Quadratic Equations: CBSE Class 10 Previous Year Questions

209 different questions from Quadratic Equations (NCERT Chapter 4) asked in CBSE Class 10 Maths board exams 2022–2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions73 questions 2 marks questions38 questions 3 marks questions23 questions 4 marks questions13 questions 5 marks questions62 questions

Most asked Quadratic Equations questions

Q21 markMCQQuadratic Equations CBSE 2026 · Basic 430/5/1

The roots of the quadratic equation $x^{2} + 9 = 0$ are

(A)real and equal
(B)not real
(C)real and negative of each other
(D)rational numbers

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Answer: (B) not real

Here $a = 1$ , $b = 0$ , $c = 9$ .
$D = b^{2} - 4 a c = 0 - 36 = - 36 < 0$ .
So the roots are not real.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q101 markMCQQuadratic Equations CBSE 2026 · Standard 30/3/1

If the roots of the quadratic equation $3 x^{2} - k x + 23 = 0$ are real and equal, then the value(s) of $k$ is/are :

(A) $\pm 24$
(B)0
(C)4
(D)– 5

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Answer: (A)

\pm 24

Equal roots: $D = b^{2} - 4 a c = 0$
$k^{2} - 4 \times 3 \times 23 = 0$ , so $k^{2} = 24$
$k = \pm 24$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q91 markMCQQuadratic Equations CBSE 2026 · Standard 30/5/1

The value of p for which roots of the quadratic equation $x^{2} - p x + 6 = 0$ are rational, is

(A)1
(B) $- 5$
(C)25
(D) $5$

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Answer: (B)

- 5

Roots are rational when $D = p^{2} - 24$ is a perfect square (and p is rational).
$p = 1$ : $D = - 23$ (no real roots).
$p = - 5$ : $D = 25 - 24 = 1 = 1^{2}$ , a perfect square; $x^{2} + 5 x + 6 = 0$ gives $x = - 2, - 3$ , which are rational.
$p = 25$ : $D = 601$ , not a perfect square. $p = 5$ : $D = - 19$ .
So $p = - 5$ .

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q212 marksVery Short AnswerQuadratic Equations CBSE 2026 · Standard 30/4/1

Verify that roots of the quadratic equation $(p - q) x^{2} + (q - r) x + (r - p) = 0$ are equal when $q + r = 2 p$ .

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Answer: Proved.

$D = (q - r)^{2} - 4 (p - q) (r - p)$
Given $q + r = 2 p$ , so $p = \frac{q + r}{2}$ .
Then $p - q = \frac{r - q}{2}$ and $r - p = \frac{r - q}{2}$ .
$D = (q - r)^{2} - 4 \times \frac{( r - q ) ^{2}}{4} = (q - r)^{2} - (r - q)^{2} = 0$
Since $D = 0$ , the roots are equal.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q293 marksShort AnswerQuadratic Equations CBSE 2026 · Standard 30/3/1

Find two consecutive negative integers, sum of whose squares is 481.

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Answer: – 16 and – 15

Let the integers be $x$ and $x + 1$ ( $x$ negative)
$x^{2} + (x + 1)^{2} = 481$ , so $2 x^{2} + 2 x - 480 = 0$ , i.e. $x^{2} + x - 240 = 0$
$(x + 16) (x - 15) = 0$ , so $x = - 16$ (negative)
Integers: $- 16$ and $- 15$ (check: $256 + 225 = 481$ )

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q364 marksCase StudyQuadratic Equations CBSE 2026 · Basic 430/4/1

Observe the figure given above. It shows six identical rectangular enclosures made by using fencing wire mesh. These enclosures are used to protect baby animals in a zoo. Dimensions of each enclosure is $x$ feet × y feet. The total length of fencing required is 152 feet and area of each enclosure is 80 square feet.
Based on the above, answer the following questions :
(i) Write an expression for length of fencing required in terms of $x$ and y. (1)
(ii) Write the area of each enclosure in terms of $x$ . (1)
(iii) (a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method. (2)
OR
(b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula. (2)

