NCERT Class 11 Mathematics — Linear Inequalities, Miscellaneous Exercise on Chapter 5. All 14 questions solved.

Two new shapes appear here, and both reduce to Exercise 5.1’s technique.

A double inequality $a \le f(x) \le b$ is two inequalities at once. Operate on all three parts simultaneously — and if you multiply through by a negative number, both signs reverse and the ends swap over.

A system of two separate inequalities is solved by finding each solution set and taking their intersection: only values satisfying both survive.

The word problems (11 to 14) all follow the same route — name the unknown, translate each phrase into an inequality, then solve.

Key insight. In a mixture problem the quantity that stays fixed is the amount of solute, and the quantity that changes is the total volume. In question 12, adding a 2% solution to 640 litres of an 8% solution changes both — so write the concentration as (total acid)/(total volume) and impose the bounds on that fraction. Every mixture question in this exercise is that one sentence.

Solve the inequalities in questions 1 to 6.

Question 1

$2 \le 3x – 4 \le 5$

Solution. Add $4$ to all three parts:

$$6 \le 3x \le 9$$

Divide all three by $3$ (positive, so no reversal):

$$2 \le x \le 3$$

Question 2

$6 \le -3(2x – 4) < 12$

Solution. Divide all three parts by $-3$. The divisor is negative, so both inequality signs reverse:

$$-2 \ge 2x – 4 > -4$$

Read from the right, this says $-4 < 2x – 4 \le -2$. Add $4$:

$$0 < 2x \le 2$$

Divide by $2$:

$$0 < x \le 1$$

Question 3

$-3 \le 4 – \dfrac{7x}{2} \le 18$

Solution. Subtract $4$ throughout:

$$-7 \le -\frac{7x}{2} \le 14$$

Multiply throughout by $-\dfrac{2}{7}$, reversing both signs:

$$2 \ge x \ge -4$$

which reads $-4 \le x \le 2$.

Question 4

$-15 < \dfrac{3(x – 2)}{5} \le 0$

Solution. Multiply throughout by $\dfrac{5}{3}$ (positive):

$$-25 < x – 2 \le 0$$

Add $2$:

$$-23 < x \le 2$$

Question 5

$-12 < 4 – \dfrac{3x}{-5} \le 2$

Solution. Simplify the middle term first — the two minus signs cancel:

$$4 – \frac{3x}{-5} = 4 + \frac{3x}{5}$$

So the inequality is

$$-12 < 4 + \frac{3x}{5} \le 2$$

Subtract $4$:

$$-16 < \frac{3x}{5} \le -2$$

Multiply by $\dfrac53$ (positive, so the signs stay):

$$-\frac{80}{3} < x \le -\frac{10}{3}$$

Simplifying the middle term at the start is what keeps this question short; leaving $-\frac{3x}{-5}$ in place invites a sign error later.

Question 6

$7 \le \dfrac{3x + 11}{2} \le 11$

Solution. Multiply throughout by $2$:

$$14 \le 3x + 11 \le 22$$

Subtract $11$:

$$3 \le 3x \le 11$$

Divide by $3$:

$$1 \le x \le \frac{11}{3}$$

Solve the inequalities in questions 7 to 10 and represent the solution graphically on a number line.

Question 7

$5x + 1 > -24$,   $5x – 1 < 24$

Solution. Solve each separately.

$$5x + 1 > -24 \quad\Longrightarrow\quad 5x > -25 \quad\Longrightarrow\quad x > -5$$

$$5x – 1 < 24 \quad\Longrightarrow\quad 5x < 25 \quad\Longrightarrow\quad x < 5$$

The solution set is the intersection $(-5, \infty) \cap (-\infty, 5)$:

$$-5 < x < 5$$

Question 8

$2(x – 1) < x + 5$,   $3(x + 2) > 2 – x$

Solution.

$$2x – 2 < x + 5 \quad\Longrightarrow\quad x < 7$$

$$3x + 6 > 2 – x \quad\Longrightarrow\quad 4x > -4 \quad\Longrightarrow\quad x > -1$$

Intersecting:

$$-1 < x < 7$$

Question 9

$3x – 7 > 2(x – 6)$,   $6 – x > 11 – 2x$

Solution.

$$3x – 7 > 2x – 12 \quad\Longrightarrow\quad x > -5$$

$$6 – x > 11 – 2x \quad\Longrightarrow\quad -x + 2x > 11 – 6 \quad\Longrightarrow\quad x > 5$$

Both must hold. Since $x > 5$ already implies $x > -5$, the second condition is the binding one and the first adds nothing:

$$x > 5$$

This is worth pausing on — the intersection of two half-lines pointing the same way is simply the more restrictive of the two.

Question 10

$5(2x – 7) – 3(2x + 3) \le 0$,   $2x + 19 \le 6x + 47$

Solution. First inequality:

$$10x – 35 – 6x – 9 \le 0 \quad\Longrightarrow\quad 4x – 44 \le 0 \quad\Longrightarrow\quad x \le 11$$

Second inequality:

$$19 – 47 \le 6x – 2x \quad\Longrightarrow\quad -28 \le 4x \quad\Longrightarrow\quad -7 \le x$$

Intersecting, and noting both endpoints are included:

$$-7 \le x \le 11$$

Question 11

A solution is to be kept between $68^\circ$F and $77^\circ$F. What is the range in temperature in degrees Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by $\mathrm{F} = \dfrac{9}{5}\mathrm{C} + 32$?

