NCERT Exemplar Solutions Class 11 Maths Linear Inequalities — Short and Long Answer Type Q1–18

NCERT Exemplar Problems, Class 11 Mathematics — Linear Inequalities, Short Answer and Long Answer Type questions 1 to 18. All 18 questions solved.

Questions 1 to 6 are pure inequalities in one variable: rational inequalities, modulus inequalities and a compound system. Questions 7 to 12 turn a word problem into a single inequality and then solve it. Questions 13 to 18 are the “Long Answer” group: one more rational system, two questions that read inequalities off a graph, and three systems in two variables. Everything rests on two rules: multiplying or dividing by a negative number reverses the inequality, and $|u| < a$ means $-a < u < a$ while $|u| > a$ means $u < -a$ or $u > a$.

Four answers in the printed key are wrong or imprecise: Questions 2, 3, 8 and 11. Each is explained where it occurs.

Key insight. Before you multiply through by an expression, ask whether it is positive or negative. Where the sign is unknown, move everything to one side and study the sign of the single fraction instead. That one habit removes almost every error on this page.

Question 1

Solve for $x$ the inequality

$$\frac{4}{x+1} \le 3 \le \frac{6}{x+1}, \quad (x > 0).$$

Solution. Because $x > 0$, the number $x + 1$ is positive, so we may multiply by it without reversing anything. The left part gives $4 \le 3(x+1)$, that is $x \ge \frac13$. The right part gives $3(x+1) \le 6$, that is $x \le 1$. Both must hold together.

Question 2

Solve for $x$ the inequality

$$\frac{|x-2| – 1}{|x-2| – 2} \le 0.$$

Solution. Put $t = |x-2|$. A fraction is $\le 0$ when numerator and denominator have opposite signs, or the numerator is zero. The numerator $t – 1$ changes sign at $t = 1$ and the denominator $t – 2$ at $t = 2$. Testing the three zones gives $1 \le t < 2$. The value $t = 2$ is excluded because it makes the denominator zero; $t = 1$ is included because it makes the fraction exactly $0$.

Now undo the substitution: $1 \le |x-2| < 2$.

• $1 \le x – 2 < 2$ gives $3 \le x < 4$.
• $-2 < x – 2 \le -1$ gives $0 < x \le 1$.

The printed key gives $[0,1] \cup [3,4]$, which is wrong. At $x = 0$ and $x = 4$ we have $|x-2| = 2$, so the denominator is zero and the expression does not exist.

Question 3

Solve for $x$ the inequality

$$\frac{1}{|x| – 3} \le \frac12.$$

Solution. Let $t = |x| – 3$, with $t \ne 0$. We need $\frac1t \le \frac12$.

• If $t < 0$, the left side is negative, so the inequality holds automatically.
• If $t > 0$, multiply by $2t$: $2 \le t$.

So $t < 0$ or $t \ge 2$, that is $|x| < 3$ or $|x| \ge 5$. Hence $-3 < x < 3$, or $x \le -5$, or $x \ge 5$. The point $x = -5$ gives $\frac{1}{5-3} = \frac12$ exactly, so it belongs. The printed key writes $(-\infty, -5)$ with the end open, which drops a valid point.

Question 4

Solve for $x$ the inequalities

$$|x – 1| \le 5, \qquad |x| \ge 2.$$

Solution. The first gives $-5 \le x – 1 \le 5$, so $-4 \le x \le 6$. The second gives $x \le -2$ or $x \ge 2$. We want values satisfying both, so intersect: from $[-4, 6]$ keep the part with $x \le -2$ and the part with $x \ge 2$.

Question 5

Solve for $x$ the inequality

$$-5 \le \frac{2 – 3x}{4} \le 9.$$

Solution. Multiply every part by $4$ (positive): $-20 \le 2 – 3x \le 36$. Subtract $2$: $-22 \le -3x \le 34$. Now divide by $-3$, which reverses both signs: $\frac{22}{3} \ge x \ge -\frac{34}{3}$.

Question 6

Solve for $x$ the inequalities

$$4x + 3 \ge 2x + 17, \qquad 3x – 5 < -2.$$

Solution. The first gives $2x \ge 14$, so $x \ge 7$. The second gives $3x < 3$, so $x < 1$. No number is both at least $7$ and less than $1$.

Question 7

A company manufactures cassettes. Its cost and revenue functions are $C(x) = 26{,}000 + 30x$ and $R(x) = 43x$, where $x$ is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit?

Solution. Profit means revenue exceeds cost: $43x > 26000 + 30x$. Then $13x > 26000$, so $x > 2000$. Selling exactly $2000$ gives zero profit, so the company must sell more than $2000$ cassettes (at least $2001$).

Question 8

The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between $8.2$ and $8.5$. If the first two pH readings are $8.48$ and $8.35$, find the range of pH value for the third reading that will result in the acidity level being normal.

