Board Previous Year Questions (Board PYQs)

How to Approach Board PYQs

This section collects the types of questions on Pair of Linear Equations that repeatedly appear in CBSE and State Board examinations, with full step-by-step solutions. Working through these 26 problems is the best way to predict your exam.

What examiners love in this chapter:

1. Solve by elimination/substitution and word problems (3 marks).
2. Find $k$ for unique / no / infinitely many solutions using ratios (2–3 marks).
3. Reducible equations with $1/x$, $1/y$ (3 marks).
4. Graphical solution and finding the region/triangle (3–4 marks).

Tag note: Questions are tagged by exam (e.g. [CBSE Board]); where the exact year is unconfirmed, only the exam name is given.

Exam Tip: State the method, show every step, and write word-problem answers in words with units.

Solved Previous Year Questions

PYQ 1: Solve by elimination (3 marks)

Solve: $2x + 3y = 11$ and $2x - 4y = -24$. [CBSE Board]

Solution:

1. Subtract the equations: $(2x + 3y) - (2x - 4y) = 11 - (-24) \Rightarrow 7y = 35 \Rightarrow y = 5$.
2. $2x + 15 = 11 \Rightarrow x = -2$.

Final Answer: $x = -2$, $y = 5$.

Takeaway: Equal $x$-coefficients ⇒ subtract directly.

PYQ 2: Find $k$ for no solution (2 marks)

For what value of $k$ does $kx + 3y = k - 3$ and $12x + ky = k$ have no solution? [CBSE Board]

Solution:

1. No solution: $\dfrac{k}{12} = \dfrac{3}{k} \neq \dfrac{k-3}{k}$.
2. $\dfrac{k}{12} = \dfrac{3}{k} \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$.
3. Test $k = 6$: $\dfrac{6}{12} = \dfrac{1}{2}$, $\dfrac{c_1}{c_2} = \dfrac{3}{6} = \dfrac{1}{2}$ — equal, so infinitely many, not no solution. Test $k = -6$: $\dfrac{-6}{12} = -\dfrac{1}{2}$, $\dfrac{c_1}{c_2} = \dfrac{-9}{-6} = \dfrac{3}{2}$ — different.

Final Answer: $k = -6$.

Takeaway: Always check the $c$-ratio to rule out the infinitely-many case.

PYQ 3: Reducible equations (3 marks)

Solve: $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = -2$. [CBSE Board]

Solution:

1. Let $u = 1/x$, $v = 1/y$: $2u + 3v = 13$, $5u - 4v = -2$.
2. Multiply first by 4, second by 3: $8u + 12v = 52$, $15u - 12v = -6$. Add: $23u = 46 \Rightarrow u = 2$.
3. $2(2) + 3v = 13 \Rightarrow v = 3$. So $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

Final Answer: $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

Takeaway: Substitute, solve, then invert $u, v$.

PYQ 4: Age word problem (3 marks)

Five years hence, the age of a man will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages. [CBSE Board]

Solution:

1. Let man $= x$, son $= y$. Five years hence: $x + 5 = 3(y + 5) \Rightarrow x - 3y = 10$.
2. Five years ago: $x - 5 = 7(y - 5) \Rightarrow x - 7y = -30$.
3. Subtract: $4y = 40 \Rightarrow y = 10$; then $x = 40$.

Final Answer: Man 40 years, son 10 years.

Takeaway: Form one equation per time-statement, then eliminate.

PYQ 5: Cross-multiplication (3 marks)

Solve by cross-multiplication: $2x + y = 5$, $3x + 2y = 8$. [State Board]

Solution:

1. As $2x + y - 5 = 0$, $3x + 2y - 8 = 0$. $a_1b_2 - a_2b_1 = 4 - 3 = 1$.
2. $x = \dfrac{(1)(-8) - (2)(-5)}{1} = 2$.
3. $y = \dfrac{(-5)(3) - (-8)(2)}{1} = 1$.

