NCERT Class 11 Mathematics — Linear Inequalities, Exercise 5.1. All 26 questions solved.

Solving a linear inequality is solving a linear equation with one extra rule:

Multiplying or dividing both sides by a negative number reverses the inequality sign.

Adding or subtracting anything, and multiplying or dividing by a positive number, leave the direction unchanged.

The second thing this exercise tests is that the solution set depends on where $x$ is allowed to live. The inequality $24x < 100$ gives $x < \tfrac{25}{6}$ whatever $x$ is; but the answer is $\{1,2,3,4\}$ over $\mathbf{N}$, $\{\ldots,-1,0,1,2,3,4\}$ over $\mathbf{Z}$ and the interval $\left(-\infty, \tfrac{25}{6}\right)$ over $\mathbf{R}$.

Key insight. Solve first, restrict afterwards. Reduce the inequality to $x < k$ or $x > k$ using ordinary algebra, and only then ask which members of $\mathbf{N}$, $\mathbf{Z}$ or $\mathbf{R}$ satisfy it. Trying to keep track of the domain while manipulating the inequality is how the natural-number cases go wrong.

Question 1

Solve $24x < 100$, when (i) $x$ is a natural number, (ii) $x$ is an integer.

Solution. Divide both sides by $24$, a positive number, so the sign is unchanged:

$$x < \frac{100}{24} = \frac{25}{6} = 4.1\overline{6}$$

(i) The natural numbers below $4.1\overline{6}$ are $1, 2, 3, 4$.

(ii) The integers below $4.1\overline{6}$ continue indefinitely downwards.

Question 2

Solve $-12x > 30$, when (i) $x$ is a natural number, (ii) $x$ is an integer.

Solution. Divide both sides by $-12$. Because the divisor is negative, the inequality reverses:

$$x < \frac{30}{-12} = -\frac{5}{2} = -2.5$$

(i) No natural number is less than $-2.5$ — natural numbers start at $1$. So there is no solution.

(ii) The integers below $-2.5$ are $-3, -4, -5, \ldots$

Question 3

Solve $5x – 3 < 7$, when (i) $x$ is an integer, (ii) $x$ is a real number.

Solution.

$$5x – 3 < 7 \quad\Longrightarrow\quad 5x < 10 \quad\Longrightarrow\quad x < 2$$

(i) The integers less than $2$ are $1, 0, -1, -2, \ldots$

(ii) Over $\mathbf{R}$ the answer is an interval, open at $2$ because $x = 2$ does not satisfy a strict inequality.

Question 4

Solve $3x + 8 > 2$, when (i) $x$ is an integer, (ii) $x$ is a real number.

Solution.

$$3x + 8 > 2 \quad\Longrightarrow\quad 3x > -6 \quad\Longrightarrow\quad x > -2$$

Solve the inequalities in questions 5 to 16 for real $x$.

Question 5

$4x + 3 < 5x + 7$

Solution. Collect $x$ on the side where its coefficient stays positive — that way no sign reversal is needed:

$$4x + 3 < 5x + 7 \quad\Longrightarrow\quad 3 – 7 < 5x – 4x \quad\Longrightarrow\quad -4 < x$$

Question 6

$3x – 7 > 5x – 1$

Solution.

$$3x – 5x > -1 + 7 \quad\Longrightarrow\quad -2x > 6$$

Dividing by $-2$ reverses the sign:

$$x < -3$$

Question 7

$3(x – 1) \le 2(x – 3)$

Solution. Expand both sides:

$$3x – 3 \le 2x – 6 \quad\Longrightarrow\quad 3x – 2x \le -6 + 3 \quad\Longrightarrow\quad x \le -3$$

The endpoint is included, so the interval is closed at $-3$.

Question 8

$3(2 – x) \ge 2(1 – x)$

Solution.

$$6 – 3x \ge 2 – 2x \quad\Longrightarrow\quad 6 – 2 \ge 3x – 2x \quad\Longrightarrow\quad 4 \ge x$$

Question 9

$x + \dfrac{x}{2} + \dfrac{x}{3} < 11$

Solution. Combine the left side over the common denominator $6$:

$$\frac{6x + 3x + 2x}{6} < 11 \quad\Longrightarrow\quad \frac{11x}{6} < 11$$

Multiplying by $\tfrac{6}{11}$ (positive):

$$x < 6$$

Question 10

$\dfrac{x}{3} > \dfrac{x}{2} + 1$

Solution. Multiply throughout by $6$:

$$2x > 3x + 6 \quad\Longrightarrow\quad -x > 6 \quad\Longrightarrow\quad x < -6$$

Question 11

$\dfrac{3(x-2)}{5} \le \dfrac{5(2-x)}{3}$

Solution. Multiply both sides by $15$:

$$9(x – 2) \le 25(2 – x) \quad\Longrightarrow\quad 9x – 18 \le 50 – 25x$$

$$9x + 25x \le 50 + 18 \quad\Longrightarrow\quad 34x \le 68 \quad\Longrightarrow\quad x \le 2$$

