NCERT Solutions for Class 10 Maths Exercise 3.5

NCERT Solutions for Class 10 Maths Exercise 3.5 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Pair of Linear Equations in Two Variables Download as PDF

NCERT Solutions for Class 10 Maths Pair of Linear Equations in Two Variables

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 − 5y = 20

6x − 10y = 40

(iv) x − 3y – 7 = 0

3x − 3y – 15 = 0

Ans. (i) x − 3y – 3 = 0

3x − 9y – 2 = 0

Comparing equation x − 3y – 3 = 0 with a₁x +b₁y + c₁ = 0 and 3x − 9y – 2 = 0 with ,

We get

Here this means that the two lines are parallel.

Therefore, there is no solution for the given equations i.e. it is inconsistent.

(ii) 2x + y = 5

3x + 2y = 8

Comparing equation 2x + y = 5 with and 3x + 2y = 8 with ,

We get

Herethis means that there is unique solution for the given equations.

⇒

⇒ x = 2 and y = 1

NCERT Solutions for Class 10 Maths Exercise 3.5

(iii) 3x − 5y = 20

6x − 10y = 40

Comparing equation 3x − 5y = 20 with and 6x − 10y = 40 with ,

We get

Here

It means lines coincide with each other.

Hence, there are infinite many solutions.

(iv) x − 3y – 7 = 0

3x − 3y – 15 = 0

Comparing equation x − 3y – 7 = 0 with and 3x − 3y – 15 = 0 with ,

We get

Here this means that we have unique solution for these equations.

⇒

⇒

⇒ x = 4 and y = –1

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NCERT Solutions for Class 10 Maths Exercise 3.5

2. (i) For which values of a and b does the following pair of linear equations 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 equations have no

solution?

3x + y = 1

(2k − 1) x + (k − 1) y = 2k + 1

Ans. (i) Comparing equation 2x + 3y – 7 = 0 with and (a − b) x + (a + b)

y − 3a – b + 2 = 0 with

We get and and

Linear equations have infinite many solutions if

⇒

⇒

⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b

⇒ a = 5b… (1) and −2a = −4b – 6… (2)

Putting (1) in (2), we get

−2 (5b) = −4b – 6

⇒ −10b + 4b = −6

⇒ −6b = –6 ⇒ b = 1

Putting value of b in (1), we get

a = 5b = 5 (1) = 5

Therefore, a = 5 and b = 1

(ii) Comparing (3x + y – 1 = 0) with and (2k − 1)x + (k − 1)y −2k – 1 = 0) with ,

We get and and c₂ = −2k − 1

Linear equations have no solution if

⇒

⇒

⇒ 3 (k − 1) = 2k – 1

⇒ 3k – 3 = 2k − 1

⇒ k = 2

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NCERT Solutions for Class 10 Maths Exercise 3.5

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

8x + 5y = 9

3x + 2y = 4

Ans. Substitution Method

8x + 5y = 9 … (1)

3x + 2y = 4 … (2)

From equation (1),

5y = 9 − 8x⇒ y =

NCERT Solutions for Class 10 Maths Exercise 3.5

Putting this in equation (2), we get

3x + 2 = 4

⇒ 3x + = 4

⇒ 3x −

⇒ 15x − 16x = 20 – 18

⇒ x = −2

Putting value of x in (1), we get

8 (−2) + 5y = 9

⇒ 5y = 9 + 16 = 25⇒ y = 5

Therefore, x = −2 and y = 5

Cross multiplication method

8x + 5y = 9 … (1)

3x + 2y = 4 … (2)

⇒

⇒

⇒ x = −2 and y = 5

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NCERT Solutions for Class 10 Maths Exercise 3.5

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 Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes 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.

Ans.(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y

According to given conditions,

x + 20y = 1000 … (1),

and x + 26y = 1180 … (2)

Subtracting equation (1) from equation (2), we get

6y = 180

⇒ y = 30

Putting value of y in (1), we get

x + 20 (30) = 1000

⇒ x = 1000 – 600 = 400

Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

(ii) Let numerator = x and let denominator = y

According to given conditions,

⇒ 3x – 3 = y … (1) 4x = y + 8 … (1)

⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)

Subtracting equation (1) from (2), we get

4x – y − (3x − y) = 8 – 3

⇒ x = 5

Putting value of x in (1), we get

3 (5) – y = 3

⇒ 15 – y = 3

⇒ y = 12

Therefore, numerator = 5 and, denominator = 12

It means fraction =

NCERT Solutions for Class 10 Maths Exercise 3.5

(iii) Let number of correct answers = x and let number of wrong answers = y

According to given conditions,

3x – y = 40 … (1)

And, 4x − 2y = 50 … (2)

From equation (1), y = 3x − 40

Putting this in (2), we get

4x – 2 (3x − 40) = 50

⇒ 4x − 6x + 80 = 50

⇒ −2x = −30

⇒ x = 15

Putting value of x in (1), we get

3 (15) – y = 40

⇒ 45 – y = 40

⇒ y = 45 – 40 = 5

Therefore, number of correct answers = x = 15and number of wrong answers = y = 5

Total questions = x + y = 15 + 5 = 20

(iv)Let speed of car which starts from part A = x km/hr

Let speed of car which starts from part B = y km/hr

According to given conditions,

(Assuming x > y)

⇒ 5x − 5y = 100

⇒ x – y = 20 … (1)

And,

⇒ x + y = 100 … (2)

Adding (1) and (2), we get

2x = 120

⇒ x = 60 km/hr

Putting value of x in (1), we get

60 – y = 20

⇒ y = 60 – 20 = 40 km/hr

Therefore, speed of car starting from point A = 60 km/hr

And, Speed of car starting from point B = 40 km/hr

(v) Let length of rectangle = x units and Let breadth of rectangle = y units

Area =xy square units. According to given conditions,

NCERT Solutions for Class 10 Maths Exercise 3.5

xy – 9 = (x − 5) (y + 3)

⇒ xy – 9 = xy + 3x − 5y – 15

⇒ 3x − 5y = 6 … (1)

And, xy + 67 = (x + 3) (y + 2)

⇒ xy + 67 = xy + 2x + 3y + 6

⇒ 2x + 3y = 61 … (2)

From equation (1),

3x = 6 + 5y

⇒ x =

Putting this in (2), we get

2 + 3y = 61

⇒ 12 + 10y + 9y = 183

⇒ 19y = 171

⇒ y = 9 units

Putting value of y in (2), we get

2x + 3 (9) = 61

⇒ 2x = 61 – 27 = 34

⇒ x = 17 units

Therefore, length = 17 units and, breadth = 9 units

NCERT Solutions for Class 10 Maths Exercise 3.5

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