Board Previous Year Questions (Board PYQs)

How to Approach Board PYQs

This section collects the types of Quadratic Equation questions 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 factorisation / formula and word problems (3 marks).
2. Nature of roots using the discriminant; find $k$ for equal/real/no roots (2–3 marks).
3. Speed–time, area, age, number word problems (3–4 marks).
4. Reciprocal and reducible equations (3 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, reject inadmissible roots, and write word-problem answers in words with units.

Solved Previous Year Questions

PYQ 1: Solve by factorisation (3 marks)

Solve $2x^2 - 5x + 3 = 0$ by factorisation. [CBSE Board]

Solution:

1. $a \times c = 6$; split $-5$ as $-2, -3$.
2. $2x^2 - 2x - 3x + 3 = (x - 1)(2x - 3) = 0$.
3. $x = 1, \dfrac{3}{2}$.

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

Takeaway: Standard middle-term split.

PYQ 2: Nature of roots (2 marks)

Find the nature of the roots of $2x^2 - 3x + 5 = 0$. [CBSE Board]

Solution:

1. $D = 9 - 40 = -31 < 0$.

Final Answer: No real roots.

Takeaway: $D < 0$ ⇒ no real roots.

PYQ 3: Equal roots — find $k$ (2 marks)

Find $k$ for which $2x^2 + kx + 3 = 0$ has two equal roots. [CBSE Board]

Solution:

1. $D = k^2 - 4(2)(3) = k^2 - 24 = 0$.
2. $k = \pm\sqrt{24} = \pm 2\sqrt6$.

Final Answer: $k = \pm 2\sqrt6$.

Takeaway: Equal roots ⇒ $D = 0$; simplify the surd.

PYQ 4: Quadratic formula (3 marks)

Solve $2x^2 + x - 4 = 0$ using the quadratic formula. [CBSE Board]

Solution:

1. $D = 1 + 32 = 33$.
2. $x = \dfrac{-1 \pm \sqrt{33}}{4}$.

Final Answer: $x = \dfrac{-1 \pm \sqrt{33}}{4}$.

Takeaway: Leave surd roots in exact form.

PYQ 5: Speed–time word problem (4 marks)

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed of the train. [CBSE Board]

Solution:

1. $\dfrac{360}{x} - \dfrac{360}{x + 5} = 1 \Rightarrow x^2 + 5x - 1800 = 0$.
2. $(x + 45)(x - 40) = 0 \Rightarrow x = 40$.

Final Answer: Speed $= 40$ km/h.

Takeaway: Reject the negative speed.

PYQ 6: Area word problem (3 marks)

The area of a right triangle is 600 cm². If the base is 10 cm more than twice the height, find the dimensions. [CBSE Board]

Solution:

1. Let height $= x$; base $= 2x + 10$. $\dfrac{1}{2}x(2x + 10) = 600 \Rightarrow x^2 + 5x - 600 = 0$.
2. $(x + 30)(x - 20) = 0 \Rightarrow x = 20$.
3. Height 20 cm, base 50 cm.

Final Answer: Height 20 cm, base 50 cm.

Takeaway: Triangle area uses $\dfrac{1}{2}$; simplify before factorising.

PYQ 7: Reciprocal/number problem (3 marks)

The sum of a number and its reciprocal is $\dfrac{26}{5}$. Find the number. [CBSE Board]

Solution:

1. $x + \dfrac{1}{x} = \dfrac{26}{5} \Rightarrow 5x^2 - 26x + 5 = 0$.
2. $(5x - 1)(x - 5) = 0 \Rightarrow x = 5$ or $\dfrac{1}{5}$.

Final Answer: The number is 5 (or $\dfrac{1}{5}$).

Takeaway: Clear the denominator, then factorise.