Solution. The condition on $\mathrm F$ is

$$68 < \frac95 \mathrm C + 32 < 77$$

Subtract $32$ throughout:

$$36 < \frac95 \mathrm C < 45$$

Multiply throughout by $\dfrac59$:

$$20 < \mathrm C < 25$$

Question 12

A solution of $8\%$ boric acid is to be diluted by adding a $2\%$ boric acid solution to it. The resulting mixture is to be more than $4\%$ but less than $6\%$ boric acid. If we have $640$ litres of the $8\%$ solution, how many litres of the $2\%$ solution will have to be added?

Solution. Let $x$ litres of the $2\%$ solution be added.

• Total volume: $640 + x$ litres.
• Total boric acid: $8\%$ of $640$ plus $2\%$ of $x$, i.e. $\dfrac{8}{100}(640) + \dfrac{2}{100}x$ litres.

The requirement is

$$4\% \text{ of } (640 + x) < \frac{8}{100}(640) + \frac{2}{100}x < 6\% \text{ of } (640+x)$$

Multiplying everything by $100$ to clear the percentages:

$$4(640 + x) < 8(640) + 2x < 6(640 + x)$$

Left inequality:

$$2560 + 4x < 5120 + 2x \quad\Longrightarrow\quad 2x < 2560 \quad\Longrightarrow\quad x < 1280$$

Right inequality:

$$5120 + 2x < 3840 + 6x \quad\Longrightarrow\quad 1280 < 4x \quad\Longrightarrow\quad x > 320$$

Both must hold, so

$$320 < x < 1280$$

The two bounds make sense: too little dilutant leaves the mixture above $6\%$, too much drops it below $4\%$.

Question 13

How many litres of water will have to be added to $1125$ litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ but less than $30\%$ acid content?

Solution. Let $x$ litres of water be added. Water contains no acid, so:

• Total volume: $1125 + x$ litres.
• Total acid: unchanged at $45\%$ of $1125 = \dfrac{45}{100}(1125) = 506.25$ litres.

This is the key difference from question 12 — here the numerator is fixed and only the denominator grows.

$$25\% \text{ of } (1125 + x) < 506.25 < 30\% \text{ of } (1125 + x)$$

Left inequality:

$$\frac{25}{100}(1125 + x) < 506.25 \quad\Longrightarrow\quad 281.25 + 0.25x < 506.25$$

$$0.25x < 225 \quad\Longrightarrow\quad x < 900$$

Right inequality:

$$506.25 < \frac{30}{100}(1125 + x) = 337.5 + 0.3x \quad\Longrightarrow\quad 168.75 < 0.3x \quad\Longrightarrow\quad x > 562.5$$

Combining:

$$562.5 < x < 900$$

Question 14

IQ of a person is given by the formula

$$\mathrm{IQ} = \frac{\mathrm{MA}}{\mathrm{CA}} \times 100$$

where MA is mental age and CA is chronological age. If $80 \le \mathrm{IQ} \le 140$ for a group of $12$-year-old children, find the range of their mental age.

Solution. The children are $12$ years old, so $\mathrm{CA} = 12$ and

$$\mathrm{IQ} = \frac{\mathrm{MA}}{12} \times 100 = \frac{100\,\mathrm{MA}}{12} = \frac{25\,\mathrm{MA}}{3}$$

The condition becomes

$$80 \le \frac{25\,\mathrm{MA}}{3} \le 140$$

Multiply throughout by $\dfrac{3}{25}$ (positive):

$$80 \times \frac{3}{25} \le \mathrm{MA} \le 140 \times \frac{3}{25}$$

$$\frac{240}{25} \le \mathrm{MA} \le \frac{420}{25}$$

$$9.6 \le \mathrm{MA} \le 16.8$$

Common mistakes

• Question 2, reversing only one of the two signs. Dividing $6 \le -3(2x-4) < 12$ by $-3$ flips both relations at once, giving $-2 \ge 2x – 4 > -4$. Flipping one produces an impossible statement.
• Question 5, not simplifying $-\dfrac{3x}{-5}$. It is $+\dfrac{3x}{5}$. Carrying the double negative through the working is the main source of error in this question.
• Questions 7 to 10, taking the union instead of the intersection. A system requires both inequalities to hold, so the answer is where the two solution sets overlap. In question 9 that overlap is $x > 5$, not $x > -5$.
• Question 9, assuming both conditions must appear in the answer. When one inequality is strictly stronger, the other is redundant. It still had to be solved — you cannot know which is stronger until you do.
• Question 12, forgetting that the total volume changes. Both the acid and the volume grow when a $2\%$ solution is added; treating the denominator as fixed at $640$ gives a wrong pair of bounds.
• Question 13, changing the amount of acid. Adding water leaves the acid at $506.25$ litres. Only the denominator moves, which is why the two questions look alike but are set up differently.
• Question 14, forgetting to substitute $\mathrm{CA} = 12$. The formula has two unknowns until the age is used; the whole question hinges on that substitution.
• Writing the endpoints of a double inequality backwards. After a sign reversal the smaller bound ends up on the right, as in question 3’s $2 \ge x \ge -4$. Rewrite it as $-4 \le x \le 2$ before quoting an interval.

Practise next

• Exercise 6.1 — the next chapter, Permutations and Combinations, which starts a new topic entirely.
• Exercise 5.1 — worth working again if the system questions felt slow, since each half of a system is exactly an Exercise 5.1 problem.