CBSE Maths › Class 10 PYQs › Quadratic Equations

Quadratic Equations: 4 marks Questions (CBSE Class 10)

13 different 4 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)

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

Q364 marksCase StudyQuadratic Equations CBSE 2025 · Standard 30/3/1

A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway.
The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres.
Based on the information given above, answer the following questions :
(i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway = $x$ m. (1)
(ii) (a) Solve the quadratic equation to find the width of the walkway ' $x$ '. (2)
OR
(b) If the cost of paving the walkway at the rate of ₹ 50 per square metre is ₹ 12,000, calculate the area of the walkway. (2)
(iii) Find the perimeter of the lawn. (1)

Diagram for CBSE 2025 Class 10 Maths question 36

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Answer: (i)

(12 + 2 x) (10 + 2 x) = 360

, i.e.

x^{2} + 11 x - 60 = 0

(ii) (a)

x = 4

m OR (b) 240 m

^{2}

(iii) 44 m

(i) Outer dimensions are $(12 + 2 x)$ m and $(10 + 2 x)$ m, so $(12 + 2 x) (10 + 2 x) = 360$ .
(i) $4 x^{2} + 44 x + 120 = 360$ , i.e. $x^{2} + 11 x - 60 = 0$ .
(ii)(a) $x^{2} + 15 x - 4 x - 60 = 0$ , so $(x + 15) (x - 4) = 0$ .
(ii)(a) Width cannot be negative, so $x = 4$ m.
(ii)(b) Area of walkway $= \frac{12000}{50} = 240$ m $^{2}$ .
(iii) Perimeter of lawn $= 2 (12 + 10) = 44$ m.

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q384 marksCase StudyQuadratic Equations CBSE 2024 · Basic 430/1/2

To keep the lawn green and cool, Sadhna uses water sprinklers which rotate in circular shape and cover a particular area.
The diagram below shows the circular areas covered by two sprinklers :
Two circles touch externally. The sum of their areas is $130 π$ sq m and the distance between their centres is 14 m.
Based on above information, answer the following questions :
(i) Obtain a quadratic equation involving R and r from above. (1)
(ii) Write a quadratic equation involving only r. (1)
(iii) (a) Find the radius r and the corresponding area irrigated. (2)
OR
(b) Find the radius R and the corresponding area irrigated. (2)

Diagram for CBSE 2024 Class 10 Maths question 38

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Answer: (i)

R^{2} + r^{2} = 130

(with

R + r = 14

) (ii)

r^{2} - 14 r + 33 = 0

(iii) (a)

r = 3

m, area

= 9 π

sq m OR (b)

R = 11

m, area

= 121 π

sq m

(i) $π R^{2} + π r^{2} = 130 π$ , so $R^{2} + r^{2} = 130$ ; also $R + r = 14$ (circles touch externally).
(ii) $R = 14 - r$ : $(14 - r)^{2} + r^{2} = 130$ , so $2 r^{2} - 28 r + 66 = 0$ , i.e. $r^{2} - 14 r + 33 = 0$ .
(iii) (a) $(r - 3) (r - 11) = 0$ , so $r = 3$ or 11. Since $R > r$ and $R + r = 14$ , $r = 3$ m. Area $= π \times 9 = 9 π$ sq m ( $\approx 28.29$ sq m).
(iii) (b) $R = 14 - 3 = 11$ m. Area $= π \times 121 = 121 π$ sq m ( $\approx 380.29$ sq m).

Also asked in: 2024 Basic 430/1/3

Q364 marksCase StudyQuadratic Equations CBSE 2024 · Standard 30/1/1

A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information. (1)
(ii) Write the corresponding quadratic equation in standard form. (1)
(iii) (a) Find the value of $x$ , the length of side of a tile by factorisation. (2)
OR (b) Solve the quadratic equation for $x$ , using quadratic formula. (2)

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Answer: (i)

200 x^{2} = 128 (x + 1)^{2}

(ii)

9 x^{2} - 32 x - 16 = 0

(iii) (a)

x = 4

units OR (b)

x = 4

(rejecting

x = - \frac{4}{9}

)

(i) Floor area is the same in both cases: $200 x^{2} = 128 (x + 1)^{2}$ .
(ii) $200 x^{2} = 128 x^{2} + 256 x + 128$ , so $72 x^{2} - 256 x - 128 = 0$ . Dividing by 8: $9 x^{2} - 32 x - 16 = 0$ .
(iii) (a) $9 x^{2} - 36 x + 4 x - 16 = 0 \Rightarrow 9 x (x - 4) + 4 (x - 4) = 0 \Rightarrow (x - 4) (9 x + 4) = 0$ .
$x = 4$ or $x = - \frac{4}{9}$ ; length cannot be negative, so $x = 4$ units.
OR (b) $D = (- 32)^{2} - 4 (9) (- 16) = 1024 + 576 = 1600$ , $D = 40$ .
$x = \frac{32 \pm 40}{18}$ , so $x = 4$ or $x = - \frac{4}{9}$ . Taking the positive value, $x = 4$ units.

