CBSE Maths › Class 10 PYQs › Quadratic Equations

Quadratic Equations: 3 marks Questions (CBSE Class 10)

23 different 3 marks questions on Quadratic Equations from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (73)2 marks (38)3 marks (23)4 marks (13)5 marks (62)

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

Q273 marksShort AnswerQuadratic Equations CBSE 2025 · Basic 430/5/1

The sum of the squares of two consecutive even numbers is 452. Find the numbers.

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Answer: 14 and 16 (or –16 and –14)

Let the numbers be $x$ and $x + 2$ .
$x^{2} + (x + 2)^{2} = 452$ , so $2 x^{2} + 4 x - 448 = 0$ , i.e. $x^{2} + 2 x - 224 = 0$ .
$(x + 16) (x - 14) = 0$ , so $x = 14$ or $x = - 16$ .
The numbers are 14 and 16 (or –16 and –14).

Q313 marksShort AnswerQuadratic Equations CBSE 2025 · Basic 430/5/2

A rectangular field is 16 m long and 10 m wide. There is a path of equal width all around it, having an area of 120 sq.m. Find the width of the path.

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Answer: 2 m

Let the width of the path be $x$ m. The outer rectangle is $(16 + 2 x)$ m by $(10 + 2 x)$ m.
Area of path $= (16 + 2 x) (10 + 2 x) - 16 \times 10 = 120$ .
$160 + 52 x + 4 x^{2} - 160 = 120$ , so $4 x^{2} + 52 x - 120 = 0$ , i.e. $x^{2} + 13 x - 30 = 0$ .
$(x + 15) (x - 2) = 0$ , so $x = 2$ (width cannot be negative).
Width of the path $= 2$ m.

Q303 marksShort AnswerQuadratic Equations CBSE 2025 · Basic 430/5/3

The sum of the squares of two consecutive odd numbers is 514. Find the numbers.

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Answer: 15 and 17 (or –17 and –15)

Let the numbers be $x$ and $x + 2$ .
$x^{2} + (x + 2)^{2} = 514$ , so $2 x^{2} + 4 x - 510 = 0$ , i.e. $x^{2} + 2 x - 255 = 0$ .
$(x + 17) (x - 15) = 0$ , so $x = 15$ or $x = - 17$ .
The numbers are 15 and 17 (or –17 and –15).

Q273 marksShort AnswerQuadratic Equations CBSE 2025 · Basic 430/6/1

Find length and breadth of a rectangular park whose perimeter is 100 m and area is 600 m $^{2}$ .

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Answer: Length = 30 m, breadth = 20 m

Let length = $x$ m. Then $2 (x + b) = 100$ gives breadth $= (50 - x)$ m.
Area: $x (50 - x) = 600$ , so $x^{2} - 50 x + 600 = 0$ .
$(x - 30) (x - 20) = 0$ , so $x = 30$ or $x = 20$ .
Taking length as the longer side: length = 30 m, breadth = 20 m.

Also asked in: 2025 Basic 430/6/3

Q313 marksShort AnswerQuadratic Equations CBSE 2025 · Basic 430/6/2

The sum of a number and its reciprocal is $\frac{13}{6}$ . Find the number.

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Answer:

\frac{3}{2}

\frac{2}{3}

Let the number be $x$ . Then $x + \frac{1}{x} = \frac{13}{6}$ .
$6 x^{2} + 6 = 13 x$ , so $6 x^{2} - 13 x + 6 = 0$ .
$6 x^{2} - 9 x - 4 x + 6 = 3 x (2 x - 3) - 2 (2 x - 3) = (3 x - 2) (2 x - 3) = 0$ .
$x = \frac{2}{3}$ or $x = \frac{3}{2}$ .

Q273 marksShort AnswerQuadratic Equations CBSE 2024 · Basic 430/3/1

The altitude of a right-angled triangle is 7 cm less than its base. If its hypotenuse is 17 cm long, then
(a) represent the above information in the form of a quadratic equation;
(b) find the length of the sides of the triangle.

