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## Class IX Chapter 4– Linear Equations in Two Variables Maths Solution

** Maths Chapter 4 Linear**

Equations in Two Variables

**Class IX Chapter 4– Linear Equations in Two Variables Maths**

Equations in Two Variables

Page 1 of 19

Exercise 4.1

Question 1:

The cost of a notebook is twice the cost of a pen. Write a linear equation in two

variables to represent this statement.

(Take the cost of a notebook to be Rs x and that of a pen to be Rs y.)

Answer:

Let the cost of a notebook and a pen be x and y respectively.

Cost of notebook = 2 × Cost of pen

x = 2y

x – 2y = 0

Question 2:

Express the following linear equations in the form ax + by + c = 0 and indicate the

values of a, b, c in each case:

(i) (ii) (iii) – 2x + 3 y = 6

(iv) x = 3y (v) 2x = – 5y (vi) 3x + 2 = 0

(vii) y – 2 = 0 (viii) 5 = 2x

Answer:

(i)

= 0

Comparing this equation with ax + by + c = 0,

a = 2, b = 3,

(ii)

Comparing this equation with ax + by + c = 0,

a = 1, b = , c = -10

(iii) – 2x + 3 y = 6

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 2 of 19

– 2x + 3 y – 6 = 0

Comparing this equation with ax + by + c = 0,

a = -2, b = 3, c = -6

(iv) x = 3y

1x – 3y + 0 = 0

Comparing this equation with ax + by + c = 0,

a = 1, b = -3, c = 0

(v) 2x = -5y

2x + 5y + 0 = 0

Comparing this equation with ax + by + c = 0,

a = 2, b = 5, c = 0

(vi) 3x + 2 = 0

3x + 0.y + 2 = 0

Comparing this equation with ax + by + c = 0,

a = 3, b = 0, c = 2

(vii) y – 2 = 0

0.x + 1.y – 2 = 0

Comparing this equation with ax + by + c = 0,

a = 0, b = 1, c = -2

(vii) 5 = 2x

– 2x + 0.y + 5 = 0

Comparing this equation with ax + by + c = 0,

a = -2, b = 0, c = 5

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 3 of 19

Exercise 4.2

Question 1:

Which one of the following options is true, and why?

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

Answer:

y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions.

As for every value of x, there will be a value of y satisfying the above equation and

vice-versa.

Hence, the correct answer is (iii).

Question 2:

Write four solutions for each of the following equations:

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

Answer:

(i) 2x + y = 7

For x = 0,

2(0) + y = 7

⇒ y = 7

Therefore, (0, 7) is a solution of this equation.

For x = 1,

2(1) + y = 7

⇒ y = 5

Therefore, (1, 5) is a solution of this equation.

For x = -1,

2(-1) + y = 7

⇒ y = 9

Therefore, (-1, 9) is a solution of this equation.

For x = 2,

2(2) + y = 7

⇒ y = 3

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 4 of 19

Therefore, (2, 3) is a solution of this equation.

(ii) πx + y = 9

For x = 0,

π(0) + y = 9

⇒ y = 9

Therefore, (0, 9) is a solution of this equation.

For x = 1,

π(1) + y = 9

⇒ y = 9 – π

Therefore, (1, 9 – π) is a solution of this equation.

For x = 2,

π(2) + y = 9

⇒ y = 9 – 2π

Therefore, (2, 9 -2π) is a solution of this equation.

For x = -1,

π(-1) + y = 9

⇒ y = 9 + π

⇒ (-1, 9 + π) is a solution of this equation.

(iii) x = 4y

For x = 0,

0 = 4y

⇒ y = 0

Therefore, (0, 0) is a solution of this equation.

For y = 1,

x = 4(1) = 4

Therefore, (4, 1) is a solution of this equation.