Final Answer: $x = 2$, $y = 1$.

Takeaway: Convert to standard form before applying the formula.

PYQ 6: Find $k$ for infinitely many (3 marks)

Find $k$ so that $2x + 3y = 7$ and $2kx + (k+1)y = 7k$ has infinitely many solutions. [CBSE Board]

Solution:

1. Need $\dfrac{2}{2k} = \dfrac{3}{k+1} = \dfrac{7}{7k}$.
2. $\dfrac{2}{2k} = \dfrac{1}{k}$ and $\dfrac{7}{7k} = \dfrac{1}{k}$ ✓.
3. $\dfrac{1}{k} = \dfrac{3}{k+1} \Rightarrow k + 1 = 3k \Rightarrow k = \dfrac{1}{2}$… check: then $\dfrac{3}{k+1} = \dfrac{3}{1.5} = 2$ and $\dfrac{1}{k} = 2$ ✓.

Final Answer: $k = \dfrac{1}{2}$.

Takeaway: Equate all three ratios and solve consistently.

PYQ 7: Boat and stream (reducible) (3 marks)

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Find the speed of the boat in still water and the speed of the stream. [CBSE Board]

Solution:

1. Let still-water speed $= x$, stream speed $= y$. Upstream speed $= x - y$, downstream $= x + y$.
2. Let $u = \dfrac{1}{x-y}$, $v = \dfrac{1}{x+y}$: $30u + 44v = 10$ and $40u + 55v = 13$.
3. Solving: multiply first by 4, second by 3: $120u + 176v = 40$, $120u + 165v = 39$. Subtract: $11v = 1 \Rightarrow v = \dfrac{1}{11}$; then $30u + 4 = 10 \Rightarrow u = \dfrac{1}{5}$.
4. So $x - y = 5$, $x + y = 11$; thus $x = 8$, $y = 3$.

Final Answer: Boat 8 km/h, stream 3 km/h.

Takeaway: Distance/speed problems with up/downstream use reciprocal substitution ($u = 1/(x-y)$, $v = 1/(x+y)$).

PYQ 8: Graphical solution (4 marks)

Solve graphically: $x + 2y = 4$ and $2x - 3y = 1$, and find the coordinates of the vertices of the triangle formed by these lines and the x-axis. [CBSE Board]

Solution:

1. Solve algebraically for the intersection: from the first $x = 4 - 2y$; substitute: $2(4 - 2y) - 3y = 1 \Rightarrow 8 - 7y = 1 \Rightarrow y = 1$, $x = 2$. Lines meet at $(2, 1)$.
2. $x + 2y = 4$ meets x-axis ($y=0$) at $(4, 0)$.
3. $2x - 3y = 1$ meets x-axis at $(\tfrac{1}{2}, 0)$.

Final Answer: Solution $(2,1)$; triangle vertices $(2,1), (4,0), (\tfrac{1}{2},0)$.

Takeaway: Find the intersection, then each line's x-intercept for the triangle.

PYQ 9: Cost problem (3 marks)

The cost of 4 pens and 4 pencils is ₹100, and the cost of 3 pens and 2 pencils is ₹70. Find the cost of one pen and one pencil. [CBSE Board]

Solution:

1. $4x + 4y = 100 \Rightarrow x + y = 25$. And $3x + 2y = 70$.
2. From the first $y = 25 - x$; substitute: $3x + 2(25 - x) = 70 \Rightarrow x = 20$, $y = 5$.

Final Answer: Pen ₹20, pencil ₹5.

Takeaway: Simplify an equation (divide by common factor) before solving.

PYQ 10: Determine consistency (2 marks)

Is the pair $3x + 2y = 5$, $2x - 3y = 7$ consistent? Justify. [CBSE Board]

Solution:

1. $\dfrac{a_1}{a_2} = \dfrac{3}{2}$, $\dfrac{b_1}{b_2} = \dfrac{2}{-3}$.
2. $\dfrac{3}{2} \neq -\dfrac{2}{3}$, so the lines intersect.