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Answer: (a)

x^{2} - 7 x - 120 = 0

(base

x

cm) (b) base 15 cm, altitude 8 cm, hypotenuse 17 cm

Let the base be $x$ cm; then the altitude is $(x - 7)$ cm.
By Pythagoras theorem, $x^{2} + (x - 7)^{2} = 1 7^{2}$ .
$2 x^{2} - 14 x + 49 - 289 = 0$ , i.e. $x^{2} - 7 x - 120 = 0$ .
$(x - 15) (x + 8) = 0$ , so $x = 15$ (length cannot be negative).
Base = 15 cm, altitude = 8 cm, hypotenuse = 17 cm.

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q283 marksShort AnswerQuadratic Equations CBSE 2024 · Basic 430/4/1

Using quadratic formula, find the real roots of the equation $2 x^{2} + 2 x + 9 = 0$ , if they exist.

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Answer: No real roots exist.

Here $a = 2$ , $b = 2$ , $c = 9$ .
$D = b^{2} - 4 a c = 4 - 72 = - 68 < 0$ .
By the quadratic formula $x = \frac{- b \pm D}{2 a}$ , and $- 68$ is not real.
Hence the equation has no real roots.

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q28 (OR) (OR)3 marksShort AnswerQuadratic Equations CBSE 2024 · Basic 430/4/1

Find the values of ‘k’ for which the quadratic equation $k x^{2} - 2 k x + 6 = 0$ has real and equal roots. Also, find the roots.

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Answer:

k = 6

; roots are 1, 1

For equal roots, $D = 0$ : $(- 2 k)^{2} - 4 (k) (6) = 0$ .
$4 k^{2} - 24 k = 0 \Rightarrow 4 k (k - 6) = 0 \Rightarrow k = 0$ or $k = 6$ .
$k = 0$ does not give a quadratic equation, so $k = 6$ .
Equation: $6 x^{2} - 12 x + 6 = 0 \Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow (x - 1)^{2} = 0$ .
Roots are 1 and 1.

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q313 marksShort AnswerQuadratic Equations CBSE 2024 · Standard 30/2/1

In a 2-digit number, the digit at the unit's place is 5 less than the digit at the ten's place. The product of the digits is 36. Find the number.

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Answer: 94

Let the ten's digit be $x$ ; then the unit's digit is $x - 5$ .
$x (x - 5) = 36$ , so $x^{2} - 5 x - 36 = 0$ .
$(x - 9) (x + 4) = 0$ , so $x = 9$ ( $x = - 4$ is not a digit).
Unit's digit $= 4$ . The number is 94.

Q283 marksShort AnswerQuadratic Equations CBSE 2024 · Standard 30/2/2

Three consecutive integers are such that sum of the square of second and product of other two is 161. Find the three integers.

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Answer: 8, 9, 10 or

- 10

- 9

- 8

Let the integers be $n - 1$ , $n$ , $n + 1$ .
$n^{2} + (n - 1) (n + 1) = 161$
$2 n^{2} - 1 = 161$ , so $n^{2} = 81$ and $n = \pm 9$ .
The integers are 8, 9, 10 or $- 10$ , $- 9$ , $- 8$ .

Q303 marksShort AnswerQuadratic Equations CBSE 2024 · Standard 30/2/3

A dealer sells an article for ₹ 75 and gains as much percent as the cost price of the article. Find the cost price of the article.

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Answer: ₹50

Let the cost price be ₹ $x$ . Then the gain is $x %$ of $x = \frac{x ^{2}}{100}$ .
SP = CP + gain: $x + \frac{x ^{2}}{100} = 75$
$x^{2} + 100 x - 7500 = 0$ , so $(x + 150) (x - 50) = 0$ .
$x = 50$ (cost price cannot be negative).
Cost price = ₹50. (Check: 50% gain on ₹50 is ₹25, SP = ₹75.)

Q263 marksShort AnswerQuadratic Equations CBSE 2023 · Basic 430/6/1

The sum of the reciprocals of Varun’s age (in years) 3 years ago and 5 years from now is $\frac{1}{3}$ . Find his present age.

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Answer: 7 years

Let the present age be $x$ years.
$\frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3}$
$3 (2 x + 2) = (x - 3) (x + 5)$ , so $6 x + 6 = x^{2} + 2 x - 15$ .
$x^{2} - 4 x - 21 = 0$ , so $(x - 7) (x + 3) = 0$ .
Age cannot be negative, so $x = 7$ . Present age = 7 years.

Also asked in: 2023 Basic 430/6/2

Q273 marksShort AnswerQuadratic Equations CBSE 2023 · Basic 430/6/3

Solve for $x$ :
$\frac{1}{x} - \frac{1}{x - 2} = 3; x \neq = 0, 2$

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Answer:

x = \frac{3 + 3}{3}

x = \frac{3 - 3}{3}

$\frac{( x - 2 ) - x}{x ( x - 2 )} = 3$ , so $\frac{- 2}{x ^{2} - 2 x} = 3$ .
$- 2 = 3 x^{2} - 6 x$ , so $3 x^{2} - 6 x + 2 = 0$ .
$x = \frac{6 \pm 36 - 24}{6} = \frac{6 \pm 2 3}{6} = \frac{3 \pm 3}{3}$ .
Both values are allowed (neither is 0 or 2), so $x = \frac{3 + 3}{3}$ or $x = \frac{3 - 3}{3}$ .

Q313 marksShort AnswerQuadratic Equations CBSE 2023 · Standard 30/4/1

Find the value of ‘p’ for which the quadratic equation $p x (x - 2) + 6 = 0$ has two equal real roots.

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Answer:

p = 6

$p x^{2} - 2 p x + 6 = 0$ , with $p \neq = 0$ .
For equal roots, $D = 0$ : $(- 2 p)^{2} - 4 (p) (6) = 0$
$4 p^{2} - 24 p = 0 \Rightarrow 4 p (p - 6) = 0$
$p = 0$ is rejected (not quadratic), so $p = 6$ .

Also asked in: 2023 Standard 30/4/2

Q263 marksShort AnswerQuadratic Equations CBSE 2023 · Standard 30/4/3

Find the value of ‘p’ for which one root of the quadratic equation $p x^{2} - 14 x + 8 = 0$ is 6 times the other.

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Answer:

p = 3

Let the roots be $α$ and $6 α$ .
Sum: $7 α = \frac{14}{p} \Rightarrow α = \frac{2}{p}$
Product: $6 α^{2} = \frac{8}{p} \Rightarrow 6 \cdot \frac{4}{p ^{2}} = \frac{8}{p} \Rightarrow 24 = 8 p \Rightarrow p = 3$
Check: $3 x^{2} - 14 x + 8 = (3 x - 2) (x - 4)$ , roots $\frac{2}{3}$ and 4, and $4 = 6 \times \frac{2}{3}$ .

Q293 marksShort AnswerQuadratic Equations CBSE 2023 · Standard 30/6/1

The sum of two numbers is 15. If the sum of their reciprocals is $\frac{3}{10}$ , find the two numbers.

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Answer: 10 and 5

Let the numbers be x and 15 – x.
$\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10}$
$\frac{15}{x ( 15 - x )} = \frac{3}{10}$
$x (15 - x) = 50$
$x^{2} - 15 x + 50 = 0$
$(x - 10) (x - 5) = 0$ , so x = 10 or 5.
The numbers are 10 and 5.

Also asked in: 2023 Standard 30/6/2

Q273 marksShort AnswerQuadratic Equations CBSE 2023 · Standard 30/6/3

A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

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Answer: 8

Let the number be x. Then $x + 12 = \frac{160}{x}$ .
$x^{2} + 12 x - 160 = 0$
$(x + 20) (x - 8) = 0$
x = 8 (x = –20 is not a natural number).

Q27 (OR) (OR)3 marksShort AnswerQuadratic Equations CBSE 2023 · Standard 30/6/3

If one root of the quadratic equation $x^{2} + 12 x - k = 0$ is thrice the other root, then find the value of k.

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Answer: k = –27

Let the roots be $α$ and $3 α$ .
Sum: $4 α = - 12$ , so $α = - 3$ ; roots are –3 and –9.
Product: $3 α^{2} = - k$ , so $27 = - k$ .
k = –27

Q73 marksShort AnswerQuadratic Equations CBSE 2022 · Basic 430/4/1

If $x = 3$ is one root of the quadratic equation $2 x^{2} + p x + 30 = 0$ , find the value of p and the other root of the quadratic equation.

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Answer:

p = - 16

; other root

= 5

Put $x = 3$ : $2 (9) + 3 p + 30 = 0 \Rightarrow 3 p = - 48 \Rightarrow p = - 16$ .
Equation: $2 x^{2} - 16 x + 30 = 0 \Rightarrow x^{2} - 8 x + 15 = 0$ .
$(x - 3) (x - 5) = 0$ , so $x = 3$ or $x = 5$ .
The other root is 5.

Also asked in: 2022 Basic 430/4/2, 2022 Basic 430/4/3

Q7 (OR) (OR)3 marksShort AnswerQuadratic Equations CBSE 2022 · Basic 430/4/1

The length of a rectangular park is 5 metres more than twice its breadth. If the area of the park is 250 sq m, find the length and breadth of the park.

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Answer: Length = 25 m, breadth = 10 m

Let the breadth be $x$ m; then length $= (2 x + 5)$ m.
$x (2 x + 5) = 250 \Rightarrow 2 x^{2} + 5 x - 250 = 0$ .
$2 x^{2} + 25 x - 20 x - 250 = 0 \Rightarrow (2 x + 25) (x - 10) = 0$ .
$x = 10$ (breadth cannot be negative).
Breadth = 10 m, length $= 2 (10) + 5 = 25$ m.

Also asked in: 2022 Basic 430/4/2, 2022 Basic 430/4/3

Q83 marksShort AnswerQuadratic Equations CBSE 2022 · Standard 30/4/1

Find the value of ‘p’ for which the quadratic equation $p (x - 4) (x - 2) + (x - 1)^{2} = 0$ has real and equal roots.

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Answer:

p = 0

p = 3

$p (x^{2} - 6 x + 8) + x^{2} - 2 x + 1 = 0$
$(p + 1) x^{2} - (6 p + 2) x + (8 p + 1) = 0$
For real and equal roots, $D = 0$ :
$(6 p + 2)^{2} - 4 (p + 1) (8 p + 1) = 0$
$36 p^{2} + 24 p + 4 - 32 p^{2} - 36 p - 4 = 0$
$4 p^{2} - 12 p = 0 \Rightarrow 4 p (p - 3) = 0$
$p = 0$ or $p = 3$ (for both, $p + 1 \neq = 0$ , so the equation stays quadratic).

Also asked in: 2022 Standard 30/4/2, 2022 Standard 30/4/3

Q8 (OR) (OR)3 marksShort AnswerQuadratic Equations CBSE 2022 · Standard 30/4/1

Had Aarush scored 8 more marks in a Mathematics test, out of 35 marks, 7 times these marks would have been 4 less than square of his actual marks. How many marks did he get in the test ?

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Answer: 12 marks

Let his actual marks be $x$ .
$7 (x + 8) = x^{2} - 4$
$x^{2} - 7 x - 60 = 0$
$(x - 12) (x + 5) = 0$
$x = 12$ (marks cannot be negative)
Aarush got 12 marks.

Also asked in: 2022 Standard 30/4/2, 2022 Standard 30/4/3

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