Question 12

$\dfrac{1}{2}\left(\dfrac{3x}{5} + 4\right) \ge \dfrac{1}{3}(x – 6)$

Solution. Multiply both sides by $30$, the lowest common multiple of $2$, $5$ and $3$:

$$15\left(\frac{3x}{5} + 4\right) \ge 10(x – 6) \quad\Longrightarrow\quad 9x + 60 \ge 10x – 60$$

$$60 + 60 \ge 10x – 9x \quad\Longrightarrow\quad 120 \ge x$$

Question 13

$2(2x + 3) – 10 < 6(x – 2)$

Solution.

$$4x + 6 – 10 < 6x – 12 \quad\Longrightarrow\quad 4x – 4 < 6x – 12$$

$$-4 + 12 < 6x – 4x \quad\Longrightarrow\quad 8 < 2x \quad\Longrightarrow\quad 4 < x$$

Question 14

$37 – (3x + 5) \ge 9x – 8(x – 3)$

Solution. Clear both brackets, watching the minus signs in front of them:

$$37 – 3x – 5 \ge 9x – 8x + 24 \quad\Longrightarrow\quad 32 – 3x \ge x + 24$$

$$32 – 24 \ge x + 3x \quad\Longrightarrow\quad 8 \ge 4x \quad\Longrightarrow\quad 2 \ge x$$

Question 15

$\dfrac{x}{4} < \dfrac{5x – 2}{3} – \dfrac{7x – 3}{5}$

Solution. Simplify the right side first, over the denominator $15$:

$$\frac{5(5x-2) – 3(7x-3)}{15} = \frac{25x – 10 – 21x + 9}{15} = \frac{4x – 1}{15}$$

So the inequality is

$$\frac{x}{4} < \frac{4x – 1}{15}$$

Multiply both sides by $60$:

$$15x < 4(4x – 1) \quad\Longrightarrow\quad 15x < 16x – 4 \quad\Longrightarrow\quad 4 < x$$

Question 16

$\dfrac{2x – 1}{3} \ge \dfrac{3x – 2}{4} – \dfrac{2 – x}{5}$

Solution. Right side over the denominator $20$:

$$\frac{5(3x – 2) – 4(2 – x)}{20} = \frac{15x – 10 – 8 + 4x}{20} = \frac{19x – 18}{20}$$

The inequality becomes

$$\frac{2x – 1}{3} \ge \frac{19x – 18}{20}$$

Multiply by $60$:

$$20(2x – 1) \ge 3(19x – 18) \quad\Longrightarrow\quad 40x – 20 \ge 57x – 54$$

$$54 – 20 \ge 57x – 40x \quad\Longrightarrow\quad 34 \ge 17x \quad\Longrightarrow\quad 2 \ge x$$

Solve the inequalities in questions 17 to 20 and show the graph of the solution in each case on a number line.

Question 17

$3x – 2 < 2x + 1$

Solution.

$$3x – 2x < 1 + 2 \quad\Longrightarrow\quad x < 3$$

On the number line, the circle at $3$ is open because $3$ itself is excluded, and the shading runs to the left.

Question 18

$5x – 3 \ge 3x – 5$

Solution.

$$5x – 3x \ge -5 + 3 \quad\Longrightarrow\quad 2x \ge -2 \quad\Longrightarrow\quad x \ge -1$$

The circle at $-1$ is filled, because $\ge$ includes the endpoint.

Question 19

$3(1 – x) < 2(x + 4)$

Solution.

$$3 – 3x < 2x + 8 \quad\Longrightarrow\quad 3 – 8 < 2x + 3x \quad\Longrightarrow\quad -5 < 5x \quad\Longrightarrow\quad -1 < x$$

Same endpoint as question 18, but the circle is open this time because the inequality is strict — the two questions are set together to make exactly that contrast.

Question 20

$\dfrac{x}{2} \ge \dfrac{5x – 2}{3} – \dfrac{7x – 3}{5}$

Solution. The right side is the same expression as in question 15:

$$\frac{5(5x – 2) – 3(7x – 3)}{15} = \frac{4x – 1}{15}$$

So

$$\frac{x}{2} \ge \frac{4x – 1}{15}$$

Multiply both sides by $30$:

$$15x \ge 2(4x – 1) \quad\Longrightarrow\quad 15x \ge 8x – 2 \quad\Longrightarrow\quad 7x \ge -2 \quad\Longrightarrow\quad x \ge -\frac{2}{7}$$

Question 21

Ravi obtained $70$ and $75$ marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least $60$ marks.

Solution. Let $x$ be the marks in the third test. “An average of at least $60$” translates to $\ge 60$:

$$\frac{70 + 75 + x}{3} \ge 60$$

Multiply by $3$:

$$145 + x \ge 180 \quad\Longrightarrow\quad x \ge 35$$

Question 22

To receive Grade ‘A’ in a course, one must obtain an average of $90$ marks or more in five examinations (each of $100$ marks). If Sunita’s marks in the first four examinations are $87$, $92$, $94$ and $95$, find the minimum marks that Sunita must obtain in the fifth examination to get grade ‘A’ in the course.

