These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5

Question 1.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0

3x – 9y – 2 = 0

(ii) 2x + y = 5

3x + 2y = 8

(iii) 3x – Sy = 20

6x – 10y = 40

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

Solution:

(iii) Equations are 3x – 5y = 20 and 6x – 10y = 40

Here,

So the linear equation has infinitely many solution.

(iv) We have

Question 2.

(i) for which values of a and b does the following pair of linear equation have an infinite number of solutions?

2x + 3y = 7

(a – b)x + (a + b)y = 3a + b – 2

(ii) For which value of K will the following pair of linear equation have no solution.

3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

Solution:

(i) We have

2x + 3y – 7 = 0 … (i)

(a – b) x + (a + b) y – (3a + b – 2) = 0 … (ii)

Here, a_{1} = 2, b_{1} = 3, c_{1} = 7

a_{2} = (a – b), b_{2} = a + b, c_{3} = – (3a + b – 2)

Substracting equation (iv) from equation (iii) we eleminute a

– 4b = – 4

b = 1

Putting this value in equation (iv)

a – 5 = 0

d = 5

Hence a = 5 and h = 1 are the values when equation gives infinite many solution.

(ii) We have

3x + y – 1 = 0 … (i)

(2k – 1)s + (k – 1)y – 2k + 1 = 0 … (ii)

a_{1} = 3, b_{2} = 1

a_{2} = (2k – 1) b_{2} = (k – 1)

For no solution

k = 2 is the value when equation has no solution.

Question 3.

Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

Equations are

By cross multiplication Method :

Equations are

Question 4.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ₹

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

(i) Let x be the fixed charge of the food and y be the charge of food per day.

According to question’s first condition

x + 20y = 1000 … (i)

According to question’s second condition

x + 26y = 1180 … (ii)

By elimination method

Subtract equation (i) from equation (ii)

We get

6y = 180

y = 30

Putting this value is equation (i), we get

x + 20 x 30 = 1000

x = 400

(ii) Let the numerator of the fraction be x and the denominator of the fraction by y.

According to equation first condition

\(\frac { x – 1 }{ y }\) = \(\frac { 1 }{ 3 }\)

3x – y = 3 … (i)

According to question’s second condition

\(\frac { x }{ y + 8}\) = \(\frac { 1 }{ 4 }\)

4x – y = 8 … (ii)

By elemination method

Substracted equation (i) from equation (ii)

x = 5

Putting this value in equation (i)

y = 12

Hence the fraction is \(\frac { 5 }{ 12 }\)

(iii) Let the number of right answers be x and the number of wrong answers by y.

According to question

3x – y = 40 … (i)

4x – 2y = 50

2x – y = 25 … (ii)

By elimination method

Substracted equation (ii) from equation (i)

x = 15

Putting this value in equation (i)

45 – y = 40

y = 5

So the right questions are 15 are wrong questions are 5. The total questions are 20.

(iv) Let the speed of one car be u km/h.

and the speed of another car be v km/h.

According to question

u – v = \(\frac { 100 }{ 5 }\)

u – v = 20

and u + v = 100

By elimination method

Adding both equations

2u = 120

u = 60 km/h

Putting this value in equation in (i)

v = 40 km/h

Hence the speed of one car is 60 km/h and another is 40 km/h.

(v) Let the length of the rectangle be x unit.

and the breadth of the rectangle be y unit.

According to question

In first case

Area of rectangle = x × y

(x – 5) (y + 3) = xy – 9

or 3x – 5y – 6 = 0

In second case

(x + 3) (y – 2) = xy + 67

or 2x + 3y – 61 = 0

By cross multiplication method.