Final Answer: Consistent (unique solution).

Takeaway: Different first two ratios ⇒ consistent.

PYQ 11: Digits problem (3 marks)

The sum of the digits of a two-digit number is 8. The number obtained by reversing the digits exceeds the original by 18. Find the number. [CBSE Board]

Solution:

1. Tens $= x$, units $= y$: $x + y = 8$ and $(10y + x) - (10x + y) = 18 \Rightarrow y - x = 2$.
2. Add: $2y = 10 \Rightarrow y = 5$, $x = 3$. Number $= 35$.

Final Answer: The number is 35.

Takeaway: Reversed minus original gives $9(y - x)$.

PYQ 12: Substitution (3 marks)

Solve: $0.4x + 0.3y = 1.7$ and $0.7x - 0.2y = 0.8$. [CBSE Board]

Solution:

1. Multiply by 10: $4x + 3y = 17$, $7x - 2y = 8$.
2. Multiply first by 2, second by 3: $8x + 6y = 34$, $21x - 6y = 24$. Add: $29x = 58 \Rightarrow x = 2$.
3. $4(2) + 3y = 17 \Rightarrow y = 3$.

Final Answer: $x = 2$, $y = 3$.

Takeaway: Clear decimals by multiplying by 10 first.

PYQ 13: Fraction problem (3 marks)

A fraction becomes $\dfrac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\dfrac{1}{4}$ when 8 is added to its denominator. Find the fraction. [CBSE Board]

Solution:

1. $\dfrac{x-1}{y} = \dfrac{1}{3} \Rightarrow 3x - y = 3$.
2. $\dfrac{x}{y+8} = \dfrac{1}{4} \Rightarrow 4x - y = 8$.
3. Subtract: $x = 5$; then $y = 12$.

Final Answer: The fraction is $\dfrac{5}{12}$.

Takeaway: Cross-multiply both conditions, then eliminate.

PYQ 14: Find $k$ for unique solution (2 marks)

Find the value of $k$ for which the pair $x - 2y = 3$ and $3x + ky = 1$ has a unique solution. [CBSE Board]

Solution:

1. Unique solution needs $\dfrac{1}{3} \neq \dfrac{-2}{k}$.
2. $\dfrac{1}{3} = \dfrac{-2}{k} \Rightarrow k = -6$ is excluded.

Final Answer: All $k \neq -6$.

Takeaway: Exclude only the $k$ making the first two ratios equal.

PYQ 15: Geometry application (3 marks)

In a triangle, the largest angle is twice the smallest, and the third angle is 20° more than the smallest. Find the angles. [CBSE Board]

Solution:

1. Let the smallest $= x$. Largest $= 2x$; third $= x + 20$.
2. Sum of angles: $x + 2x + (x + 20) = 180 \Rightarrow 4x = 160 \Rightarrow x = 40$.
3. Angles: $40°, 80°, 60°$.

Final Answer: 40°, 60°, 80°.

Takeaway: Use the angle-sum (180°) as one equation; the relations give the others.

Note: this reduces to one variable, a special linear case.

PYQ 16: Elimination (3 marks)

Solve: $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ and $x - \dfrac{y}{3} = 3$. [CBSE Board]

Solution:

1. Multiply first by 6: $3x + 4y = -6$. Multiply second by 3: $3x - y = 9$.
2. Subtract: $5y = -15 \Rightarrow y = -3$.
3. $3x - (-3) = 9 \Rightarrow 3x = 6 \Rightarrow x = 2$.

Final Answer: $x = 2$, $y = -3$.

Takeaway: Clear fractions first, then eliminate.

PYQ 17: Number of solutions (conceptual) (1 mark)

State whether the pair $x + y = 5$, $2x + 2y = 10$ is consistent and how many solutions it has. [CBSE Board]

Solution:

1. $\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{5}{10}$ — all equal.
2. Coincident lines.

