CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar 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
26 NCERT Exemplar questions from Chapter 3
Exercise 3.1
Exemplar Ex 3.1 Q21 mark
The pair of equations x + 2y + 5 = 0 and −3x − 6y + 1 = 0 has:
The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father are, respectively:
Step 1Let son's age = y and father's age = x. Then x = 6y …(1).
Step-by-step solution
Let son's age = y and father's age = x. Then x = 6y …(1).
Four years hence: x + 4 = 4(y + 4) → x = 4y + 12 …(2).
There are some students in two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A, the number in A becomes double the number in B. Find the number of students in each hall.
Step 1Let number of students in Hall A = x and in Hall B = y.
Step-by-step solution
Let number of students in Hall A = x and in Hall B = y.
Condition 1 (10 from A to B makes counts equal): x − 10 = y + 10 → x − y = 20 …(1).
Condition 2 (20 from B to A makes A's count double B's): x + 20 = 2(y − 20) → x − 2y = −60 …(2).
Subtract (2) from (1): (x − y) − (x − 2y) = 20 − (−60) → y = 80.
From (1): x = 80 + 20 = 100. Hall A = 100 students, Hall B = 80 students.
A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days and an additional charge for each day thereafter. Latika paid ₹22 for a book kept for six days, while Anand paid ₹16 for a book kept for four days. Find the fixed charge and the charge for each extra day.
Step 1Let fixed charge for the first two days = ₹x and per-extra-day charge = ₹y.
Step-by-step solution
Let fixed charge for the first two days = ₹x and per-extra-day charge = ₹y.
Latika kept 6 days = 2 + 4 extra: x + 4y = 22 …(1).
Anand kept 4 days = 2 + 2 extra: x + 2y = 16 …(2).
Subtract (2) from (1): 2y = 6 → y = 3.
Substitute in (2): x + 6 = 16 → x = 10. Fixed charge ₹10, extra-day charge ₹3.
In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
Step 1Let number of correct answers = x and number of wrong answers = y.
Step-by-step solution
Let number of correct answers = x and number of wrong answers = y.
Total questions attempted: x + y = 120 …(1).
Marks scored: 1·x + (−1/2)·y = 90 → 2x − y = 180 …(2).
Add (1) and (2): 3x = 300 → x = 100.
From (1): y = 20. So Jayanti answered 100 correctly and 20 incorrectly.
The cost of 4 pens and 4 pencil boxes is ₹100. Three times the cost of a pen is ₹15 more than the cost of a pencil box. Form the pair of linear equations and find the cost of a pen and of a pencil box.
Step 1Let cost of one pen = ₹x and cost of one pencil box = ₹y.
Step-by-step solution
Let cost of one pen = ₹x and cost of one pencil box = ₹y.
Total cost: 4x + 4y = 100 → x + y = 25 …(1).
Relation: 3x = y + 15 → 3x − y = 15 …(2).
Add (1) and (2): 4x = 40 → x = 10.
From (1): y = 25 − 10 = 15. Pen costs ₹10, pencil box ₹15.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems; the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.