NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 – Two Variables, One Line help students understand the concept of linear equations in two variables in a simple and structured way. This chapter explains how two variables can be related through an equation and represented as a straight line on a graph.

The solutions provide step-by-step explanations for all the textbook questions, making it easier for students to understand key concepts, practise problem-solving, and prepare effectively for exams.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.1

1. Write a linear equation in two variables in which a = 3, b = 0 and c = −1/5.

Answer:

The standard form of a linear equation in two variables is:

ax + by + c = 0

Given:
a = 3, b = 0, c = −1/5

Substituting these values:

3x + 0y − 1/5 = 0

Therefore, the required linear equation is:

3x − 1/5 = 0

2. Complete the following table after expressing the given linear equations in standard form.

Answer:

The standard form of a linear equation in two variables is:

ax + by + c = 0

Thus, each equation has been rewritten in the form ax + by + c = 0, and the corresponding coefficients of x, y, and the constant term have been identified.

3. (i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to be ₹t and that of a pen to be ₹p. Charlie wrote t = 2p, whereas Meera wrote p = 2t. Which of these two representations is correct? (ii) In a one-day International Cricket match between India and Sri Lanka played in Nagpur, two Indian batsmen together scored 176 runs. Manisha expressed this situation as x + y = 176, where the number of runs scored by one batsman is x, and the number of runs scored by the other is y. Is this a correct representation?

Answer:

(i)

The cost of a notebook is twice the cost of a pen.

Let:

• Cost of a notebook = ₹t
• Cost of a pen = ₹p

Therefore,

t = 2p

So, Charlie’s representation is correct.

Meera’s equation p = 2t would mean that the cost of a pen is twice the cost of a notebook, which is not given.

(ii)

Let:

• Runs scored by one batsman = x
• Runs scored by the other batsman = y

Together, they scored 176 runs.

Therefore,

x + y = 176

Hence, Manisha’s representation is correct.

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.2

1. Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.

Answer:

To verify whether (4, 3) is a solution of 5x − 6y = 2, substitute x = 4 and y = 3 into the equation.

5(4) − 6(3) = 20 − 18 = 2

The left-hand side is equal to the right-hand side.

Therefore, (4, 3) is a solution of 5x − 6y = 2.

2. Find any two solutions for each of the following equations: (i) 7x − 3y = 21 (ii) 2x + 3y = 5

Answer:

(i) 7x − 3y = 21

For x = 3:

7(3) − 3y = 21
21 − 3y = 21
y = 0

So, one solution is (3, 0).

For x = 0:

7(0) − 3y = 21
−3y = 21
y = −7

So, another solution is (0, −7).

Therefore, two solutions are:

(3, 0) and (0, −7)

(ii) 2x + 3y = 5

For x = 1:

2(1) + 3y = 5
2 + 3y = 5
y = 1

So, one solution is (1, 1).

For x = 4:

2(4) + 3y = 5
8 + 3y = 5
y = −1

So, another solution is (4, −1).

Therefore, two solutions are:

(1, 1) and (4, −1).

3. In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny = −10. If (2, −1) is the solution of both equations, find the values of m and n.

Answer:

Since (2, −1) is a solution of both equations, substitute x = 2 and y = −1 into each equation.

For the first equation:

2mx + 3y = 7

Substituting:

2m(2) + 3(−1) = 7

4m − 3 = 7

4m = 10

m = 5/2

For the second equation:

4x + ny = −10

Substituting:

4(2) + n(−1) = −10

8 − n = −10

−n = −18

n = 18

Therefore,

m = 5/2 and n = 18.

(i) 5x + 3y = 7

For x = −1:

5(−1) + 3y = 7
−5 + 3y = 7
3y = 12
y = 4

So, (−1, 4) is a solution and lies in Quadrant II.

For x = 2:

5(2) + 3y = 7
10 + 3y = 7
3y = −3
y = −1

So, (2, −1) is a solution and lies in Quadrant IV.