PYQ 8: Consecutive integers (3 marks)

The sum of the squares of two consecutive odd positive integers is 290. Find them. [CBSE Board]

Solution:

1. Let them be $x$ and $x + 2$: $x^2 + (x+2)^2 = 290 \Rightarrow 2x^2 + 4x - 286 = 0 \Rightarrow x^2 + 2x - 143 = 0$.
2. $(x + 13)(x - 11) = 0 \Rightarrow x = 11$.
3. The integers are 11 and 13.

Final Answer: 11 and 13.

Takeaway: Consecutive odd integers differ by 2.

PYQ 9: Find $k$ for real roots (2 marks)

Find the values of $k$ for which $kx^2 - 4x + 1 = 0$ has real roots. [CBSE Board]

Solution:

1. Real roots ⇒ $D \geq 0$: $16 - 4k \geq 0$.
2. $k \leq 4$ (and $k \neq 0$ for it to be quadratic).

Final Answer: $k \leq 4$, $k \neq 0$.

Takeaway: 'Real roots' allows $D \geq 0$ (equal or distinct).

PYQ 10: Completing the square (3 marks)

Solve $4x^2 + 4\sqrt3 x + 3 = 0$ by completing the square. [CBSE Board]

Solution:

1. Divide by 4: $x^2 + \sqrt3 x + \dfrac{3}{4} = 0 \Rightarrow x^2 + \sqrt3 x = -\dfrac{3}{4}$.
2. Half of $\sqrt3$ is $\dfrac{\sqrt3}{2}$, square $\dfrac{3}{4}$: $\left(x + \dfrac{\sqrt3}{2}\right)^2 = 0$.
3. $x = -\dfrac{\sqrt3}{2}$ (equal roots).

Final Answer: $x = -\dfrac{\sqrt3}{2}$ (repeated).

Takeaway: A zero right side means equal roots.

PYQ 11: Boat and stream (4 marks)

A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. [CBSE Board]

Solution:

1. $\dfrac{24}{18 - x} - \dfrac{24}{18 + x} = 1 \Rightarrow x^2 + 48x - 324 = 0$.
2. $(x + 54)(x - 6) = 0 \Rightarrow x = 6$.

Final Answer: Stream speed $= 6$ km/h.

Takeaway: Upstream/downstream speeds are $18 \mp x$.

PYQ 12: Age word problem (3 marks)

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48. Find their present ages. [CBSE Board]

Solution:

1. Let ages be $x$ and $20 - x$. Four years ago: $(x - 4)(16 - x) = 48$.
2. $16x - x^2 - 64 + 4x = 48 \Rightarrow -x^2 + 20x - 112 = 0 \Rightarrow x^2 - 20x + 112 = 0$.
3. $D = 400 - 448 = -48 < 0$, so no real solution — such a situation is impossible.

Final Answer: No real solution (the situation cannot occur).

Takeaway: A negative discriminant shows the scenario is impossible.

PYQ 13: Solve by factorisation (surd) (3 marks)

Solve $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$. [CBSE Board]

Solution:

1. $a \times c = 21$; split 10 as $3 + 7$.
2. $(x + \sqrt3)(\sqrt3 x + 7) = 0$.
3. $x = -\sqrt3, -\dfrac{7\sqrt3}{3}$.

Final Answer: $x = -\sqrt3, -\dfrac{7\sqrt3}{3}$.

Takeaway: Surd factorisation; rationalise the final roots.

PYQ 14: Two pipes (4 marks)

Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time each takes. [CBSE Board]

Solution:

1. Let smaller take $x$ hours, larger $x - 10$. Combined: $\dfrac{1}{x} + \dfrac{1}{x - 10} = \dfrac{8}{75}$.
2. $75(2x - 10) = 8x(x - 10) \Rightarrow 8x^2 - 230x + 750 = 0 \Rightarrow 4x^2 - 115x + 375 = 0$.
3. $x = \dfrac{115 \pm \sqrt{13225 - 6000}}{8} = \dfrac{115 \pm 85}{8} = 25$ or $3.75$. Reject 3.75 (then $x - 10 < 0$).