Solution. Let $x$ be the fifth mark. The first four sum to

$$87 + 92 + 94 + 95 = 368$$

The condition is

$$\frac{368 + x}{5} \ge 90 \quad\Longrightarrow\quad 368 + x \ge 450 \quad\Longrightarrow\quad x \ge 82$$

Since each paper is out of $100$, the achievable range is $82 \le x \le 100$, so the target is genuinely attainable.

Question 23

Find all pairs of consecutive odd positive integers both of which are smaller than $10$ such that their sum is more than $11$.

Solution. Let the smaller integer be $x$, so the pair is $x$ and $x + 2$ with $x$ odd and positive. Two conditions apply.

Both smaller than $10$: the larger one is the binding constraint,

$$x + 2 < 10 \quad\Longrightarrow\quad x < 8$$

Sum more than $11$:

$$x + (x + 2) > 11 \quad\Longrightarrow\quad 2x > 9 \quad\Longrightarrow\quad x > 4.5$$

Combining: $4.5 < x < 8$, and $x$ must be odd, so $x = 5$ or $x = 7$.

Question 24

Find all pairs of consecutive even positive integers, both of which are larger than $5$, such that their sum is less than $23$.

Solution. Let the pair be $x$ and $x + 2$ with $x$ even and positive.

Both larger than $5$: the smaller one binds,

$$x > 5$$

Sum less than $23$:

$$x + (x+2) < 23 \quad\Longrightarrow\quad 2x < 21 \quad\Longrightarrow\quad x < 10.5$$

So $5 < x < 10.5$ with $x$ even: $x = 6, 8, 10$.

Question 25

The longest side of a triangle is $3$ times the shortest side and the third side is $2$ cm shorter than the longest side. If the perimeter of the triangle is at least $61$ cm, find the minimum length of the shortest side.

Solution. Let the shortest side be $x$ cm. Then

$$\text{longest} = 3x, \qquad \text{third} = 3x – 2$$

The perimeter condition:

$$x + 3x + (3x – 2) \ge 61 \quad\Longrightarrow\quad 7x – 2 \ge 61 \quad\Longrightarrow\quad 7x \ge 63 \quad\Longrightarrow\quad x \ge 9$$

So the shortest side is at least $9$ cm. Checking $x = 9$: the sides are $9$, $27$ and $25$, with perimeter $61$ ✓, and the triangle inequality holds since $9 + 25 = 34 > 27$.

Question 26

A man wants to cut three lengths from a single piece of board of length $91$ cm. The second length is to be $3$ cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5$ cm longer than the second?

Solution. Let the shortest piece be $x$ cm. Then the second is $x + 3$ and the third is $2x$. There are two conditions.

The three pieces must come out of $91$ cm:

$$x + (x + 3) + 2x \le 91 \quad\Longrightarrow\quad 4x + 3 \le 91 \quad\Longrightarrow\quad 4x \le 88 \quad\Longrightarrow\quad x \le 22$$

The third is at least $5$ cm longer than the second:

$$2x \ge (x + 3) + 5 \quad\Longrightarrow\quad 2x \ge x + 8 \quad\Longrightarrow\quad x \ge 8$$

Both must hold at once, so

$$8 \le x \le 22$$

Note the two conditions pull in opposite directions: one caps the length, the other sets a floor, and the answer is where they overlap.

Common mistakes

• Forgetting to reverse the sign after dividing by a negative. In question 2, $-12x > 30$ becomes $x < -\tfrac52$, not $x > -\tfrac52$. This single rule accounts for most wrong answers in the whole chapter.
• Question 2(i), writing $\varphi$ as a list. There genuinely is no natural number below $-2.5$, so “no solution” is the answer — not $\{0\}$, since $0$ is not a natural number and does not satisfy the inequality anyway.
• Confusing an open interval with a closed one. $<$ and $>$ give round brackets; $\le$ and $\ge$ give square ones. Questions 18 and 19 have the same endpoint $-1$ and differ only in this.
• Question 14, not distributing the minus sign. $37 – (3x + 5)$ is $32 – 3x$; writing $37 – 3x + 5$ changes the constant by $10$.
• Questions 15 and 16, subtracting the fractions carelessly. The numerator of the second fraction must be multiplied out before the minus sign is applied: $-3(7x – 3) = -21x + 9$.
• Question 23, using the wrong endpoint as binding. “Both smaller than $10$” is a condition on the larger member $x + 2$, so it gives $x < 8$, not $x < 10$. Question 24 is the mirror image — “both larger than $5$” binds on the smaller.
• Questions 23 and 24, forgetting the parity. The interval $4.5 < x < 8$ contains $5, 6, 7$, but only the odd values are admissible.
• Question 26, using only one of the two conditions. The board length gives an upper bound and the “5 cm longer” clause gives a lower one; the answer is the overlap $[8, 22]$, not either bound alone.

Practise next

• Miscellaneous Exercise on Chapter 5 — double inequalities of the form $a \le f(x) \le b$ and systems of two inequalities solved together, both built directly on the technique here.
• Exercise 1.3 — worth revising for interval notation, since every answer from question 3 onwards is written as an interval.