MathsCBSE Class 10 · Basic · Board exam 2026, set 430/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2026: question 36

Set in the CBSE Class 10 board exam of 2026, paper 430/5/1, 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-basedPair of Linear Equations in Two Variables

Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories.

To find the number of calories burned per minute on each machine, answer the following:

(i)
Represent the above situation in terms of a pair of linear equations in two variables.
1 mark
(ii)
Show that the equations have unique solution.
1 mark
(iii)(a)
Solve both equations to find the values of the variables using elimination method.
2 marks
OR
(iii)(b)
Solve both equations to find the values of the variables using substitution method.
2 marks

Figure in the paper: Photograph of outdoor exercise machines in a park, including an exercise bicycle and a double cross walker, set on a paved surface. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/5/1.

Hints

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

Hint 1

Let $x$ and $y$ be the calories burned per minute on each machine, and turn each sentence into 'minutes × rate' added together.

Hint 2

For a unique solution, compare the ratios of the coefficients of $x$ and of $y$ in the two equations.

Hint 3

Simplify the equations first by dividing by 15 and by 10. Then make the $y$ -coefficients equal to eliminate $y$ , or write $x$ in terms of $y$ to substitute.

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) Let Seema burn $x$ calories per minute on the exercise bicycle and $y$ calories per minute on the double cross walker. The two situations give:
$15 x + 30 y = 435$ ...(1)
$30 x + 40 y = 690$ ...(2)
1 mark
(ii) Here $a_{1} = 15, b_{1} = 30, a_{2} = 30, b_{2} = 40$ .
$\frac{a _{1}}{a _{2}} = \frac{15}{30} = \frac{1}{2}$ and $\frac{b _{1}}{b _{2}} = \frac{30}{40} = \frac{3}{4}$ .
Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect at exactly one point, so the pair of equations has a unique solution.
1 mark
(iii)(a) Elimination method: dividing (1) by 15 gives $x + 2 y = 29$ ...(3). Dividing (2) by 10 gives $3 x + 4 y = 69$ ...(4).
Multiply (3) by 2: $2 x + 4 y = 58$ ...(5).
Subtract (5) from (4): $(3 x + 4 y) - (2 x + 4 y) = 69 - 58$ , so $x = 11$ .
1 mark
Put $x = 11$ in (3): $11 + 2 y = 29 \Rightarrow 2 y = 18 \Rightarrow y = 9$ .
So 11 calories per minute are burned on the bicycle and 9 calories per minute on the walker. Check: $15 (11) + 30 (9) = 165 + 270 = 435$ ✓.
1 mark
(iii)(b) Substitution method: from (1), $x + 2 y = 29 \Rightarrow x = 29 - 2 y$ ...(6).
Substitute in (2) written as $3 x + 4 y = 69$ : $3 (29 - 2 y) + 4 y = 69 \Rightarrow 87 - 6 y + 4 y = 69 \Rightarrow - 2 y = - 18 \Rightarrow y = 9$ .
1 mark
From (6): $x = 29 - 2 (9) = 11$ .
So $x = 11$ and $y = 9$ . The bicycle burns 11 calories per minute and the walker burns 9 calories per minute. Check: $30 (11) + 40 (9) = 330 + 360 = 690$ ✓.
1 mark

Answer: (i) $15 x + 30 y = 435$ , $30 x + 40 y = 690$ ; (ii) $\frac{15}{30} \neq = \frac{30}{40}$ , so the solution is unique; (iii) $x = 11$ , $y = 9$ (11 cal/min on the bicycle, 9 cal/min on the walker) by either method

Where the marks are

Defines the variables and writes both equations $15 x + 30 y = 435$ and $30 x + 40 y = 690$ (1 mark)
Compares $a_{1} / a_{2}$ with $b_{1} / b_{2}$ , shows they are unequal and concludes the solution is unique (1 mark)
(iii)(a) Eliminates one variable correctly to get $x = 11$ (1 mark)
(iii)(a) Back-substitutes to get $y = 9$ (1 mark)
(iii)(b) Expresses one variable in terms of the other and substitutes to get $y = 9$ (1 mark)
(iii)(b) Finds $x = 11$ (1 mark)

Revise the chapter

Chapter: Pair of Linear Equations in Two Variables (notes, formulas, where marks are lost and every past question)
Topic: Word Problems on Linear Equations

Practise this chapterBoard-style questions from Pair of Linear Equations in Two Variables, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 430/5/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2026: question 36

Hints

Step-by-step solution

Revise the chapter

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