For y = -1,

x = 4(-1)

⇒ x = -4

Therefore, (-4, -1) is a solution of this equation.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 5 of 19

For x = 2,

2 = 4y

⇒

Therefore, | is a solution of this equation. |

Question 3: | |

Check which of the following are solutions of the equation x – 2y = 4 and which are | |

not: | |

(i) (0, 2 (ii) (2, 0) (iii) (4, 0) |

(iv) (v) (1, 1)

Answer:

(i) (0, 2)

Putting x = 0 and y = 2 in the L.H.S of the given equation,

x – 2y = 0 – 2×2 = – 4 ≠ 4

L.H.S ≠ R.H.S

Therefore, (0, 2) is not a solution of this equation.

(ii) (2, 0)

Putting x = 2 and y = 0 in the L.H.S of the given equation,

x – 2y = 2 – 2 × 0 = 2 ≠ 4

L.H.S ≠ R.H.S

Therefore, (2, 0) is not a solution of this equation.

(iii) (4, 0)

Putting x = 4 and y = 0 in the L.H.S of the given equation,

x – 2y = 4 – 2(0)

= 4 = R.H.S

Therefore, (4, 0) is a solution of this equation.

(iv)

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 6 of 19

Putting and in the L.H.S of the given equation,

L.H.S ≠ R.H.S

Therefore, is not a solution of this equation.

(v) (1, 1)

Putting x = 1 and y = 1 in the L.H.S of the given equation,

x – 2y = 1 – 2(1) = 1 – 2 = – 1 ≠ 4

L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution of this equation.

Question 4:

Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer:

Putting x = 2 and y = 1 in the given equation,

2x + 3y = k

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

⇒ k = 7

Therefore, the value of k is 7.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 7 of 19

Exercise 4.3

Question 1:

Draw the graph of each of the following linear equations in two variables:

(i) (ii) (iii) y = 3x (iv) 3 = 2x + y

Answer:

(i)

It can be observed that x = 0, y = 4 and x = 4, y = 0 are solutions of the above

equation. Therefore, the solution table is as follows.

x | 0 | 4 |

y | 4 | 0 |

The graph of this equation is constructed as follows.

(ii)

It can be observed that x = 4, y = 2 and x = 2, y = 0 are solutions of the above

equation. Therefore, the solution table is as follows.

x | 4 | 2 |

y | 2 | 0 |

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 8 of 19

The graph of the above equation is constructed as follows.

(iii) y = 3x

It can be observed that x = -1, y = -3 and x = 1, y = 3 are solutions of the above

equation. Therefore, the solution table is as follows.

x | – 1 | 1 |

y | – 3 | 3 |

The graph of the above equation is constructed as follows.

(iv) 3 = 2x + y

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 9 of 19

It can be observed that x = 0, y = 3 and x = 1, y = 1 are solutions of the above

equation. Therefore, the solution table is as follows.

x | 0 | 1 |

y | 3 | 1 |

The graph of this equation is constructed as follows.

Question 2:

Give the equations of two lines passing through (2, 14). How many more such lines

are there, and why?

Answer:

It can be observed that point (2, 14) satisfies the equation 7x – y = 0 and

x – y + 12 = 0.

Therefore, 7x – y = 0 and x – y + 12 = 0 are two lines passing through point (2,

14).

As it is known that through one point, infinite number of lines can pass through,

therefore, there are infinite lines of such type passing through the given point.

Question 3:

If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Answer:

Putting x = 3 and y = 4 in the given equation,

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 10 of 19

3y = ax + 7

3 (4) = a (3) + 7

5 = 3a

Question 4:

The taxi fare in a city is as follows: For the first kilometre, the fares is Rs 8 and for

the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and

total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer:

Total distance covered = x km

Fare for 1st kilometre = Rs 8

Fare for the rest of the distance = Rs (x – 1) 5

Total fare = Rs [8 + (x – 1) 5]

y = 8 + 5x – 5

y = 5x + 3

5x – y + 3 = 0

It can be observed that point (0, 3) and satisfies the above equation.

Therefore, these are the solutions of this equation.

x | 0 | |

y | 3 | 0 |

The graph of this equation is constructed as follows.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 11 of 19

Here, it can be seen that variable x and y are representing the distance covered and

the fare paid for that distance respectively and these quantities may not be negative.