Solutions: (−1, 4) – Quadrant II and (2, −1) – Quadrant IV

(ii) 5x − 3y = 7

For x = 2:

5(2) − 3y = 7
10 − 3y = 7
−3y = −3
y = 1

So, (2, 1) lies in Quadrant I.

For x = −1:

5(−1) − 3y = 7
−5 − 3y = 7
−3y = 12
y = −4

So, (−1, −4) lies in Quadrant III.

Solutions: (2, 1) – Quadrant I and (−1, −4) – Quadrant III

(iii) −5x + 3y = 7

For x = 1:

−5(1) + 3y = 7
−5 + 3y = 7
3y = 12
y = 4

So, (1, 4) lies in Quadrant I.

For x = −2:

−5(−2) + 3y = 7
10 + 3y = 7
3y = −3
y = −1

So, (−2, −1) lies in Quadrant III.

Solutions: (1, 4) – Quadrant I and (−2, −1) – Quadrant III

(iv) −5x − 3y = 7

For x = −2:

−5(−2) − 3y = 7
10 − 3y = 7
−3y = −3
y = 1

So, (−2, 1) lies in Quadrant II.

For x = 1:

−5(1) − 3y = 7
−5 − 3y = 7
−3y = 12
y = −4

So, (1, −4) lies in Quadrant IV.

Solutions: (−2, 1) – Quadrant II and (1, −4) – Quadrant IV

Summary

Plotting the two points obtained for each equation and joining them gives the corresponding straight line, verifying that both points satisfy the equation.

5. Consider the graph of the equation 3x − 7y = 21 shown below. Does the point C (2, 3) lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation?

Answer:

No, C(2, 3) does not lie on the line.

At C,

3x − 7y = 3(2) − 7(3) = 6 − 21 = −15

Since −15 ≠ 21, C does not satisfy the equation.

The line represents exactly those points whose coordinates satisfy 3x − 7y = 21. Therefore, a point that does not lie on the line cannot satisfy the equation.

To draw the line, we can use its intercepts A(7, 0) and B(0, −3).

(i) Compare the solutions of 3x + 4y = 7 and 6x + 8y = 14.

The second equation is obtained by multiplying the first equation by 2:

3x + 4y = 7

Multiplying both sides by 2:

6x + 8y = 14

So, any ordered pair (x, y) that satisfies 3x + 4y = 7 will also satisfy 6x + 8y = 14.

Similarly, dividing 6x + 8y = 14 by 2 gives:

3x + 4y = 7

Therefore, every solution of one equation is also a solution of the other.

Hence, both equations have the same set of solutions.

(ii) Show that ax + by = c and kax + kby = kc, where k ≠ 0, have the same set of solutions.

Suppose (x, y) satisfies:

ax + by = c

Multiplying both sides by k:

k(ax + by) = kc

So,

kax + kby = kc

Hence, every solution of ax + by = c is also a solution of kax + kby = kc.

Conversely, if:

kax + kby = kc

then, since k ≠ 0, dividing both sides by k gives:

ax + by = c

Therefore, every solution of the second equation is also a solution of the first.

Hence, ax + by = c and kax + kby = kc have the same set of solutions.

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.3

The slope of a ramp is calculated as:

Slope = Rise / Run

Given:

Rise = 2.5 m
Run = 4 m

So,

Slope = 2.5 / 4 = 0.625 = 5/8

Therefore, the slope of the ramp is 5/8 or 0.625.

Answer:

The ramp guideline is:

1 cm vertical rise : 12 cm horizontal length

Given the total vertical rise is 18 cm:

Horizontal length = 18 × 12 = 216 cm

Therefore, the required horizontal length of the ramp is:

216 cm, or 2.16 m.

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.4

Let:

₹x = cost of one movie ticket
₹y = cost of one snack box

(i)

Cost of 2 movie tickets and 3 snack boxes is ₹850.

Therefore,

2x + 3y = 850

(ii)

Cost of 4 movie tickets and 1 snack box is ₹1100.