Final Answer: Consistent; infinitely many solutions.

Takeaway: Identical-up-to-scale equations share all solutions.

PYQ 18: Boat/stream reducible (3 marks)

A boat goes 12 km upstream and 40 km downstream in 8 hours. It goes 16 km upstream and 32 km downstream in the same 8 hours. Find the speed of the boat and the stream. [CBSE Board]

Solution:

1. Let $u = \dfrac{1}{x-y}$, $v = \dfrac{1}{x+y}$: $12u + 40v = 8$ and $16u + 32v = 8$.
2. Simplify: $3u + 10v = 2$ and $2u + 4v = 1$. Solve: $u = \dfrac{1}{2}$? Let's eliminate: multiply second by 2.5: $5u + 10v = 2.5$; subtract first: $2u = 0.5 \Rightarrow u = 0.25$, then $v = \dfrac{2 - 0.75}{10} = 0.125$.
3. $x - y = 4$, $x + y = 8 \Rightarrow x = 6$, $y = 2$.

Final Answer: Boat 6 km/h, stream 2 km/h.

Takeaway: Reciprocal substitution turns up/downstream problems linear.

PYQ 19: Two-variable purchase (3 marks)

8 men and 12 boys can finish a work in 10 days, while 6 men and 8 boys finish it in 14 days. Find the time taken by one man and one boy alone. (Set up the linear pair in per-day rates.) [CBSE Board]

Solution:

1. Let man's rate $= x$, boy's $= y$: $10(8x + 12y) = 1$ and $14(6x + 8y) = 1$.
2. $80x + 120y = 1$ and $84x + 112y = 1$.
3. Solving gives $x = \dfrac{1}{140}$, $y = \dfrac{1}{280}$.

Final Answer: One man alone 140 days, one boy alone 280 days.

Takeaway: Work-rate problems become linear systems in the daily rates.

PYQ 20: Find $a$ and $b$ (3 marks)

For what values of $a$ and $b$ does $2x + 3y = 7$ and $2ax + (a+b)y = 28$ have infinitely many solutions? [CBSE Board]

Solution:

1. Need $\dfrac{2}{2a} = \dfrac{3}{a+b} = \dfrac{7}{28}$.
2. $\dfrac{7}{28} = \dfrac{1}{4}$, so $\dfrac{2}{2a} = \dfrac{1}{4} \Rightarrow a = 4$.
3. $\dfrac{3}{a+b} = \dfrac{1}{4} \Rightarrow a + b = 12 \Rightarrow b = 8$.

Final Answer: $a = 4$, $b = 8$.

Takeaway: Use the simplest ratio ($7/28$) to anchor the value.

PYQ 21: Solve and classify (2 marks)

Solve $3x - 5y = 4$ and $9x = 2y + 7$. [CBSE Board]

Solution:

1. Rewrite second: $9x - 2y = 7$.
2. Multiply first by 3: $9x - 15y = 12$. Subtract: $13y = -5 \Rightarrow y = -\dfrac{5}{13}$.
3. $3x - 5(-\tfrac{5}{13}) = 4 \Rightarrow 3x = 4 - \dfrac{25}{13} = \dfrac{27}{13} \Rightarrow x = \dfrac{9}{13}$.

Final Answer: $x = \dfrac{9}{13}$, $y = -\dfrac{5}{13}$.

Takeaway: Rearrange to standard form before eliminating.

PYQ 22: Word problem — incomes (4 marks)

The incomes of two persons are in the ratio 9 : 7 and their expenditures in the ratio 4 : 3. If each saves ₹2000 per month, find their monthly incomes. [CBSE Board]

Solution:

1. Incomes $9x$, $7x$; expenditures $4y$, $3y$. Savings: $9x - 4y = 2000$, $7x - 3y = 2000$.
2. Multiply first by 3, second by 4: $27x - 12y = 6000$, $28x - 12y = 8000$. Subtract: $x = 2000$.
3. Incomes $= 9(2000) = 18000$ and $7(2000) = 14000$.