Final Answer: Smaller 25 hours, larger 15 hours.

Takeaway: $9\tfrac{3}{8} = \tfrac{75}{8}$, so combined rate $= \tfrac{8}{75}$.

PYQ 15: Find $k$ for equal roots (parameter in $a$) (2 marks)

Find $k$ such that $(k + 1)x^2 - 2(k - 1)x + 1 = 0$ has equal roots. [CBSE Board]

Solution:

1. $D = [2(k-1)]^2 - 4(k+1)(1) = 0$.
2. $4(k^2 - 2k + 1) - 4(k + 1) = 0 \Rightarrow k^2 - 3k = 0 \Rightarrow k(k - 3) = 0$.
3. $k = 0$ or $k = 3$.

Final Answer: $k = 0$ or $k = 3$.

Takeaway: Expand $D = 0$; both values are valid (each keeps $a \neq 0$).

PYQ 16: Number problem (3 marks)

The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the larger. Find the numbers. [CBSE Board]

Solution:

1. Let larger $= x$, smaller $= y$: $x^2 - y^2 = 180$ and $y^2 = 8x$.
2. Substitute: $x^2 - 8x - 180 = 0 = (x - 18)(x + 10)$; $x = 18$.
3. $y^2 = 8(18) = 144 \Rightarrow y = 12$.

Final Answer: The numbers are 18 and 12.

Takeaway: Substitute the second condition to get a quadratic in one variable.

PYQ 17: Solve fractional equation (3 marks)

Solve $\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6}$, $x \neq 3, -5$. [CBSE Board]

Solution:

1. $\dfrac{(x+5) - (x-3)}{(x-3)(x+5)} = \dfrac{1}{6} \Rightarrow \dfrac{8}{x^2 + 2x - 15} = \dfrac{1}{6}$.
2. $48 = x^2 + 2x - 15 \Rightarrow x^2 + 2x - 63 = 0 = (x + 9)(x - 7)$.
3. $x = 7$ or $x = -9$.

Final Answer: $x = 7, -9$.

Takeaway: Combine the fractions, cross-multiply, and factorise.

PYQ 18: Show equal roots condition (3 marks)

If the roots of $(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$ are equal, show that $ad = bc$. [CBSE Board]

Solution:

1. $D = 0$: $4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$.
2. Expand: $(ac + bd)^2 = (a^2+b^2)(c^2+d^2)$.
3. This simplifies to $(ad - bc)^2 = 0 \Rightarrow ad = bc$.

Final Answer: $ad = bc$.

Takeaway: Setting $D = 0$ and expanding yields a perfect-square condition.

PYQ 19: Word problem — marks (3 marks)

In a class test, the sum of the marks obtained by P in Maths and Science is 28. Had he got 3 more in Maths and 4 less in Science, the product of the marks would have been 180. Find his original marks in each. [CBSE Board]

Solution:

1. Let Maths $= x$, Science $= 28 - x$. Condition: $(x + 3)(28 - x - 4) = 180 \Rightarrow (x + 3)(24 - x) = 180$.
2. $24x - x^2 + 72 - 3x = 180 \Rightarrow -x^2 + 21x - 108 = 0 \Rightarrow x^2 - 21x + 108 = 0$.
3. $(x - 9)(x - 12) = 0 \Rightarrow x = 9$ or $12$. So (Maths, Science) is (9, 19) or (12, 16).

Final Answer: Maths 9, Science 19 (or Maths 12, Science 16).

Takeaway: Two valid solutions can both be reported.

PYQ 20: Discriminant decides feasibility (2 marks)

Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its dimensions. [CBSE Board]

Solution:

1. Half-perimeter $= 40$; sides are roots of $t^2 - 40t + 400 = 0$.
2. $D = 1600 - 1600 = 0$, so equal roots: $t = 20$.
3. Yes — it is a square of side 20 m.

Final Answer: Yes; a 20 m × 20 m square.