Therefore,

4x + y = 1100

Hence, the required pair of linear equations is:

2x + 3y = 850
4x + y = 1100

Let:

₹x = fixed charge for one taxi trip
₹y = additional charge per kilometre

For Sahil:

x + 6y = 122

For Meena:

x + 8y = 160

Hence, the required pair of linear equations is:

x + 6y = 122
x + 8y = 160

3. In a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100 each. A total of 200 people attended the meet, and the total amount collected from ticket sales was ₹25,000. (Let the number of adult tickets sold be x and the number of children’s tickets sold be y.) Frame a pair of linear equations in x and y to represent this situation.

Answer:

Let:

x = number of adult tickets sold
y = number of children’s tickets sold

Since a total of 200 people attended:

x + y = 200

The total amount collected was ₹25,000:

150x + 100y = 25,000

Hence, the required pair of linear equations is:

x + y = 200
150x + 100y = 25,000

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.5

(i) The sum of two integers is +5 and their difference is −21.

Let the two integers be x and y.

x + y = 5
x − y = −21

Adding the equations:

2x = −16
x = −8

Substituting in x + y = 5:

−8 + y = 5
y = 13

Therefore, the two integers are −8 and 13.

(ii) The difference between two numbers is 26 and one number is three times the other.

Let the larger number be x and the smaller number be y.

x − y = 26
x = 3y

Substituting x = 3y:

3y − y = 26
2y = 26
y = 13

Therefore,

x = 3 × 13 = 39

Hence, the numbers are 39 and 13.

(iii) The coach buys 7 bats and 6 balls for ₹8880 and 3 bats and 5 balls for ₹4000.

Let the cost of one bat be ₹x and one ball be ₹y.

7x + 6y = 8880
3x + 5y = 4000

Multiplying the first equation by 5:

35x + 30y = 44400

Multiplying the second equation by 6:

18x + 30y = 24000

Subtracting:

17x = 20400

x = 1200

Substituting in 3x + 5y = 4000:

3600 + 5y = 4000

5y = 400

y = 80

Therefore, the cost of each bat is ₹1200 and each ball is ₹80.

(iv) Taxi charges

Let the fixed charge be ₹x and the charge per kilometre be ₹y.

For 10 km:

x + 10y = 155

For 15 km:

x + 15y = 220

Subtracting the first equation from the second:

5y = 65

y = 13

Substituting in x + 10y = 155:

x + 130 = 155

x = 25

Therefore:

Fixed charge = ₹25
Charge per km = ₹13

For 25 km:

Fare = 25 + 25(13)
= 25 + 325 = ₹350

Hence, the fare for travelling 25 km is ₹350.

(v) Find the fraction.

Let the numerator be x and the denominator be y.

According to the first condition:

(x + 2)/(y + 2) = 9/11

Cross-multiplying:

11x + 22 = 9y + 18

11x − 9y = −4 ...(1)

According to the second condition:

(x + 3)/(y + 3) = 5/6

Cross-multiplying:

6x + 18 = 5y + 15

6x − 5y = −3 ...(2)

Solving (1) and (2), we get:

x = 7, y = 9

Therefore, the fraction is 7/9.

(vi) Find the fraction.

Let the numerator be x and the denominator be y.

From the first condition:

(x + 1)/(y − 1) = 1

Therefore,

x + 1 = y − 1

x − y = −2 ...(1)

From the second condition:

x/(y + 1) = 1/2

Therefore,

2x = y + 1

2x − y = 1 ...(2)

Subtracting (1) from (2):

x = 3

Substituting in (1):

3 − y = −2

y = 5

Therefore, the fraction is 3/5.

(vii) Ages of Nuri and Sonu

Let Nuri's present age be x years and Sonu's present age be y years.

Five years ago:

x − 5 = 3(y − 5)

x − 3y = −10 ...(1)

Ten years later:

x + 10 = 2(y + 10)

x − 2y = 10 ...(2)

Subtracting (1) from (2):

y = 20

Substituting in (2):

x − 40 = 10

x = 50

Therefore, Nuri is 50 years old and Sonu is 20 years old.

