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. (ii) Write the area of each enclosure in terms of x. (iii) (a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method. (iii) (b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula.

(i) 8x + 9y = 152; (ii) (x(152-8x))/(9) = 80; (iii) x² - 19x + 90 = 0, so x = 10, y = 8 (or x = 9, y = 80/9): each enclosure is 10 ft × 8 ft (or 9 ft × 80/9 ft)

MathsCBSE Class 10 · Basic · Board exam 2026, set 430/4/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2026: question 36

Set in the CBSE Class 10 board exam of 2026, paper 430/4/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedQuadratic Equations

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 $\times$ $y$ 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 mark
(ii)
Write the area of each enclosure in terms of $x$ .
1 mark
(iii)
(a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method.
2 marks
OR
(iii)
(b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula.
2 marks

Figure in the paper: Left: photo of an animal behind wire mesh. Right: a rectangle of width x (top, marked with arrows) and height y (right side), divided into six identical rectangular enclosures arranged in 2 columns and 3 rows, with fencing lines and tick marks on the sides. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Count how many horizontal and how many vertical fence lines appear in the figure, and how long each one is.

Hint 2

There are 4 horizontal lines of length $2 x$ and 3 vertical lines of length $3 y$ . Make $y$ the subject and substitute it into $x y = 80$ .

Hint 3

Clear the fraction and divide by 8 to get a quadratic in $x$ , then factorise it or apply the quadratic formula.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) The six enclosures sit in 2 columns and 3 rows. There are 4 horizontal fence lines, each of length $2 x$ , which give $8 x$ . There are 3 vertical fence lines, each of length $3 y$ , which give $9 y$ .
Length of fencing: $8 x + 9 y = 152$
1 mark
(ii) From (i), $y = \frac{152 - 8 x}{9}$ .
Area of each enclosure $= x y = \frac{x ( 152 - 8 x )}{9}$ , so $\frac{x ( 152 - 8 x )}{9} = 80$ .
1 mark
(iii)(a) Rearrange into quadratic form:
$152 x - 8 x^{2} = 720 \Rightarrow 8 x^{2} - 152 x + 720 = 0 \Rightarrow x^{2} - 19 x + 90 = 0$
½ mark
Factorise by splitting the middle term:
$x^{2} - 10 x - 9 x + 90 = 0 \Rightarrow (x - 10) (x - 9) = 0$
½ mark
So $x = 10$ or $x = 9$ .
½ mark
Find $y$ from $y = \frac{80}{x}$ :
- When $x = 10$ , $y = 8$ .
- When $x = 9$ , $y = \frac{80}{9}$ .
Dimensions of each enclosure: $10$ ft $\times$ $8$ ft, or $9$ ft $\times \frac{80}{9}$ ft.
½ mark
(iii)(b) The quadratic is $x^{2} - 19 x + 90 = 0$ , with $a = 1$ , $b = - 19$ , $c = 90$ .
Discriminant: $D = b^{2} - 4 a c = 361 - 360 = 1$
½ mark
By the quadratic formula:
$x = \frac{- b \pm D}{2 a} = \frac{19 \pm 1}{2} = \frac{19 \pm 1}{2}$
½ mark
So $x = 10$ or $x = 9$ .
½ mark
Then $y = \frac{80}{x}$ gives $y = 8$ or $y = \frac{80}{9}$ .
Dimensions: $10$ ft $\times$ $8$ ft, or $9$ ft $\times \frac{80}{9}$ ft.
½ mark

Answer: (i) $8 x + 9 y = 152$ ; (ii) $\frac{x ( 152 - 8 x )}{9} = 80$ ; (iii) $x^{2} - 19 x + 90 = 0$ , so $x = 10$ , $y = 8$ (or $x = 9$ , $y = \frac{80}{9}$ ): each enclosure is 10 ft × 8 ft (or 9 ft × 80/9 ft)

Where the marks are

Writes the fencing equation $8 x + 9 y = 152$ (1 mark)
Writes the area as $\frac{x ( 152 - 8 x )}{9} = 80$ (1 mark)
Forms $x^{2} - 19 x + 90 = 0$ (½ mark)
Factorises as $(x - 10) (x - 9) = 0$ in (a), or substitutes correctly into the quadratic formula in (b) (½ mark)
Gets $x = 10$ or $9$ (½ mark)
Gets $y = 8$ or $\frac{80}{9}$ (½ mark)

Revise the chapter

Chapter: Quadratic Equations (notes, formulas, where marks are lost and every past question)
Topic: Word Problems on Quadratic Equations

Practise this chapterBoard-style questions from Quadratic Equations, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 430/4/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2026: question 36

Hints

Step-by-step solution

Revise the chapter

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