Takeaway: $D = 0$ ⇒ a unique (here square) solution.

PYQ 21: Solve $a^2x^2 - 3abx + 2b^2 = 0$ (3 marks)

Solve for $x$: $a^2x^2 - 3abx + 2b^2 = 0$. [CBSE Board]

Solution:

1. Split $-3ab$ as $-ab - 2ab$: $a^2x^2 - abx - 2abx + 2b^2 = ax(ax - b) - 2b(ax - b) = (ax - b)(ax - 2b) = 0$.
2. $x = \dfrac{b}{a}$ or $x = \dfrac{2b}{a}$.

Final Answer: $x = \dfrac{b}{a}, \dfrac{2b}{a}$.

Takeaway: Treat $a$, $b$ as constants and factorise normally.

PYQ 22: Train word problem (distance fixed) (4 marks)

The time taken by a person to cover 150 km was 2.5 hours more than the time taken in the return journey. If he returned at a speed 10 km/h more than the onward speed, find the onward speed. [CBSE Board]

Solution:

1. Onward speed $= x$, return $= x + 10$. $\dfrac{150}{x} - \dfrac{150}{x + 10} = 2.5$.
2. $150(10) = 2.5x(x + 10) \Rightarrow 1500 = 2.5x^2 + 25x \Rightarrow x^2 + 10x - 600 = 0$.
3. $(x + 30)(x - 20) = 0 \Rightarrow x = 20$.

Final Answer: Onward speed $= 20$ km/h.

Takeaway: Set up the time difference; reject the negative speed.

PYQ 23: Nature with a parameter (2 marks)

For what value of $k$ does $x^2 + 5kx + 16 = 0$ have no real roots? [CBSE Board]

Solution:

1. No real roots ⇒ $D < 0$: $(5k)^2 - 64 < 0$.
2. $25k^2 < 64 \Rightarrow -\dfrac{8}{5} < k < \dfrac{8}{5}$.

Final Answer: $-\dfrac{8}{5} < k < \dfrac{8}{5}$.

Takeaway: $D < 0$ gives an interval for $k$.

PYQ 24: Solve by formula (decimals) (3 marks)

Solve $0.2x^2 + 0.3x - 1 = 0$ (or equivalently $2x^2 + 3x - 10 = 0$). [State Board]

Solution:

1. Multiply by 5: $x^2 + 1.5x - 5 = 0$, or directly use $2x^2 + 3x - 10 = 0$: $D = 9 + 80 = 89$.
2. $x = \dfrac{-3 \pm \sqrt{89}}{4}$.

Final Answer: $x = \dfrac{-3 \pm \sqrt{89}}{4}$.

Takeaway: Clear decimals first, then apply the formula.

PYQ 25: Rectangular field (3 marks)

The diagonal of a rectangular field is 60 m more than the shorter side. The longer side is 30 m more than the shorter side. Find the sides. [CBSE Board]

Solution:

1. Let shorter side $= x$. Longer $= x + 30$, diagonal $= x + 60$.
2. Pythagoras: $x^2 + (x + 30)^2 = (x + 60)^2$.
3. $x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600 \Rightarrow x^2 - 60x - 2700 = 0 = (x - 90)(x + 30)$; $x = 90$.
4. Sides: 90 m and 120 m (diagonal 150 m).

Final Answer: Shorter 90 m, longer 120 m.

Takeaway: Set up Pythagoras with all sides in terms of $x$.

PYQ 26: Solve and find both roots' use (3 marks)

Solve $x^2 - (\sqrt3 + 1)x + \sqrt3 = 0$. [CBSE Board]

Solution:

1. Split: $x^2 - \sqrt3 x - x + \sqrt3 = x(x - \sqrt3) - 1(x - \sqrt3) = (x - \sqrt3)(x - 1) = 0$.
2. $x = \sqrt3$ or $x = 1$.

Final Answer: $x = \sqrt3, 1$.

Takeaway: Group the surd terms when splitting the middle term.