Hence, only those values of x and y which are lying in the 1st quadrant will be

considered.

Question 5:

From the choices given below, choose the equation whose graphs are given in the

given figures.

For the first figure For the second figure

(i) | y = x | (i) | y = x +2 |

(ii) | x + y = 0 | (ii) | y = x – 2 |

(iii) | y = 2x | (iii) | y = – x + 2 |

(iv) | 2 + 3y = 7x | (iv) | x + 2y = 6 |

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 12 of 19

Answer:

Points on the given line are (-1, 1), (0, 0), and (1, -1).

It can be observed that the coordinates of the points of the graph satisfy the

equation x + y = 0. Therefore, x + y = 0 is the equation corresponding to the graph

as shown in the first figure.

Hence, (ii) is the correct answer.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 13 of 19

Points on the given line are (-1, 3), (0, 2), and (2, 0). It can be observed that the

coordinates of the points of the graph satisfy the equation y = – x + 2.

Therefore, y = – x + 2 is the equation corresponding to the graph shown in the

second figure.

Hence, (iii) is the correct answer.

Question 6:

If the work done by a body on application of a constant force is directly proportional

to the distance travelled by the body, express this in the form of an equation in two

variables and draw the graph of the same by taking the constant force as 5 units.

Also read from the graph the work done when the distance travelled by the body is

(i) 2 units (ii) 0 units

Answer:

Let the distance travelled and the work done by the body be x and y respectively.

Work done ∝ distance travelled

y ∝ x

y = kx

Where, k is a constant

If constant force is 5 units, then work done y = 5x

It can be observed that point (1, 5) and (-1, -5) satisfy the above equation.

Therefore, these are the solutions of this equation. The graph of this equation is

constructed as follows.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 14 of 19

(i)From the graphs, it can be observed that the value of y corresponding to x = 2 is

10. This implies that the work done by the body is 10 units when the distance

travelled by it is 2 units.

(ii) From the graphs, it can be observed that the value of y corresponding to x = 0 is

0. This implies that the work done by the body is 0 units when the distance travelled

by it is 0 unit.

Question 7:

Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100

towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a

linear equation which satisfies this data. (You may take their contributions as Rs x

and Rs y.) Draw the graph of the same.

Answer:

Let the amount that Yamini and Fatima contributed be x and y respectively towards

the Prime Minister’s Relief fund.

Amount contributed by Yamini + Amount contributed by Fatima = 100

x + y = 100

It can be observed that (100, 0) and (0, 100) satisfy the above equation. Therefore,

these are the solutions of the above equation. The graph is constructed as follows.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 15 of 19

Here, it can be seen that variable x and y are representing the amount contributed

by Yamini and Fatima respectively and these quantities cannot be negative. Hence,

only those values of x and y which are lying in the 1st quadrant will be considered.

Question 8:

In countries like USA and Canada, temperature is measured in Fahrenheit, whereas

in countries like India, it is measured in Celsius. Here is a linear equation that

converts Fahrenheit to Celsius:

(i) Draw the graph of the linear equation above using Celsius for x-axis and

Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the

temperature is 0°F, what is the temperature in Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and

Celsius? If yes, find it.

Answer:

(i)

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 16 of 19

It can be observed that points (0, 32) and (-40, -40) satisfy the given equation.

Therefore, these points are the solutions of this equation.

The graph of the above equation is constructed as follows.

(ii) Temperature = 30°C

Therefore, the temperature in Fahrenheit is 86°F.

(iii) Temperature = 95°F

Therefore, the temperature in Celsius is 35°C.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 17 of 19

(iv)

If C = 0°C, then

Therefore, if C = 0°C, then F = 32°F

If F = 0°F, then

Therefore, if F = 0°F, then C = -17.8°C

(v)

Here, F = C

Yes, there is a temperature, -40°, which is numerically the same in both Fahrenheit

and Celsius.

