Important Questions for Chapter 3 - Pair of Linear Equations in Two Variables

Our free set of important questions carries step-by-step explanation and solution. Our math subject experts with years of experience have studied the CBSE exam patterns very carefully and come up with these set of questions based on the current CBSE syllabus.

Important questions based on NCERT syllabus for Chapter 3 - Pair of Linear Equations in Two Variables:

Question 1: Check whether the pair of equations x−2y=2,4x−2y=5 is consistent.
Solution:
Given equations are,
x − 2y = 2,
⇒ x − 2y − 2 = 0……(i)
a1 = 1,b1 = −2,c1 = −2
4x − 2y = 5
⇒ 4x − 2y − 5 = 0……(ii)
a2 = 4, b2 = −2, c2 = −5
For pair of equations to be consistent,
a1a2 ≠ b1b2 OR a1/a2 = b1/b2 = c1/c2
Here,
a1/a2 ≠ b1/b2
Since , 1/4 ≠ −2/−2
⇒ 1/4 ≠ −1
⇒ Pair of equations have unique solution
∴ Pair of equations is consistent.

Question 2: 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?
Solution:
Let Nuri’s age be x and Sonu’s age be y
Five years ago,
x–5 = 3(y−5)
x = 3y − 10……(i)

Ten years later their age,
x+10 = 2(y+10)
x = 2y + 10……(ii)

From (i) and (ii) we get,

3y − 10 = 2y + 10
y = 20
x = 3y − 10
x = 60 − 10 = 50
x = 50, y = 20

Question 3: Three chairs and two tables cost Rs. 1850. Five chairs and three tables cost Rs.2850. Then find the total cost of one chair and one table.

Solution:

Let the cost of a chair be x and that of a table be y
Then according to given condition,
3x + 2y = 1850……(i)

5x + 3y = 2850……(ii)

Multiplying (i) by 3 we get,

9x + 6y = 5550……(iii)

Multiplying (ii) by 2

10x + 6y = 5700……(iv)

Equation (iv) – (iii) gives

x = 150, y = 700

∴ The total cost of one chair and one table is
x + y = Rs.150 + Rs.700 = Rs.850