(viii) Find the two-digit number.

Let the tens digit be x and the units digit be y.

The sum of the digits is 9:

x + y = 9 ...(1)

The number is 10x + y, and the reversed number is 10y + x.

According to the question:

9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

88x − 11y = 0

8x − y = 0 ...(2)

From (2):

y = 8x

Substituting in (1):

x + 8x = 9

9x = 9

x = 1, y = 8

Therefore, the number is 18.

(ix) ₹50 and ₹100 notes

Let the number of ₹50 notes be x and the number of ₹100 notes be y.

Total number of notes:

x + y = 25 ...(1)

Total amount:

50x + 100y = 2000

Dividing by 50:

x + 2y = 40 ...(2)

Subtracting (1) from (2):

y = 15

Therefore,

x = 25 − 15 = 10

Hence, Meena received 10 notes of ₹50 and 15 notes of ₹100.

(x) Lending library charges

Let the fixed charge for the first three days be ₹x and the additional charge per extra day be ₹y.

Saritha kept the book for 7 days, so she paid for 4 extra days:

x + 4y = 27 ...(1)

Susy kept the book for 5 days, so she paid for 2 extra days:

x + 2y = 21 ...(2)

Subtracting (2) from (1):

2y = 6

y = 3

Substituting in (2):

x + 6 = 21

x = 15

Therefore, the fixed charge is ₹15 and the charge for each extra day is ₹3.

Let the number of boys be x and the number of girls be y.

Since there are 10 students in total:

x + y = 10 ...(1)

Since the number of girls is 4 more than the number of boys:

y = x + 4

or,

y − x = 4 ...(2)

To represent the equations graphically:

• Equation (1): x + y = 10
• Equation (2): y − x = 4

The two lines intersect at the point (3, 7).

For a pair of linear equations

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

we use:

• If a₁/a₂ ≠ b₁/b₂, the lines intersect at one point.
• If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
• If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident.

(i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0

Here,

a₁/a₂ = 5/7

b₁/b₂ = −4/6 = −2/3

Since,

5/7 ≠ −2/3

the lines intersect at one point.

(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0

Here,

a₁/a₂ = 9/18 = 1/2

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = 12/24 = 1/2

Since,

a₁/a₂ = b₁/b₂ = c₁/c₂

the lines are coincident.

(iii) 6x − 3y + 10 = 0 and 2x − y + 9 = 0

Here,

a₁/a₂ = 6/2 = 3

b₁/b₂ = −3/−1 = 3

c₁/c₂ = 10/9

Since,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

the lines are parallel.

Therefore:

(i) Intersecting lines
(ii) Coincident lines
(iii) Parallel lines

(i) x + y = 5 and 2x + 2y = 10

Dividing the second equation by 2:

x + y = 5

Thus, both equations represent the same line. Therefore, the lines coincide and have infinitely many common points.

Hence, the pair has infinitely many solutions.

(ii) x − y = 8 and 3x − 3y = 16

The equations can be written as:

y = x − 8

and

y = x − 16/3

Both lines have the same slope but different intercepts. Therefore, they are parallel and never intersect.

Hence, the pair has no solution.

(iii) 2x + y − 6 = 0 and 4x − 2y − 4 = 0

The equations are:

2x + y = 6

4x − 2y = 4

From the first equation:

y = 6 − 2x

From the second:

y = 2x − 2

The two lines intersect at (2, 2).

Therefore, the pair has a unique solution:

x = 2, y = 2

(iv) 2x − 2y − 2 = 0 and 4x − 4y − 5 = 0

The first equation becomes:

x − y = 1

The second equation becomes:

x − y = 5/4

The two lines have the same slope but different intercepts. Therefore, they are parallel and do not intersect.

Hence, the pair has no solution.

Pair	Graphical relationship	Solution
(i)	Coincident lines	Infinitely many solutions
(ii)	Parallel lines	No solution
(iii)	Intersecting lines	(2, 2)
(iv)	Parallel lines	No solution

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:

Let the width of the garden be x m.

