NCERT Solutions for Class 10 Maths Exercise 3.5 Chapter 3 Linear Equations in Two Variables are practice guides for students. All the solutions to the questions provided in the textbook can be found in NCERT Solutions Class 10 Maths. The students can refer to these in case of any queries. It will be helpful to them while preparing for the Maths board examinations.

NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables is an important chapter from the board examination perspective. The solutions to the questions mentioned in the chapter are provided in NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.5. NCERT Solutions help the students to evaluate themselves. They can solve the exercises after completing a chapter and analyse their shortcomings. This helps them improve on their weaknesses before the examinations begin.

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### Access Other Exercise Solutions of Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Exercise 3.1 Solutions – 3 Questions

Exercise 3.2 Solutions – 7 Questions

Exercise 3.3 Solutions – 3 Questions

Exercise 3.4 Solutions – 2 Questions

Exercise 3.6 Solutions – 2 Questions

Exercise 3.7 Solutions – 8 Questions

### Access Answers to NCERT Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.5

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

**(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8**

**(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0**

**Solutions: **

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

a_{1}/a_{2}=1/3 , b_{1}/b_{2}= -3/-9 =1/3, c_{1}/c_{2}=-3/-2 = 3/2

(a_{1}/a_{2}) = (b_{1}/b_{2}) ≠ (c_{1}/c_{2})

Since the given set of lines is parallel to each other, they will not intersect each other, and therefore, there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8

a_{1}/a_{2} = 2/3 , b_{1}/b_{2} = 1/2 , c_{1}/c_{2} = -5/-8

(a_{1}/a_{2}) ≠ (b_{1}/b_{2})

Since they intersect at a unique point, these equations will have a unique solution by cross multiplication method.

x/(b_{1}c_{2}-c_{1}b_{2}) = y/(c_{1}a_{2} – c_{2}a1) = 1/(a_{1}b_{2}-a_{2}b_{1})

x/(-8-(-10)) = y/(-15-(-16)) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a_{1}/a_{2}) = 3/6 = 1/2

(b_{1}/b_{2}) = -5/-10 = 1/2

(c_{1}/c_{2}) = 20/40 = 1/2

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Since the given sets of lines overlap each other, there will be an infinite number of solutions for this pair of equations.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a_{1}/a_{2}) = 1/3

(b_{1}/b_{2}) = -3/-3 = 1

(c_{1}/c_{2}) = -7/-15

a_{1}/a_{2} ≠ b_{1}/b_{2}

Since this pair of lines intersect each other at a unique point, there will be a unique solution.

By cross-multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1

**2. (i) For which values of a and b do 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**

**Solution:**

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a_{1}/a_{2} = 2/(a-b) , b_{1}/b_{2} = 3/(a+b) , c_{1}/c_{2} = -7/-(3a + b -2)

For infinitely many solutions,

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Thus, 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus, at a = 5 and b = 1, the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x + (k-1)y – 2k -1 = 0

a_{1}/a_{2} = 3/(2k -1) , b_{1}/b_{2} = 1/(k-1), c_{1}/c_{2} = -1/(-2k -1) = 1/( 2k +1)

For no solutions,

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

3/(2k-1) = 1/(k -1) _{ }≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

k =2

Therefore, for k = 2, the given pair of linear equations will have no solution.

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

**8x + 5y = 9 **

**3x + 2y = 4**

**Solution:**

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2), we get

x = (4 – 2y )/ 3 ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5

Now, using the Cross-multiplication method,

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =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 a mess. When student A takes food for 20 days, she has to pay Rs.1,000 as hostel charges, whereas student B, who takes food for 26 days, pays Rs.1,180 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 on 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 its 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.**

**Solutions:**

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

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii), we get

6y = 180

y = Rs.30

Using this value in equation (ii), we get

x = 1180 -26 x 30

x= Rs.400.

Therefore, the fixed charge is Rs.400, and the charge per day is Rs.30.

**(ii) **Let the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

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

x = 5 ………………………………………….(3)

Using this value in equation (2), we get

(4×5)– y = 8

y= 12

Therefore, the fraction is 5/12.

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

According to the given question,

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

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

x = 15 ….….(3)

Putting this in equation (2), we obtain

30 – y = 25

Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, the total number of questions = 20

**(iv)** Let x km/h be the speed of the car from point A and y km/h be the speed of the car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equations (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get

60 – y = 20

y = 40 km/h

Therefore, the speed of the car from point A = 60 km/h

Speed of car from point B = 40 km/h

(v) Let,

The length of the rectangle = x unit

And the breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

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

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9

Hence, the length of the rectangle = 17 units

And the breadth of the rectangle = 9 units

This exercise contains 4 questions. The solutions provided by the subject experts in NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equation in Two Variables Exercise 3.5 are accurate and easy-to-understand format. The students can refer to these pages for reference. This will surely help them score well.

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

- The answers to the questions are accurate.
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