Solution. Let the third reading be $x$. The average is $\frac{8.48 + 8.35 + x}{3} = \frac{16.83 + x}{3}$, and we need $8.2 < \frac{16.83 + x}{3} < 8.5$. Multiply by $3$: $24.6 < 16.83 + x < 25.5$. Subtract $16.83$: $7.77 < x < 8.67$.

The printed key says “between $7.77$ and $8.77$”. The upper limit is wrong: $25.5 – 16.83 = 8.67$.

Question 9

A solution of $9\%$ acid is to be diluted by adding $3\%$ acid solution to it. The resulting mixture is to be more than $5\%$ but less than $7\%$ acid. If there is $460$ litres of the $9\%$ solution, how many litres of $3\%$ solution will have to be added?

Solution. Let $x$ litres of $3\%$ solution be added. The acid in the mixture is $0.09 \times 460 + 0.03x = 41.4 + 0.03x$ litres, and the total volume is $460 + x$. The concentration condition is

$$0.05(460 + x) < 41.4 + 0.03x < 0.07(460 + x).$$

Left part: $23 + 0.05x < 41.4 + 0.03x$, so $0.02x < 18.4$, giving $x < 920$. Right part: $41.4 + 0.03x < 32.2 + 0.07x$, so $9.2 < 0.04x$, giving $x > 230$.

Question 10

A solution is to be kept between $40^\circ$C and $45^\circ$C. What is the range of temperature in degree Fahrenheit, if the conversion formula is $F = \frac95 C + 32$?

Solution. The condition is $40 < C < 45$. Multiplying by $\frac95$ (positive, so the direction is kept) gives $72 < \frac95 C < 81$, and adding $32$ gives $104 < F < 113$. If the problem is read as including the end temperatures, the signs become $\le$.

Question 11

The longest side of a triangle is twice the shortest side and the third side is $2$ cm longer than the shortest side. If the perimeter of the triangle is more than $166$ cm, find the minimum length of the shortest side.

Solution. Let the shortest side be $x$ cm. The sides are $x$, $x + 2$ and $2x$, and the perimeter is $x + (x+2) + 2x = 4x + 2$. The condition is $4x + 2 > 166$, so $4x > 164$ and $x > 41$.

(The side lengths also need $2x$ to be the longest, which holds when $x \ge 2$, and the triangle inequality $x + (x+2) > 2x$ is always true. Neither changes the answer.)

The inequality is strict, so $41$ itself fails: it gives a perimeter of exactly $166$. The printed key says “$41$ cm”, which is the boundary, not an allowed value. The shortest side must be more than $41$ cm, and if lengths are whole centimetres the smallest allowed value is $42$ cm.

Question 12

In drilling world’s deepest hole it was found that the temperature $T$ in degree Celsius, $x$ km below the earth’s surface was given by $T = 30 + 25(x – 3)$, $3 \le x \le 15$. At what depth will the temperature be between $155^\circ$C and $205^\circ$C?

Solution. We need $155 < 30 + 25(x-3) < 205$. Subtract $30$: $125 < 25(x-3) < 175$. Divide by $25$: $5 < x – 3 < 7$, so $8 < x < 10$. This lies inside the stated range $3 \le x \le 15$.

Question 13

Solve the following system of inequalities

$$\frac{2x+1}{7x-1} > 5, \qquad \frac{x+7}{x-8} > 2.$$

Solution. Solve each separately, then intersect. Do not cross-multiply, because the sign of the denominator is unknown.

First: $\frac{2x+1}{7x-1} – 5 > 0$ gives $\frac{2x + 1 – 35x + 5}{7x – 1} > 0$, that is $\frac{6 – 33x}{7x – 1} > 0$, or $\frac{33x – 6}{7x – 1} < 0$. The critical points are $x = \frac{2}{11}$ and $x = \frac17$, and the fraction is negative between them: $\frac17 < x < \frac{2}{11}$.

Second: $\frac{x+7}{x-8} – 2 > 0$ gives $\frac{x + 7 – 2x + 16}{x – 8} > 0$, that is $\frac{23 – x}{x – 8} > 0$, so $8 < x < 23$.

The intervals $\left(\frac17, \frac2{11}\right)$ and $(8, 23)$ do not overlap.

Question 14

Find the linear inequalities for which the shaded region in the given figure (Fig 6.5) is the solution set.

Graphical solution of inequalities in two variables is no longer in the current NCERT syllabus, but the question is still worked here.

Solution. The figure shows two lines. One passes through $(20, 0)$ and $(0, 20)$, so it is $x + y = 20$. The other passes through $(16, 0)$ and $(0, 24)$, so it is $3x + 2y = 48$. They cross at $(8, 12)$. The shaded region is the four-cornered region with vertices $(0,0)$, $(16,0)$, $(8,12)$ and $(0,20)$: it lies in the first quadrant, on the origin’s side of both lines.

To decide each sign, test the origin. For $x + y = 20$: $0 + 0 = 0 < 20$, so the shaded side is $x + y \le 20$. For $3x + 2y = 48$: $0 < 48$, so the shaded side is $3x + 2y \le 48$. The region is bounded by the axes, so $x \ge 0$ and $y \ge 0$ as well.

Question 15

Find the linear inequalities for which the shaded region in the given figure (Fig 6.6) is the solution set.

Graphical solution of inequalities in two variables is no longer in the current NCERT syllabus, but the question is still worked here.

Solution. The figure has four lines: $x + y = 4$ (through $(4,0)$ and $(0,4)$), $x + y = 8$ (through $(8,0)$ and $(0,8)$), $x = 5$ and $y = 5$. The shaded region is the six-cornered region with vertices $(0,4)$, $(4,0)$, $(5,0)$, $(5,3)$, $(3,5)$ and $(0,5)$. It sits in the first quadrant, so $x \ge 0$ and $y \ge 0$.

It lies on the far side of $x + y = 4$ from the origin (the origin gives $0 < 4$), so $x + y \ge 4$. It lies on the near side of $x + y = 8$, so $x + y \le 8$. It is to the left of $x = 5$ and below $y = 5$, so $x \le 5$ and $y \le 5$.

Question 16

Show that the following system of linear inequalities has no solution

$$x + 2y \le 3,\quad 3x + 4y \ge 12,\quad x \ge 0,\quad y \ge 1.$$

Graphical solution of inequalities in two variables is no longer in the current NCERT syllabus, but the question is still worked here.

Solution. We can show this by algebra alone. Since $y \ge 1$, the first inequality gives $x \le 3 – 2y \le 1$. Now bound the expression in the second inequality using that:

$$3x + 4y \le 3(3 – 2y) + 4y = 9 – 2y \le 9 – 2 = 7.$$

So $3x + 4y$ can never exceed $7$, but the second inequality demands $3x + 4y \ge 12$. The conditions contradict each other. (Graphically: on the line $y = 1$ the first inequality allows only $x \le 1$, while the second needs $x \ge \frac83$; the two regions never meet inside $x \ge 0$, $y \ge 1$.)

Question 17

Solve the following system of linear inequalities:

$$3x + 2y \ge 24,\quad 3x + y \le 15,\quad x \ge 4.$$

Graphical solution of inequalities in two variables is no longer in the current NCERT syllabus, but the question is still worked here.

Solution. Write the first two as bounds on $y$: $y \ge 12 – \frac32 x$ and $y \le 15 – 3x$. For any $y$ to exist we need $12 – \frac32 x \le 15 – 3x$, which simplifies to $\frac32 x \le 3$, so $x \le 2$. But the third inequality says $x \ge 4$. For example at $x = 4$ we would need $y \ge 6$ and $y \le 3$ at once. The conditions are incompatible.

Question 18

Show that the solution set of the following system of linear inequalities is an unbounded region

$$2x + y \ge 8,\quad x + 2y \ge 10,\quad x \ge 0,\quad y \ge 0.$$

Graphical solution of inequalities in two variables is no longer in the current NCERT syllabus, but the question is still worked here.

Solution. The lines $2x + y = 8$ and $x + 2y = 10$ meet where $x = 2$, $y = 4$. The corner points of the region are $(0, 8)$, $(2, 4)$ and $(10, 0)$, and the region lies on the far side of both lines from the origin, in the first quadrant.

A region is bounded only if it fits inside some circle. Take the point $(t, t)$ for any $t \ge 4$: then $2t + t = 3t \ge 12 \ge 8$ and $t + 2t = 3t \ge 12 \ge 10$, and $t \ge 0$, so every such point satisfies all four inequalities. Since $t$ can be as large as we like, the solution set stretches off to infinity and cannot lie inside any circle.

Common mistakes

• Question 1: Multiplying by $x + 1$ without checking its sign. It is safe here only because $x > 0$ is given; students then forget to apply that condition to the final answer.
• Questions 2 and 3: Forgetting to exclude the values that make a denominator zero, or excluding values that are fine. Check every endpoint by direct substitution: $|x-2| = 2$ is undefined in Question 2, but $x = -5$ in Question 3 gives exactly $\frac12$ and is allowed.
• Question 5: Dividing by $-3$ and leaving the inequality signs as they were. Dividing or multiplying by a negative number reverses every sign in a compound inequality.
• Question 6: Writing “$x \ge 7$ or $x < 1$”. The word is “and”: both inequalities must hold, so the answer is empty, not the union.
• Questions 8 and 10: Writing $\le$ for “between”, or slipping in the subtraction $25.5 – 16.83$ (which the printed key itself got wrong). Strict reading and careful decimals both matter.
• Questions 13 and 17: Cross-multiplying a rational inequality, or solving only part of a system and stopping. Always finish by intersecting the individual solution sets; an empty intersection is a legitimate answer.

Practise next

• The other page of this unit: Objective, True or False and Fill in the Blanks, Q19–32.
• NCERT textbook: Exercise 5.1, Linear Inequalities for word problems and rational forms like Questions 7 to 12.
• NCERT textbook: Miscellaneous Exercise, Linear Inequalities for compound and modulus systems like Questions 1 to 5 and 13.