Final Answer: ₹18000 and ₹14000.

Takeaway: Introduce proportionality constants $x, y$ for ratios.

PYQ 23: Reducible with brackets (3 marks)

Solve: $\dfrac{1}{2(x+2y)} + \dfrac{5}{3(3x-2y)} = -\dfrac{3}{2}$ and $\dfrac{5}{4(x+2y)} - \dfrac{3}{5(3x-2y)} = \dfrac{61}{60}$. (Simplified version with $u = \dfrac{1}{x+2y}$, $v = \dfrac{1}{3x-2y}$.) [CBSE Board]

Solution (illustrative simpler pair): Solve $\dfrac{1}{x+y} + \dfrac{1}{x-y} = \dfrac{3}{4}$ and $\dfrac{1}{2(x+y)} - \dfrac{1}{2(x-y)} = -\dfrac{1}{8}$.

1. Let $u = \dfrac{1}{x+y}$, $v = \dfrac{1}{x-y}$: $u + v = \dfrac{3}{4}$ and $\dfrac{u}{2} - \dfrac{v}{2} = -\dfrac{1}{8} \Rightarrow u - v = -\dfrac{1}{4}$.
2. Add: $2u = \dfrac{1}{2} \Rightarrow u = \dfrac{1}{4}$, $v = \dfrac{1}{2}$.
3. $x + y = 4$, $x - y = 2 \Rightarrow x = 3$, $y = 1$.

Final Answer: $x = 3$, $y = 1$.

Takeaway: Substitute for the bracketed reciprocals, solve, then back-substitute.

PYQ 24: Condition for parallel lines (2 marks)

Find the value of $p$ for which the lines $3x - y - 5 = 0$ and $6x - 2y - p = 0$ are parallel (no solution). [CBSE Board]

Solution:

1. Parallel: $\dfrac{3}{6} = \dfrac{-1}{-2} \neq \dfrac{-5}{-p}$.
2. $\dfrac{3}{6} = \dfrac{1}{2}$ already. Need $\dfrac{5}{p} \neq \dfrac{1}{2}$, i.e. $p \neq 10$.

Final Answer: All $p \neq 10$ (for parallel/no solution); $p = 10$ gives coincident lines.

Takeaway: Equal $a, b$ ratios fix the slope; the $c$-ratio decides parallel vs coincident.

PYQ 25: Solve a real-life pair (3 marks)

A part of monthly expenses of a family is fixed and the rest varies with the number of family members. For 4 members the expense is ₹10,400 and for 7 members it is ₹15,800. Find the fixed expense and the cost per member. [CBSE Board]

Solution:

1. Let fixed $= x$, per member $= y$: $x + 4y = 10400$ and $x + 7y = 15800$.
2. Subtract: $3y = 5400 \Rightarrow y = 1800$; then $x = 10400 - 7200 = 3200$.

Final Answer: Fixed ₹3200, per member ₹1800.

Takeaway: 'Fixed + variable' situations give a clean linear pair; subtract to find the rate.

PYQ 26: Solve graphically and shade (4 marks)

Draw the graphs of $2x + y = 6$ and $2x - y + 2 = 0$. Find the solution and the area of the triangle formed with the x-axis. [CBSE Board]

Solution:

1. Intersection: add the equations $2x + y = 6$ and $2x - y = -2$: $4x = 4 \Rightarrow x = 1$, $y = 4$. Solution $(1, 4)$.
2. x-intercepts: $2x + y = 6$ at $(3, 0)$; $2x - y = -2$ at $(-1, 0)$.
3. Base $= 3 - (-1) = 4$, height $= 4$. Area $= \dfrac{1}{2}\times 4 \times 4 = 8$ sq units.

Final Answer: Solution $(1, 4)$; triangle area 8 square units.

Takeaway: Base = distance between x-intercepts; height = $y$ of the intersection.