Diagram for CBSE 2026 Class 10 Maths question 36

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Answer: (i)

8 x + 9 y = 152

(ii) Area =

x (\frac{152 - 8 x}{9}) = \frac{152 x - 8 x ^{2}}{9}

sq feet (iii)

x^{2} - 19 x + 90 = 0

; each enclosure is 10 feet × 8 feet (x = 9 feet, y =

\frac{80}{9}

feet also satisfies)

(i) The figure has 4 horizontal fences each of length 2x and 3 vertical fences each of length 3y.
Fencing = $8 x + 9 y = 152$ .
(ii) From (i), $y = \frac{152 - 8 x}{9}$ , so area = $x y = \frac{152 x - 8 x ^{2}}{9}$ .
(iii) (a) $\frac{152 x - 8 x ^{2}}{9} = 80 \Rightarrow 8 x^{2} - 152 x + 720 = 0 \Rightarrow x^{2} - 19 x + 90 = 0$ .
$(x - 10) (x - 9) = 0$ , so $x = 10$ or $x = 9$ .
If $x = 10$ , $y = \frac{152 - 80}{9} = 8$ ; if $x = 9$ , $y = \frac{80}{9}$ .
Each enclosure is 10 feet × 8 feet (the other solution gives 9 feet × $\frac{80}{9}$ feet).
(iii) (b) $x = \frac{19 \pm 361 - 360}{2} = \frac{19 \pm 1}{2} = 10$ or 9, giving the same dimensions.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q345 marksLong AnswerQuadratic Equations CBSE 2026 · Basic 430/5/1

ABCD is a rectangle of dimensions 80 cm $\times$ 60 cm. Another rectangle PQRS is drawn inside ABCD leaving space of equal width $x$ cm along the edges of ABCD. If area PQRS is half of the area ABCD, then find the value of $x$ .

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Answer:

x = 10

Dimensions of PQRS: $(80 - 2 x)$ cm and $(60 - 2 x)$ cm.
$(80 - 2 x) (60 - 2 x) = \frac{1}{2} \times 80 \times 60 = 2400$
$4800 - 280 x + 4 x^{2} = 2400 \Rightarrow x^{2} - 70 x + 600 = 0$
$(x - 10) (x - 60) = 0$ , so $x = 10$ or $x = 60$
$x = 60$ makes $60 - 2 x$ negative, so $x = 10$ .

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Basic 430/5/1

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the same journey. Find the original speed of the train.

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Answer: 45 km/h

Let the original speed be x km/h.
$\frac{90}{x} - \frac{90}{x + 15} = \frac{1}{2}$
$\frac{90 \times 15}{x ( x + 15 )} = \frac{1}{2} \Rightarrow x^{2} + 15 x - 2700 = 0$
$(x + 60) (x - 45) = 0$
Speed cannot be negative, so x = 45 km/h.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q335 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/1/1

A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.

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Answer: Slower train 40 km/hr, faster train 50 km/hr

Let the speed of the slower train be $x$ km/hr; faster train $(x + 10)$ km/hr.
$\frac{200}{x} - \frac{200}{x + 10} = 1$
$200 (x + 10) - 200 x = x (x + 10)$
$x^{2} + 10 x - 2000 = 0$
$(x + 50) (x - 40) = 0$ , so $x = 40$ (speed cannot be negative).
Slower train: 40 km/hr; faster train: 50 km/hr.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q33 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/1/1

The sum of the areas of two squares is 640 m $^{2}$ . If the difference in their perimeters is 64 m, find the sides of the two squares.

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Answer: 24 m and 8 m

Let the sides be $x$ m and $y$ m with $x > y$ .
$4 x - 4 y = 64$ , so $x = y + 16$ .
$x^{2} + y^{2} = 640$ : $(y + 16)^{2} + y^{2} = 640$
$2 y^{2} + 32 y + 256 - 640 = 0$ , i.e. $y^{2} + 16 y - 192 = 0$
$(y + 24) (y - 8) = 0$ , so $y = 8$ (side cannot be negative).
$x = 24$ . The sides are 24 m and 8 m.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

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