Then the length is:

x + 4 m

Half the perimeter of a rectangle is:

Length + Width

So,

x + (x + 4) = 36

2x + 4 = 36

2x = 32

x = 16

Therefore,

Width = 16 m

Length = 16 + 4 = 20 m

Hence, the dimensions of the garden are 20 m × 16 m.

Given the equation:

2x + 3y − 8 = 0

Here, a₁ = 2, b₁ = 3, c₁ = −8.

(i) Intersecting lines

For intersecting lines:

a₁/a₂ ≠ b₁/b₂

One suitable equation is:

x + y − 4 = 0

Here,

2/1 ≠ 3/1

Therefore, the lines intersect at one point.

(ii) Parallel lines

For parallel lines:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

One suitable equation is:

4x + 6y − 10 = 0

Here,

2/4 = 3/6 = 1/2

but

−8/−10 = 4/5

Therefore, the lines are parallel.

(iii) Coincident lines

For coincident lines:

a₁/a₂ = b₁/b₂ = c₁/c₂

Multiplying the given equation by 2 gives:

4x + 6y − 16 = 0

Here,

2/4 = 3/6 = −8/−16 = 1/2

Therefore, the lines are coincident.

Hence, one possible set of answers is:

(i) x + y − 4 = 0
(ii) 4x + 6y − 10 = 0
(iii) 4x + 6y − 16 = 0

Let the price of one citron be x and the price of one fragrant wood-apple be y.

The equations are:

9x + 7y = 107 ...(1)
7x + 9y = 101 ...(2)

Since the coefficients of x and y are interchanged, add and subtract the equations.

Adding (1) and (2):

16x + 16y = 208

Dividing by 16:

x + y = 13 ...(3)

Subtracting (2) from (1):

2x − 2y = 6

Dividing by 2:

x − y = 3 ...(4)

Now, adding (3) and (4):

2x = 16

x = 8

Substituting in x + y = 13:

8 + y = 13

y = 5

Therefore, the price of each citron is 8 and the price of each fragrant wood-apple is 5.

8. 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.

Answer:

NCERT Class 9 Maths Ganita Manjari Chapter 13 End-of-Chapter Exercises

For the conversion,

F = (9/5)C + 32

where C is temperature in Celsius and F is temperature in Fahrenheit.

(i) Graph of the equation

Taking Celsius on the x-axis and Fahrenheit on the y-axis, the graph is a straight line.

Some convenient points are:

Plot these points and join them to obtain the straight line.

(ii) Temperature when C = 30°C

Substituting C = 30:

F = (9/5)(30) + 32

F = 54 + 32 = 86

Therefore,

30°C = 86°F

(iii) Temperature when F = 95°F

Substituting F = 95:

95 = (9/5)C + 32

63 = (9/5)C

C = 63 × 5/9 = 35

Therefore,

95°F = 35°C

(iv) Temperatures corresponding to 0°C and 0°F

When C = 0:

F = (9/5)(0) + 32 = 32

So,

0°C = 32°F

When F = 0:

0 = (9/5)C + 32

(9/5)C = −32

C = −32 × 5/9 = −160/9

C ≈ −17.8°C

So,

0°F ≈ −17.8°C

(v) Temperature that is numerically the same in Celsius and Fahrenheit

Let the common temperature be T.

Then,

T = (9/5)T + 32

T − (9/5)T = 32

−4/5 T = 32

T = −40

Therefore, the temperature that has the same numerical value on both scales is:

−40°C = −40°F

3. Solve the following system of equations graphically: 2x + y = 6, 2x − y − 2 = 0.

Answer:

Rewrite the equations as:

y = 6 − 2x

y = 2x − 2

Some points for plotting are:

Plotting these points and drawing the two straight lines, we see that they intersect at (2, 2).

Therefore, the graphical solution is:

x = 2, y = 2.

4. Find the point of intersection of the lines shown on the cover page.

Answer:

5. Give a formula to find the x-intercept of the line y = mx + c.

Answer: