NCERT Exemplar Solutions Class 10 Maths Pair of Linear Equations — Exercise 3.3

NCERT Exemplar Problems, Class 10 Mathematics — Pair of Linear Equations in Two Variables, Exercise 3.3. All 22 questions solved.

This is the long middle exercise of the unit and it runs in three distinct stretches. Questions 1 to 6 are about parameters — given a condition on the solutions, find the letter in the coefficients. Questions 7 to 13 are solving, including several pairs that are not linear until you substitute. Questions 14 to 22 are word problems, where the whole difficulty is forming the equations rather than solving them.

Two techniques recur and are worth naming:

• Substitution to linearise. Questions 9(iii), 9(iv) and 9(vii) are not linear in $x$ and $y$, but become linear in $\frac1x$ and $\frac1y$. Set $u = \frac1x$, $v = \frac1y$, solve the linear pair, then invert — and remember to invert at the end, which is where the marks go.
• Add and subtract. Where the coefficients are swapped between the two equations, as in Question 9(v) with $43$ and $67$, adding and subtracting is far quicker than elimination. The sum gives $x + y$ and the difference gives $x – y$.

Key insight. In the parameter questions, write out all three ratios before solving anything. Very often two of them contain no parameter at all, and comparing those two settles the question by itself — that is what happens in Question 4(v) and in several questions of Exercise 3.2.

Question 1

For which value(s) of $\lambda$ do the pair of linear equations $\lambda x + y = \lambda^2$ and $x + \lambda y = 1$ have (i) no solution? (ii) infinitely many solutions? (iii) a unique solution?

Solution. In standard form: $\lambda x + y – \lambda^2 = 0$ and $x + \lambda y – 1 = 0$. The three ratios are

$$\frac{a_1}{a_2} = \lambda, \qquad \frac{b_1}{b_2} = \frac{1}{\lambda}, \qquad \frac{c_1}{c_2} = \frac{-\lambda^2}{-1} = \lambda^2$$

The first two agree when $\lambda = \frac1\lambda$, that is $\lambda^2 = 1$, so $\lambda = \pm 1$. Everything now turns on what the third ratio does at those two values.

• $\lambda = 1$: the ratios are $1$, $1$, $1$ — all equal, so the lines are coincident and there are infinitely many solutions.
• $\lambda = -1$: the ratios are $-1$, $-1$, $1$ — the first two agree, the third does not, so the lines are parallel and there is no solution.
• $\lambda \ne \pm 1$: the first two ratios differ, so the lines intersect and the solution is unique.

Question 2

For which value(s) of $k$ will the pair of equations $kx + 3y = k – 3$ and $12x + ky = k$ have no solution?

Solution. No solution means $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$. From the first equality,

$$\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$$

Now test the constant ratio at each, comparing $\frac{k-3}{k}$ with $\frac{k}{12}$:

• $k = 6$: $\frac{k-3}{k} = \frac{3}{6} = \frac12$ and $\frac{k}{12} = \frac12$. All three ratios equal — coincident lines, infinitely many solutions. Not what is wanted.
• $k = -6$: $\frac{k-3}{k} = \frac{-9}{-6} = \frac32$ and $\frac{k}{12} = -\frac12$. These differ — parallel lines, no solution. ✓

Question 3

For which values of $a$ and $b$ will the following pair of linear equations have infinitely many solutions? $\ x + 2y = 1$ and $(a-b)x + (a+b)y = a + b – 2$.

Solution. Infinitely many solutions needs all three ratios equal:

$$\frac{1}{a-b} = \frac{2}{a+b} = \frac{1}{a+b-2}$$

Take the second and third first, since they share a denominator pattern:

$$\frac{2}{a+b} = \frac{1}{a+b-2} \implies 2(a+b-2) = a+b \implies a+b = 4$$

Now the first and second:

$$\frac{1}{a-b} = \frac{2}{a+b} \implies a+b = 2(a-b) \implies 4 = 2a – 2b \implies a – b = 2$$

Solving $a + b = 4$ and $a – b = 2$ gives $a = 3$, $b = 1$.

Question 4

Find the value(s) of $p$ in (i) to (iv) and $p$ and $q$ in (v) for the following pairs of equations.

(i) $3x – y – 5 = 0$ and $6x – 2y – p = 0$, if the lines represented by these equations are parallel

$$\frac{a_1}{a_2} = \frac{3}{6} = \frac12, \qquad \frac{b_1}{b_2} = \frac{-1}{-2} = \frac12$$

The first two ratios are equal automatically, so the lines can never intersect. Parallel (as opposed to coincident) requires the constant ratio to differ:

$$\frac{-5}{-p} = \frac{5}{p} \ne \frac12 \implies p \ne 10$$

So every real $p$ except $10$ gives parallel lines; at $p = 10$ the two equations describe the same line.

(ii) $-x + py = 1$ and $px – y = 1$, if the pair of equations has no solution

Standard form: $-x + py – 1 = 0$ and $px – y – 1 = 0$.

$$\frac{a_1}{a_2} = \frac{-1}{p}, \qquad \frac{b_1}{b_2} = \frac{p}{-1}, \qquad \frac{c_1}{c_2} = \frac{-1}{-1} = 1$$

Setting the first two equal: $(-1)(-1) = p \cdot p$, so $p^2 = 1$ and $p = \pm 1$. Test the constant ratio:

• $p = 1$: the ratios are $-1$, $-1$, $1$ — parallel, no solution ✓
• $p = -1$: the ratios are $1$, $1$, $1$ — coincident, infinitely many solutions ✗

(iii) $-3x + 5y = 7$ and $2px – 3y = 1$, if the lines are intersecting at a unique point

A unique point means the first two ratios must differ:

$$\frac{-3}{2p} \ne \frac{5}{-3}$$

Cross-multiplying, $(-3)(-3) \ne 5(2p)$, that is $9 \ne 10p$, so $p \ne \frac{9}{10}$.

(iv) $2x + 3y – 5 = 0$ and $px – 6y – 8 = 0$, if the pair has a unique solution

$$\frac{2}{p} \ne \frac{3}{-6} \implies 2(-6) \ne 3p \implies -12 \ne 3p \implies p \ne -4$$

(v) $2x + 3y = 7$ and $2px + py = 28 – qy$, if the pair has infinitely many solutions

First collect the $y$ terms of the second equation, which is the step the question is really testing:

$$2px + py + qy = 28 \implies 2px + (p+q)y = 28$$

Infinitely many solutions needs

$$\frac{2}{2p} = \frac{3}{p+q} = \frac{7}{28} = \frac14$$

From the first and last, $\frac{1}{p} = \frac14$, so $p = 4$. From the middle and last, $p + q = 12$, so $q = 8$.

Question 5

Two straight paths are represented by the equations $x – 3y = 2$ and $-2x + 6y = 5$. Check whether the paths cross each other or not.

Solution. “Cross” means the lines intersect, so compare the first two ratios.

$$\frac{a_1}{a_2} = \frac{1}{-2}, \qquad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac12$$

They are equal, so the paths are not intersecting. Checking the constant ratio to distinguish parallel from coincident:

$$\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5} \ne -\frac12$$

The paths are therefore parallel and distinct — they never meet, and they are not the same path either.

Question 6

Write a pair of linear equations which has the unique solution $x = -1$, $y = 3$. How many such pairs can you write?

Solution. Pick any two equations satisfied by $(-1, 3)$, then check that they are not multiples of one another. For example:

$$x – y = -1 – 3 = -4 \qquad\text{and}\qquad 2x + 3y = -2 + 9 = 7$$

so the pair is $x – y = -4$ and $2x + 3y = 7$. Uniqueness: $\frac{1}{2} \ne \frac{-1}{3}$, so the lines intersect at exactly one point, which must be $(-1,3)$.

How many such pairs? Infinitely many. Through any point there are infinitely many lines, and any two non-parallel ones among them form such a pair — take $ax + by = -a + 3b$ for any $a$, $b$ you like, twice over with different ratios $a : b$.

Question 7

If $2x + y = 23$ and $4x – y = 19$, find the values of $5y – 2x$ and $\dfrac{y}{x} – 2$.

Solution. Adding the two equations eliminates $y$ at once:

$$(2x + y) + (4x – y) = 23 + 19 \implies 6x = 42 \implies x = 7$$

Then $y = 23 – 2(7) = 9$. Substituting into the two requested expressions:

$$5y – 2x = 45 – 14 = 31$$ $$\frac{y}{x} – 2 = \frac97 – 2 = \frac{9 – 14}{7} = -\frac57$$

Question 8

Find the values of $x$ and $y$ in the following rectangle.

Solution. In a rectangle opposite sides are equal, and that is the only geometry needed. Matching top with bottom and left with right:

$$x + 3y = 13, \qquad 3x + y = 7$$

Multiply the second by $3$ and subtract the first:

$$9x + 3y = 21$$ $$\underline{\;x + 3y = 13\;}$$ $$8x = 8 \implies x = 1$$

Then $y = 7 – 3(1) = 4$.

Check both sides: $x + 3y = 1 + 12 = 13$ ✓ and $3x + y = 3 + 4 = 7$ ✓.

Question 9

Solve the following pairs of equations.

(i) $x + y = 3.3$ and $\dfrac{0.6}{3x – 2y} = -1$, where $3x – 2y \ne 0$

Clear the fraction first: $0.6 = -(3x – 2y)$, so $3x – 2y = -0.6$. Now a straightforward pair. Multiply the first by $2$ and add:

$$2x + 2y = 6.6, \qquad 3x – 2y = -0.6 \implies 5x = 6 \implies x = 1.2$$

Then $y = 3.3 – 1.2 = 2.1$.

(ii) $\dfrac{x}{3} + \dfrac{y}{4} = 4$ and $\dfrac{5x}{6} – \dfrac{y}{8} = 4$

Clear denominators: multiply the first by $12$ and the second by $24$:

$$4x + 3y = 48, \qquad 20x – 3y = 96$$

Adding, $24x = 144$, so $x = 6$, and then $3y = 48 – 24 = 24$, giving $y = 8$.

(iii) $4x + \dfrac{6}{y} = 15$ and $6x – \dfrac{8}{y} = 14$, where $y \ne 0$

Put $v = \dfrac1y$. The pair becomes linear:

$$4x + 6v = 15, \qquad 6x – 8v = 14$$

Multiply the first by $4$ and the second by $3$ to match the $v$ terms:

$$16x + 24v = 60, \qquad 18x – 24v = 42 \implies 34x = 102 \implies x = 3$$

Then $6v = 15 – 12 = 3$, so $v = \frac12$ and $y = 2$.

(iv) $\dfrac{1}{2x} – \dfrac{1}{y} = -1$ and $\dfrac{1}{x} + \dfrac{1}{2y} = 8$, where $x, y \ne 0$

Put $u = \dfrac1x$, $v = \dfrac1y$:

$$\frac{u}{2} – v = -1, \qquad u + \frac{v}{2} = 8$$

From the first, $u = 2v – 2$. Substituting,

$$2v – 2 + \frac{v}{2} = 8 \implies \frac{5v}{2} = 10 \implies v = 4$$

so $y = \frac14$, and $u = 2(4) – 2 = 6$, giving $x = \frac16$.

(v) $43x + 67y = -24$ and $67x + 43y = 24$

The coefficients are swapped, so add and subtract rather than eliminating.

Adding: $110x + 110y = 0$, so $x + y = 0$. Subtracting: $-24x + 24y = -48$, so $y – x = -2$.

From these, $x = 1$ and $y = -1$.

(vi) $\dfrac{x}{a} + \dfrac{y}{b} = a + b$ and $\dfrac{x}{a^2} + \dfrac{y}{b^2} = 2$, where $a, b \ne 0$

The shape of the first equation suggests the answer: if $x = a^2$ and $y = b^2$ then $\frac{x}{a} + \frac{y}{b} = a + b$ exactly. Check the second: $\frac{a^2}{a^2} + \frac{b^2}{b^2} = 1 + 1 = 2$ ✓.

To derive rather than guess, multiply the second equation by $b$ and subtract it from the first:

$$\frac{x}{a} – \frac{bx}{a^2} = a + b – 2b \implies \frac{x(a – b)}{a^2} = a – b$$

so $x = a^2$ provided $a \ne b$, and then $y = b^2$.

(vii) $\dfrac{2xy}{x+y} = \dfrac32$ and $\dfrac{xy}{2x-y} = -\dfrac{3}{10}$, where $x + y \ne 0$ and $2x – y \ne 0$

Invert both equations, which turns them into something linear in $\frac1x$ and $\frac1y$:

$$\frac{x+y}{2xy} = \frac23 \implies \frac{1}{2y} + \frac{1}{2x} = \frac23 \implies u + v = \frac43$$

$$\frac{2x-y}{xy} = -\frac{10}{3} \implies \frac{2}{y} – \frac{1}{x} = -\frac{10}{3} \implies 2v – u = -\frac{10}{3}$$

writing $u = \frac1x$, $v = \frac1y$. Adding the two, $3v = \frac43 – \frac{10}{3} = -2$, so $v = -\frac23$ and $y = -\frac32$. Then $u = \frac43 + \frac23 = 2$, so $x = \frac12$.

Question 10

Find the solution of the pair of equations $\dfrac{x}{10} + \dfrac{y}{5} – 1 = 0$ and $\dfrac{x}{8} + \dfrac{y}{6} = 15$. Hence, find $\lambda$ if $y = \lambda x + 5$.

Solution. Clear the denominators. Multiplying the first by $10$ and the second by $24$:

$$x + 2y = 10, \qquad 3x + 4y = 360$$

From the first, $x = 10 – 2y$. Substituting into the second:

$$3(10 – 2y) + 4y = 360 \implies 30 – 2y = 360 \implies y = -165$$

and then $x = 10 – 2(-165) = 340$.

Now impose $y = \lambda x + 5$ at this point:

$$-165 = 340\lambda + 5 \implies 340\lambda = -170 \implies \lambda = -\frac12$$

Question 11

By the graphical method, find whether the following pairs of equations are consistent or not. If consistent, solve them.

(i) $3x + y + 4 = 0$ and $6x – 2y + 4 = 0$

$$\frac{a_1}{a_2} = \frac{3}{6} = \frac12, \qquad \frac{b_1}{b_2} = \frac{1}{-2}$$

Different, so the lines intersect: consistent, with a unique solution. From the first, $y = -3x – 4$; substituting,

$$6x – 2(-3x – 4) + 4 = 12x + 12 = 0 \implies x = -1, \quad y = -1$$

Graphically, the first line passes through $(0,-4)$ and $(-2,2)$, the second through $(0,2)$ and $(-2,-4)$, and they cross at $(-1,-1)$.

(ii) $x – 2y = 6$ and $3x – 6y = 0$

$$\frac{a_1}{a_2} = \frac13, \qquad \frac{b_1}{b_2} = \frac{-2}{-6} = \frac13, \qquad \frac{c_1}{c_2} = \frac{-6}{0}$$

The first two agree; the constant terms are $-6$ and $0$, which are certainly not in the ratio $\frac13$ (the second line passes through the origin and the first does not). The lines are parallel: inconsistent, no solution.

(iii) $x + y = 3$ and $3x + 3y = 9$

The second equation is exactly $3$ times the first, so all three ratios are $\frac13$: the two equations describe the same line. The pair is consistent with infinitely many solutions, every point of $y = 3 – x$ being one — $(0,3)$, $(1,2)$, $(3,0)$ and so on.

Question 12

Draw the graph of the pair of equations $2x + y = 4$ and $2x – y = 4$. Write the vertices of the triangle formed by these lines and the $y$-axis. Also find the area of this triangle.

Solution. Find the three vertices as the three pairwise intersections.

• The two lines meet where $2x + y = 2x – y$, so $y = 0$ and $2x = 4$: the point $(2, 0)$.
• $2x + y = 4$ meets the $y$-axis at $x = 0$, giving $y = 4$: the point $(0, 4)$.
• $2x – y = 4$ meets the $y$-axis at $x = 0$, giving $-y = 4$: the point $(0, -4)$.

The triangle has vertices $(2,0)$, $(0,4)$ and $(0,-4)$. Its base is the segment of the $y$-axis from $(0,-4)$ to $(0,4)$, of length $8$, and the height is the horizontal distance from that axis to $(2,0)$, which is $2$.

$$\text{Area} = \frac12 \times 8 \times 2 = 8 \text{ square units}$$

Question 13

Write an equation of a line passing through the point representing the solution of the pair of linear equations $x + y = 2$ and $2x – y = 1$. How many such lines can we find?

Solution. First find the point. Adding the two equations, $3x = 3$, so $x = 1$ and then $y = 1$: the solution is $(1, 1)$.

Any line through $(1,1)$ will do. The simplest is

$$x = y$$

which clearly passes through it. Others are $x + y = 2$ itself, $y = 3x – 2$, $2x + 5y = 7$, and so on.

How many? Infinitely many — through any single point there pass infinitely many lines, one for each direction.

Question 14

If $x + 1$ is a factor of $2x^3 + ax^2 + 2bx + 1$, then find the values of $a$ and $b$ given that $2a – 3b = 4$.

Solution. Two conditions, two unknowns. The first comes from the factor theorem: if $x+1$ is a factor then $p(-1) = 0$.

$$2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 \implies -2 + a – 2b + 1 = 0 \implies a – 2b = 1$$

Together with the given $2a – 3b = 4$, this is an ordinary linear pair. From the first, $a = 1 + 2b$; substituting,

$$2(1 + 2b) – 3b = 4 \implies 2 + b = 4 \implies b = 2, \quad a = 5$$

Check: $2x^3 + 5x^2 + 4x + 1$ at $x = -1$ gives $-2 + 5 – 4 + 1 = 0$ ✓, and $2(5) – 3(2) = 4$ ✓.

Question 15

The angles of a triangle are $x$, $y$ and $40^\circ$. The difference between the two angles $x$ and $y$ is $30^\circ$. Find $x$ and $y$.

Solution. The angle sum of a triangle gives the first equation, the stated difference the second:

$$x + y + 40 = 180 \implies x + y = 140, \qquad x – y = 30$$

Adding, $2x = 170$, so $x = 85^\circ$, and then $y = 140 – 85 = 55^\circ$.

Question 16

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

Solution. Let Salim be $s$ and his daughter $d$ years old now, and translate each clause with the ages shifted.

$$\text{Two years ago: } s – 2 = 3(d – 2) \implies s = 3d – 4$$ $$\text{Six years later: } s + 6 = 2(d + 6) + 4 \implies s = 2d + 10$$

Setting the two expressions for $s$ equal:

$$3d – 4 = 2d + 10 \implies d = 14, \quad s = 38$$

Check: two years ago they were $36$ and $12$, and $36 = 3 \times 12$ ✓. In six years they will be $44$ and $20$, and $44 = 2(20) + 4$ ✓.

Question 17

The age of the father is twice the sum of the ages of his two children. After $20$ years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Solution. The individual children’s ages never matter — only their sum does, so use a single letter for it. Let the father be $F$ and let $S$ be the sum of the two children’s ages now.

$$F = 2S$$

After $20$ years the father is $F + 20$, and each child is $20$ years older, so their sum is $S + 40$:

$$F + 20 = S + 40$$

Substituting $F = 2S$: $2S + 20 = S + 40$, so $S = 20$ and $F = 40$.

The $40$ rather than $20$ in the second equation is the whole question. Two children age forty years between them in twenty years.

Question 18

Two numbers are in the ratio $5 : 6$. If $8$ is subtracted from each of the numbers, the ratio becomes $4 : 5$. Find the numbers.

Solution. A ratio of $5 : 6$ means the numbers are $5k$ and $6k$ for some $k$ — one unknown rather than two, which is the efficient setup.

$$\frac{5k – 8}{6k – 8} = \frac45 \implies 5(5k – 8) = 4(6k – 8) \implies 25k – 40 = 24k – 32 \implies k = 8$$

The numbers are $5(8) = 40$ and $6(8) = 48$.

Check: $\frac{40 – 8}{48 – 8} = \frac{32}{40} = \frac45$ ✓.

Question 19

There are some students in the two examination halls A and B. To make the number of students equal in each hall, $10$ students are sent from A to B. But if $20$ students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.

Solution. Let A and B hold $a$ and $b$ students. Each transfer changes both counts, which is the step that is easy to half-do.

$$\text{10 from A to B: } a – 10 = b + 10 \implies a = b + 20$$ $$\text{20 from B to A: } a + 20 = 2(b – 20)$$

Substituting the first into the second:

$$b + 20 + 20 = 2b – 40 \implies b = 80, \quad a = 100$$

Check: $100 – 10 = 90$ and $80 + 10 = 90$ ✓; $100 + 20 = 120$ and $2(80 – 20) = 120$ ✓.

Question 20

A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹22 for a book kept for six days, while Anand paid ₹16 for the book kept for four days. Find the fixed charges and the charge for each extra day.

Solution. Let the fixed charge (covering the first two days) be ₹$f$ and the charge per extra day be ₹$e$. A book kept $n$ days costs $f + (n-2)e$, so:

$$\text{Latika, 6 days: } f + 4e = 22$$ $$\text{Anand, 4 days: } f + 2e = 16$$

Subtracting, $2e = 6$, so $e = 3$ and then $f = 16 – 6 = 10$.

The “$-2$” is the point of the question: six days means four extra days, not six.

Question 21

In a competitive examination, one mark is awarded for each correct answer while $\frac12$ mark is deducted for every wrong answer. Jayanti answered $120$ questions and got $90$ marks. How many questions did she answer correctly?

Solution. Let $c$ be the number correct and $w$ the number wrong. She attempted every question, so

$$c + w = 120, \qquad c – \frac{w}{2} = 90$$

From the first, $w = 120 – c$. Substituting into the second and multiplying by $2$:

$$2c – (120 – c) = 180 \implies 3c = 300 \implies c = 100$$

So she answered $100$ correctly and $20$ wrongly — and indeed $100 – 10 = 90$ ✓.

Question 22

The angles of a cyclic quadrilateral ABCD are $\angle A = (6x + 10)^\circ$, $\angle B = (5x)^\circ$, $\angle C = (x + y)^\circ$, $\angle D = (3y – 10)^\circ$. Find $x$ and $y$, and hence the values of the four angles.

Solution. In a cyclic quadrilateral opposite angles are supplementary, so $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$. That is the only fact needed, and it gives exactly two equations.

$$(6x + 10) + (x + y) = 180 \implies 7x + y = 170$$ $$5x + (3y – 10) = 180 \implies 5x + 3y = 190$$

From the first, $y = 170 – 7x$. Substituting:

$$5x + 3(170 – 7x) = 190 \implies 5x + 510 – 21x = 190 \implies -16x = -320 \implies x = 20$$

and then $y = 170 – 140 = 30$. The angles are

$$\angle A = 6(20) + 10 = 130^\circ, \quad \angle B = 5(20) = 100^\circ$$ $$\angle C = 20 + 30 = 50^\circ, \quad \angle D = 3(30) – 10 = 80^\circ$$

Two checks: $130 + 50 = 180$ and $100 + 80 = 180$ ✓, and the four angles sum to $360^\circ$ as any quadrilateral’s must ✓.

Common mistakes

• Stopping at $\lambda^2 = 1$ in Question 1. Both roots satisfy the first condition; only the constant ratio separates “no solution” from “infinitely many”. The same structure decides Question 2 ($k = \pm6$) and Question 4(ii) ($p = \pm1$), and in each case one of the two roots is the wrong one.
• Not collecting the $y$ terms in Question 4(v). $2px + py = 28 – qy$ has a $y$ term on each side; until they are gathered, $b_2$ is not $p$ but $p + q$.
• Forgetting to invert in Question 9(iii), (iv) and (vii). Solving for $u$ and $v$ is most of the work, but the answer asked for is $x$ and $y$. Writing $u = \frac1x$ at the top of the working is the cheapest way to remember.
• Reading “six days” as six extra days in Question 20. The fixed charge covers the first two, so six days means four extra. The pair still solves, to a different and wrong answer.
• Changing only one hall’s count in Question 19. Sending ten students from A to B takes ten off A and puts ten on B; $a – 10 = b$ loses the second half.
• Ageing only the father in Question 17. Two children gain forty years between them in twenty years, not twenty. This single factor is what the question is built on.
• Using the angle sum instead of opposite angles in Question 22. The four angles do sum to $360^\circ$, but that is one equation for two unknowns. It is cyclic that supplies the second, through supplementary opposite angles.
• Assuming $a \ne b$ silently in Question 9(vi). The division by $a – b$ needs it. Checking the proposed answer $x = a^2$, $y = b^2$ directly in both equations avoids the issue entirely, and takes one line.

Practise next

• Exemplar Exercise 3.4 — the long word problems, where the equation-forming of Questions 16 to 22 above is set at full length.
• Exemplar Exercises 3.1 and 3.2 for the consistency table that Questions 1 to 5 above rest on.
• NCERT textbook Exercise 3.2 and Exercise 3.3 — substitution and elimination on the book’s own problems, including the ages and fractions this exercise extends.
• Exemplar Exercises 2.3 and 2.4 for the factor theorem used in Question 14.