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q364 marksCase StudyQuadratic Equations CBSE 2023 · Standard 30/1/1

While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by x units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based on the above information, answer the following questions :
(I) Write an algebraic equation depicting the above information. (1)
(II) Write the corresponding quadratic equation in standard form. (1)
(III) What should be the new dimensions of the enlarged photo ? (2)
OR
Can any rational value of x make the new area equal to 220 cm $^{2}$ ?

Diagram for CBSE 2023 Class 10 Maths question 36

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Answer: (I)

(18 + x) (12 + x) = 2 \times 18 \times 12

(II)

x^{2} + 30 x - 216 = 0

(III) 24 cm × 18 cm; OR: No, since

x^{2} + 30 x - 4 = 0

has discriminant 916, not a perfect square, so x is irrational.

(I) Original area = 18 × 12 = 216 cm $^{2}$ . New area = $(18 + x) (12 + x) = 2 \times 216 = 432$ .
(II) $x^{2} + 30 x + 216 = 432$ , i.e. $x^{2} + 30 x - 216 = 0$ .
(III) $(x + 36) (x - 6) = 0$ ; x > 0, so x = 6. New dimensions: 18 + 6 = 24 cm by 12 + 6 = 18 cm.
OR: $(18 + x) (12 + x) = 220$ gives $x^{2} + 30 x - 4 = 0$ .
$D = 900 + 16 = 916$ , which is not a perfect square, so x is irrational. No rational x gives area 220 cm $^{2}$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q124 marksLong AnswerQuadratic Equations CBSE 2022 · Basic 430/2/1

The sum of the ages of a boy and his sister (in years) is 25 and product of their ages is 150. Find their present ages.

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Answer: 15 years and 10 years

Let one age be $x$ years; the other is $25 - x$ .
$x (25 - x) = 150 \Rightarrow x^{2} - 25 x + 150 = 0$ .
$(x - 10) (x - 15) = 0 \Rightarrow x = 10$ or $x = 15$ .
So their ages are 15 years and 10 years.

Also asked in: 2022 Basic 430/2/2

Q114 marksLong AnswerQuadratic Equations CBSE 2022 · Basic 430/2/3

The sum of two numbers is 45. If 5 is subtracted from each of them, the product of these numbers becomes 124. Find the numbers.

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Answer: 9 and 36

Let the numbers be $x$ and $45 - x$ .
$(x - 5) (45 - x - 5) = 124 \Rightarrow (x - 5) (40 - x) = 124$ .
$- x^{2} + 45 x - 200 = 124 \Rightarrow x^{2} - 45 x + 324 = 0$ .
$(x - 9) (x - 36) = 0 \Rightarrow x = 9$ or $36$ .
The numbers are 9 and 36 (check: $4 \times 31 = 124$ ).

Q134 marksCase StudyQuadratic Equations CBSE 2022 · Basic 430/3/1

The tradition of pottery making in India is very old. In fact, it is older than Indus Valley Civilization. The shaping and baking of clay articles has continued through the ages. The picture of a potter is shown below :
A potter makes a certain number of pottery articles in a day. It was observed on a particular day the cost of production of each article (in ₹) was one more than twice the number of articles produced on that day. The total cost of production on that day was ₹ 210.
(a) Taking number of articles produced on that day as x, form a quadratic equation in x. (2)
(b) Find the number of articles produced and the cost of each article. (2)

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Answer: (a)

2 x^{2} + x - 210 = 0

(b) 10 articles; cost of each article ₹ 21

(a) Cost of each article $= ₹ (2 x + 1)$ , so $x (2 x + 1) = 210$ , i.e. $2 x^{2} + x - 210 = 0$ .
(b) $2 x^{2} + 21 x - 20 x - 210 = 0 \Rightarrow x (2 x + 21) - 10 (2 x + 21) = 0 \Rightarrow (2 x + 21) (x - 10) = 0$
$x = 10$ (rejecting $x = - \frac{21}{2}$ since the number of articles cannot be negative)
Cost of each article $= 2 (10) + 1 = ₹ 21$

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

Q124 marksLong AnswerQuadratic Equations CBSE 2022 · Standard 30/2/1

A 2-digit number is such that the product of its digits is 24. If 18 is subtracted from the number, the digits interchange their places. Find the number.

