x – 3)
$\Rightarrow$ 6x – 8 $\ge$ x – 3
$\Rightarrow$ 5x $\ge$ 5
$\Rightarrow$ x $\ge$ 1
The graphical representation of solutions is given in the figure

View full question & answer→

Question 232 Marks

Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.

Answer

We have 7x + 3 < 5x + 9
$\Rightarrow$ 7x + 3 - 3 < 5x + 9 - 3
$\Rightarrow$ 7x < 5x + 6
$\Rightarrow$ 2x < 6
$\Rightarrow$ x < 3
The graphical representation of the solutions are given in the figure

View full question & answer→

Question 242 Marks

Solve $\frac{5-2 x}{3} \leq \frac{x}{6}-5$.

Answer

We have
$\frac{5-2 x}{3} \leq \frac{x}{6}-5$
$\Rightarrow$ 2 (5 – 2x) $\leq$ x – 30
$\Rightarrow$ 10 – 4x $\leq$ x – 30
$\Rightarrow$ – 5x $\leq$ – 40,
$\Rightarrow$ x $\ge$ 8
i.e., x $\in$ [8, $\infty$).
Hence, Solution set = [8, $\infty$).

View full question & answer→

Question 252 Marks

Solve 4x + 3 < 6x +7.

Answer

We have,
4x + 3 < 6x + 7
$\Rightarrow$ 4x + 3 - 3 < 6x + 7 - 3
$\Rightarrow$ 4x < 6x + 4
$\Rightarrow$ 4x – 6x < 6x + 4 – 6x
$\Rightarrow$ – 2x < 4 or x > – 2
i.e., all the real numbers which are greater than –2, are the solutions of the given inequality.
Hence, the solution set is (–2, $\infty$).

View full question & answer→

Question 262 Marks

Solve - 5 $\leq \frac{5-3 x}{2} \leq$ 8.

Answer

We have - 5 $\leq \frac{5-3 x}{2} \leq 8$
$\Rightarrow$ –10 $\le$ 5 – 3x $\le$ 16
$\Rightarrow$ –10 - 5 $\le$ 5 – 3x - 5 $\le$ 16 -5
$\Rightarrow$ – 15 $\le$ – 3x $\le$ 11
$\Rightarrow$ 5 $\ge$ x $\ge$ - $\frac{11}{3}$
which can be written as $\frac{-11}{3} \leq x \leq$ 5
$\Rightarrow$ x $\in$ [ - $\frac{11}{3}$, 5]

View full question & answer→

Question 272 Marks

Solve – 8 $\le$ 5x – 3 < 7.

Answer

Given that, – 8 $\le$ 5x –3 < 7
$\Rightarrow$ – 8+3 $\le$ 5x –3+3< 7+3
$\Rightarrow$ –5 $\le$ 5x < 10
$\Rightarrow$ –1 $\le$ x < 2
$\Rightarrow$ x $\in$ [-1,2)

View full question & answer→

Question 282 Marks

Solve the following system of inequalities graphically

x + 2y $\le$ 8
2x + y $\le$ 8
x $\ge$ 0
y $\ge$ 0

Answer

Given, x + 2y $\le$ 8 ... (1)
2x + y $\le$ 8... (2)
x $\ge$ 0 ... (3)
y $\ge$ 0 ... (4)
We draw the graphs of the lines x + 2y = 8 and 2x + y = 8.
The inequality (1) and (2) represent the region below the two lines, including the point on the respective lines.
Since x $\ge$ 0, y $\ge$ 0, every point in the shaded region in the first quadrant represent a solution of the given system of inequalities

View full question & answer→

Question 292 Marks

Solve the following system of inequalities

8x + 3y $\le$ 100
x $\ge$ 0
y $\ge$ 0

Answer

Given, 8x + 3y $\leq$ 100 ..... (1)
We draw the graph of the line
8x + 3y = 100
The inequality 8x + 3y ≤ 100 represents the shaded region below the line, including the points on the line 8x +3y =100

Since, x $\ge$ 0, y $\ge$ 0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequalities.

View full question & answer→

Question 302 Marks

Solve the following system of linear inequalities graphically.

x + y ≥ 5
x - y $\le$ 3

Answer

Given,
x + y $\geq$ 5..... (1)
x - y $\leq$ 3..... (2)
The graph of linear equation
x + y = 5 is drawn in the figure

We note that the solution of inequality (1) is represented by the shaded region above the line x + y = 5, including the points on the line.
On the same set of axes, we draw the graph of the equation x – y = 3 as shown in the figure. Then we note that inequality (2) represents the shaded region above the line x – y = 3, including the points on the line.
Clearly, the double shaded region, common to the above two shaded regions is the required solution region of the given system of inequalities.

View full question & answer→

Question 312 Marks

Solve y < 2 graphically.

Answer

Graph of y = 2 is given in the figure

Line y = 2 divides XY-plane in two half. plane I and plane $\parallel$ Let us select a point, (0, 0) in lower half-plane I and putting y = 0 in the given inequality, we see that
1 × 0 < 2 or 0 < 2 which is true.
Thus, the solution region is the shaded region below the line y = 2.
Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.

View full question & answer→

Question 322 Marks

Solve 3x - 6 $\geq$ 0 graphically in two dimensional plane.

Answer

Graph of 3x – 6 = 0 is given in the Fig.
We select a point, say (0, 0) and substituting it in ṁgiven inequality, we see that:
3 (0) - 6 $\geq$ 0 or -6 $\geq$ 0 which is false.
Thus, the solution region is the shaded region on the right hand side of the line x = 2

View full question & answer→

MATHS (Each question 2 marks)

Take a timed test