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 18 of 19

Exercise 4.4

Question 1:

Give the geometric representation of y = 3 as an equation

(I) in one variable

(II) in two variables

Answer:

In one variable, y = 3 represents a point as shown in following figure.

In two variables, y = 3 represents a straight line passing through point (0, 3) and

parallel to x-axis. It is a collection of all points of the plane, having their y-coordinate

as 3.

Question 2:

Class IX Chapter 4– Linear Equations in Two Variables Maths

Page 19 of 19

Give the geometric representations of 2x + 9 = 0 as an equation

(1) in one variable

(2) in two variables

Answer:

(1) In one variable, 2x + 9 = 0 represents a point as shown in the

following figure.

(2) In two variables, 2x + 9 = 0 represents a straight line passing through point

(-4.5, 0) and parallel to y-axis. It is a collection of all points of the plane, having

their x-coordinate as 4.5.

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## CBSE Class 10 Mathematics Solved Sample Paper 2019

In this article you will get CBSE Class 10 Mathematics: Important 1 Mark Questions. All the questions have been provided with proper solution to help you make an easy and quick preparation.

In order to master a subject, on must have a set of questions based on various topic and concepts discussed in that subject. Solving different questions helps to assess your preparedness and familiarise with the different ways in which a concept can be tested. This ultimately helps you to be more prepared and confidents for the exams.

Here, we are providing a set of most important 1 mark questions to prepare for CBSE class 10 Mathematics board exam 2019. All these questions have been provided with detailed solutions explained by the subject experts.

Students may download the whole question bank for CBSE Class 10 Maths Exam 2019 and practice the same while revising the Maths syllabus. Solving various questions will also help to increase your speed and accuracy.

**Given below are some sample questions from the set of CBSE Class 10 Mathematics: Important 1 Mark Questions:**

**Q.** If the HCF of 65 and 117 is expressible in the form of 65 m – 117, then find the value of m.

**Sol.**

We have, 65 = 13 × 5

And 117 = 13 × 9

Hence, HCF = 13

According to question 65m – 117 = 13

⇒ 65m = 13 + 117 = 130

⇒ m = 130/65 = 2

**Q. **If the common difference of an A.P. is 3, then find a_{20 }– a_{15}.

**Sol.**

Let the first term of the AP be a.

a_{n }= a(n − 1)d

a_{20 }– a_{15} = [a + (20 – 1)d] – [a + (15 – 1)d]

= 19d – 14d

= 5d

= 5 × 3

**Q.** Find the value of a so that the point (3, a) lies on the line represented by 2x – 3y = 5.

**Sol. **

Since point (3, a) lies on line 2x – 3y = 5.

Then, 2 x 3 – 3 x a = 5

⟹ 6 – 5 = 3a

⟹ a = 1/3

**Q.** For what value of *k* will *k* + 9, 2 *k* ‒ 1 and 2*k* + 7 are the consecutive terms of an A.P.?

**Sol. **

If three terms *x*, *y* and *z *are in A.P. then, 2 *y* = *x* + *z*

Since *k* + 9, 2* k* ‒ 1 and 2 *k* + 7 are in A.P.

∴ 2(2*k* − 1) = (*k* + 9) + (2*k* + 7)

⟹ 4*k* – 2 = 3*k* + 16

⟹ *k* = 18

**Q.** Find the median using an empirical relation, when it is given that mode and mean are 8 and 9 respectively.

**Sol. **

The relation between Mean, Median and Mode of the given data is:

Mode = 3Median − 2Mean

⟹ 8 = 3Median − 2 × 9

⟹ 3Median = 8 + 18

⟹ Median = 26/3

⟹ Median = 8.67

In CBSE Class 10 Mathematics Board Exam 2019, Section – A will comprise 6 questions of 1 mark each.

Students must practice the important questions given here. This will help them to track their preparedness for the final exam and also get an idea about the important topics of class 10 Maths to be prepared for the exam. Moreover, the solutions provided here will give students an idea about how to write proper solution to each question in the board exam so as to score optimum marks