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Answer: 64

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$x y = 24$ and $10 x + y - 18 = 10 y + x \Rightarrow 9 x - 9 y = 18 \Rightarrow x = y + 2$ .
$(y + 2) y = 24 \Rightarrow y^{2} + 2 y - 24 = 0 \Rightarrow (y + 6) (y - 4) = 0$ .
$y = 4$ (a digit cannot be $- 6$ ), so $x = 6$ .
The number is 64. Check: $6 \times 4 = 24$ and $64 - 18 = 46$ .

Also asked in: 2022 Standard 30/2/3

Q12 (OR) (OR)4 marksLong AnswerQuadratic Equations CBSE 2022 · Standard 30/2/1

The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.

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Answer: 18 and 12 (or 18 and

- 12

)

Let the greater number be $x$ ; then (smaller) $^{2} = 8 x$ .
$x^{2} - 8 x = 180 \Rightarrow x^{2} - 8 x - 180 = 0$ .
$(x - 18) (x + 10) = 0 \Rightarrow x = 18$ or $x = - 10$ .
$x = - 10$ gives (smaller) $^{2} = - 80$ , impossible; so $x = 18$ .
(Smaller) $^{2} = 144 \Rightarrow$ smaller $= 12$ or $- 12$ .
The numbers are 18 and 12 (or 18 and $- 12$ ).

Also asked in: 2022 Standard 30/2/3

Q114 marksLong AnswerQuadratic Equations CBSE 2022 · Standard 30/2/2

The sum of two numbers is 34. If 3 is subtracted from one number and 2 is added to another, the product of these two numbers becomes 260. Find the numbers.

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Answer: 23 and 11, or 16 and 18

Let one number be $x$ ; the other is $34 - x$ .
$(x - 3) (34 - x + 2) = 260 \Rightarrow (x - 3) (36 - x) = 260$ .
$- x^{2} + 39 x - 108 = 260 \Rightarrow x^{2} - 39 x + 368 = 0$ .
$(x - 23) (x - 16) = 0 \Rightarrow x = 23$ or $x = 16$ .
Numbers: 23 and 11, or 16 and 18.
Check: $(23 - 3) (11 + 2) = 260$ and $(16 - 3) (18 + 2) = 260$ .

Q11 (OR) (OR)4 marksLong AnswerQuadratic Equations CBSE 2022 · Standard 30/2/2

The hypotenuse (in cm) of a right angled triangle is 6 cm more than twice the length of the shortest side. If the length of third side is 6 cm less than thrice the length of shortest side, then find the dimensions of the triangle.

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Answer: 10 cm, 24 cm and 26 cm

Let the shortest side be $x$ cm. Hypotenuse $= 2 x + 6$ , third side $= 3 x - 6$ .
$x^{2} + (3 x - 6)^{2} = (2 x + 6)^{2}$
$10 x^{2} - 36 x + 36 = 4 x^{2} + 24 x + 36$
$6 x^{2} - 60 x = 0 \Rightarrow 6 x (x - 10) = 0 \Rightarrow x = 10$ ( $x \neq = 0$ ).
Sides: 10 cm, 24 cm, 26 cm. Check: $100 + 576 = 676 = 2 6^{2}$ .

Q134 marksCase StudyQuadratic Equations CBSE 2022 · Standard 30/3/1

In the picture given below, one can see a rectangular in-ground swimming pool installed by a family in their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq. m.
(a) Based on the information given above, form a quadratic equation in terms of x. (2)
(b) Find the width of the sidewalk around the pool. (2)

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Answer: (a)

(12 - 2 x) (7 - 2 x) = 36

, i.e.

2 x^{2} - 19 x + 24 = 0

(b) 1.5 m

(a) Pool dimensions are $(12 - 2 x)$ m by $(7 - 2 x)$ m.
$(12 - 2 x) (7 - 2 x) = 36 \Rightarrow 84 - 38 x + 4 x^{2} = 36 \Rightarrow 4 x^{2} - 38 x + 48 = 0 \Rightarrow 2 x^{2} - 19 x + 24 = 0$ .
(b) $2 x^{2} - 16 x - 3 x + 24 = 0 \Rightarrow (x - 8) (2 x - 3) = 0 \Rightarrow x = 8$ or $x = \frac{3}{2}$ .
$x = 8$ is impossible since $7 - 2 x$ would be negative. So the sidewalk is 1.5 m wide.

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

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