Important Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
Important Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
Free PDF Download of CBSE Class 10 Maths Chapter 12 Areas Related to Circles Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Areas Related to Circles MCQs with Answers to know their preparation level.
Important Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
Question 1.
How many solutions does the pair of equations y = 0 and y = -5 have? (2013)
Solution:
y = 0 and y = -5 are Parallel lines, hence no solution.
Question 2.
The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number. (2015)
Solution:
Let unit and tens digit be x and y.
∴ Original number = 1x + 10y ...(i)
Reversed number = 10x + 1y
According to question,
x + y = 8
⇒ y = 8 - x ...(ii)
Also, 1x + 10Oy - (10x + y) = 18
⇒ x + 10y - 10x - y = 18
⇒ 9y - 9x = 18
⇒ y - x = 2 ...[Dividing both sides by 9
⇒ 8 - x - x = 2 ...[From (it)
⇒ 8 - 2 = 2x
⇒ 2x = 6
From (it), y = 8 - 3 = 5
From (i), Original number = 3 + 10(5) = 53
Pair of Linear Equations in Two Variables Class 10 Important Questions Long Answer (4 Marks)
Question 3.
The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father. (2012)
Solution:
Let the present ages of his children be x years and y years.
Then the present age of the father = 2(x + y) ...(i)
After 20 years, his children’s ages will be
(x + 20) and (y + 20) years
After 20 years, father’s age will be 2(x + y) + 20
According to the Question,
⇒ 2(x + y) + 20 = x + 20 + y + 20
⇒ 2x + 2y + 20 = x + y + 40
⇒ 2x + 2y - x - y = 40 - 20
⇒ x + y = 20 ...[From (i)
∴ Present age of father = 2(20) = 40 years
Question 4.
A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number. (2013)
Solution:
Let unit’s place digit be x and ten’s place digit bey.
Then original number = x + 10y
and reversed number = 10x + y
According to the Question,
x + 10y = 7(x + y)
x + 10y = 7x + 7y
⇒ 10y - 7y = 7x - x
⇒ 3y = 6x ⇒ y = 2x ...(i)
(x + 10y) - (10x + y) = 18
x + 10y - 10x - y = 18
⇒ 9y - 9x = 180
⇒ y - x = 2 ...[Dividing by 9
⇒ 2x - x = 2 ...[From (i)
∴ x = 2
Putting the value of ‘x’ in (i), we get y = 2(2) = 4
∴ Required number = x + 10y
= 2 + 10(4) = 42
Question 5.
The owner of a taxi company decides to run all the taxis on CNG fuel instead of petrol/diesel. The taxi charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 13 km, the charge paid is ₹129 and for a journey of 22 km, the charge paid is ₹210.
What will a person have to pay for travelling a distance of 32 km? (2014 )
Solution:
Let fixed charge be ₹x and the charge for the distance = ₹y per km
According to the Question,
For a journey of 13 km,
x + 13y = 129 ⇒ x = 129 - 13y ...(/)
For a journey of 22 km, x + 22y = 210 ...(ii)
⇒ 129 - 13y + 22y = 210 ...[From (i)
⇒ 9y = 210 - 129 = 81
⇒ 9y = 81 ⇒ y = 9
From (i), x = 129 - 13(9)
= 129 - 117 = 12
∴ Fixed charge, x = ₹12
∴ The charge for the distance, y = ₹9 per km
To pay for travelling a distance of 32 km
= x + 32y = 12 + 32(9) = 12 + 288 = ₹300