The difference between two numbers is 12. The greater number is 6 less than twice the smaller one. (i) Representing the above situation, frame two linear equations in two variables. (ii) Show that the equations have unique solution. (iii) Solve the equations and hence find the numbers.

(i) x-y=12, x-2y=-6; (ii) (a_1)/(a_2)≠(b_1)/(b_2), unique solution; (iii) numbers are 30 and 18

MathsCBSE Class 10 · Basic · Board exam 2026, set 430/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2026: question 35(a)

Set in the CBSE Class 10 board exam of 2026, paper 430/4/1, section D, question 35(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(a) · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables

The difference between two numbers is $12$ . The greater number is $6$ less than twice the smaller one.

(i)
Representing the above situation, frame two linear equations in two variables.
0 marks
(ii)
Show that the equations have unique solution.
0 marks
(iii)
Solve the equations and hence find the numbers.
0 marks

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

Let the greater number be x and the smaller be y, and translate each sentence into an equation.

Hint 2

For a unique solution, compare $\frac{a _{1}}{a _{2}}$ with $\frac{b _{1}}{b _{2}}$ .

Hint 3

Subtract one equation from the other to eliminate x.

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 the greater number be x and the smaller be y. Difference is 12: $x - y = 12$ ...(1)
1 mark
Greater is 6 less than twice the smaller: $x = 2 y - 6$ , i.e. $x - 2 y = - 6$ ...(2)
1 mark
(ii) $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ , $\frac{b _{1}}{b _{2}} = \frac{- 1}{- 2} = \frac{1}{2}$ . Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect, so the pair has a unique solution.
1 mark
(iii) Subtracting (2) from (1): $(x - y) - (x - 2 y) = 12 - (- 6) \Rightarrow y = 18$
1 mark
From (1): $x = 12 + 18 = 30$ . The numbers are 30 and 18.
1 mark

Answer: (i) $x - y = 12$ , $x - 2 y = - 6$ ; (ii) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , unique solution; (iii) numbers are 30 and 18

Where the marks are

Forms $x - y = 12$ (1 mark)
Forms $x - 2 y = - 6$ (1 mark)
Compares ratios $\frac{1}{1} \neq = \frac{- 1}{- 2}$ to show unique solution (1 mark)
Finds $y = 18$ (1 mark)
Finds $x = 30$ (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 430/4/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2026: question 35(a)

Hints

Step-by-step solution

Revise the chapter

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