CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
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Boards begin February 2027
Pre-boards are in December
Pair of Linear Equations in Two Variables is in Algebra, which carries 20 of the 80 marks in the 2026-27 CBSE Maths paper
16 NCERT textbook questions from Chapter 3
Exercise 3.1
NCERT Ex 3.1 Q2(i)1 mark
On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, the pair of linear equations 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0 represents lines that are:
Step 1For the two equations a₁ = 5, b₁ = −4, c₁ = 8 and a₂ = 7, b₂ = 6, c₂ = −9.
Step-by-step solution
For the two equations a₁ = 5, b₁ = −4, c₁ = 8 and a₂ = 7, b₂ = 6, c₂ = −9.
Compute a₁/a₂ = 5/7 and b₁/b₂ = −4/6 = −2/3.
Since 5/7 ≠ −2/3, i.e., a₁/a₂ ≠ b₁/b₂, the lines are intersecting (unique solution).
10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Step 1Let number of boys = x and number of girls = y.
Step-by-step solution
Let number of boys = x and number of girls = y.
From the total: x + y = 10.
From the relation: y − x = 4 (girls are 4 more than boys).
Adding the two equations: 2y = 14 → y = 7; then x = 10 − 7 = 3.
A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
Step 1Let Jacob's present age = x years and son's present age = y years.
Step-by-step solution
Let Jacob's present age = x years and son's present age = y years.
Five years hence: x + 5 = 3(y + 5) → x − 3y = 10 …(1).
Five years ago: x − 5 = 7(y − 5) → x − 7y = −30 …(2).
Subtract (2) from (1): 4y = 40 → y = 10.
Substitute in (1): x = 10 + 30 = 40. Jacob is 40 and his son is 10.
Form the pair of linear equations and solve by the elimination method: Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Step 1Let Nuri's present age = x and Sonu's present age = y.
Step-by-step solution
Let Nuri's present age = x and Sonu's present age = y.
Five years ago: x − 5 = 3(y − 5) → x − 3y = −10 …(1).
Ten years later: x + 10 = 2(y + 10) → x − 2y = 10 …(2).
Subtract (1) from (2): (x − 2y) − (x − 3y) = 10 − (−10) → y = 20.
Substitute in (2): x = 10 + 2(20) = 50.
Hence Nuri is 50 years old and Sonu is 20 years old. (Check: 5 years ago, 45 = 3 × 15 ✓; 10 years later, 60 = 2 × 30 ✓.)
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly incomes.
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Step 1Let the tens digit = x and units digit = y. Number = 10x + y; reversed = 10y + x.
Step-by-step solution
Let the tens digit = x and units digit = y. Number = 10x + y; reversed = 10y + x.
Sum condition: (10x + y) + (10y + x) = 66 → 11(x + y) = 66 → x + y = 6 …(1).
Digit difference condition: x − y = 2 …(2a) OR y − x = 2 …(2b).
Case 1: (1) and (2a) → 2x = 8 → x = 4, y = 2 → number = 42.
Case 2: (1) and (2b) → 2y = 8 → y = 4, x = 2 → number = 24. Hence two numbers exist: 42 and 24.
Answer:Two numbers — 42 and 24 (both satisfy the conditions).
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Case-based4 marks
Akhila went to a village fair with ₹20. She wanted to enjoy rides on the Giant Wheel and play Hoopla. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each Giant-Wheel ride costs ₹3 and a game of Hoopla costs ₹4, and she spent the entire ₹20. Let x = number of Giant-Wheel rides and y = number of times she played Hoopla. (i) Form a pair of linear equations representing the situation. (ii) Solve graphically (describe lines and intersection). (iii) How many rides and games did she have? (iv) Verify algebraically.
Step 1Translate to equations: y = x/2 ⇒ x − 2y = 0 …(1); total spend 3x + 4y = 20 …(2).
Step-by-step solution
Translate to equations: y = x/2 ⇒ x − 2y = 0 …(1); total spend 3x + 4y = 20 …(2).
Graphical: Eq (1) passes through (0, 0), (2, 1), (4, 2); Eq (2) passes through (0, 5), (4, 2). Lines meet at (4, 2).
Algebraic check: from (1), x = 2y. Substitute in (2): 3(2y) + 4y = 20 → 10y = 20 → y = 2; x = 4.
Conclusion: 4 rides on the Giant Wheel and 2 games of Hoopla.
Verification: 4 − 2(2) = 0 ✓; 3(4) + 4(2) = 20 ✓.
Answer:Akhila had 4 Giant-Wheel rides and 2 games of Hoopla.
The taxi fare in a city consists of a fixed charge together with a charge proportional to the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. Let the fixed charge be ₹x and the per-km charge be ₹y. (i) Form the pair of linear equations. (ii) Find the fixed charge and the per-km rate. (iii) How much does a person have to pay for travelling 25 km?
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for 7 days, while Susy paid ₹21 for a book kept for 5 days. Let the fixed charge be ₹x and the additional per-day charge (after day 3) be ₹y. (i) Form the pair of equations. (ii) Find x and y. (iii) How much would a person pay for a book kept for 10 days?
Step 1Let fixed charge for the first 3 days = ₹x; additional per-day charge = ₹y.
Step-by-step solution
Let fixed charge for the first 3 days = ₹x; additional per-day charge = ₹y.
Saritha (7 days): x + (7 − 3)y = 27 → x + 4y = 27 …(1).
Susy (5 days): x + (5 − 3)y = 21 → x + 2y = 21 …(2).
Subtract (2) from (1): 2y = 6 → y = 3. From (2): x = 21 − 6 = 15.
For 10 days → 7 extra days: charge = 15 + 7(3) = ₹36.
Answer:Fixed ₹15, per-extra-day ₹3; charge for 10 days = ₹36.
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Questions from the NCERT Class 10 Mathematics textbook (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.