Two Variables One Line Class 9 Solutions Maths Ganita Manjari Chapter 13 - #NCSOLVE 📚

Solving questions with the help of Ganita Manjari Class 9 Solutions and Part 2 Class 9 Maths Chapter 13 Two Variables One Line NCERT Solutions improves confidence.

Ganita Manjari Class 9 Chapter 13 Solutions

Class 9 Ganita Manjari Chapter 13 Solutions

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

Think and Reflect (NCERT Textbook Page No. 92)

Question 1.
Four friends-Ranju, Meena, Farhan, and Toshi- are solving problems.
Ranju: I noticed something! If c = 0 in the standard form of a line ax + by + c = 0, then the line must pass through the origin. Look, if I substitute x = 0 and y = 0 in the equation ax + by = 0, the equation is satisfied. So, the origin lies on the line! We can also say that the line passes through the origin.
Farhan: Let us try for the equation 2x + 3y = 0. Here a = 2, b = 3 but c = 0. If we substitute x = 0, then we get 3y = 0, or y = 0. This means (0, 0) lies on the line. So yes, this line passes through the origin.
Toshi: Suppose our equation has b = c = 0, say, 5x = 0. This becomes x = 0, which is the equation of the y-axis. And the y-axis passes through the origin.
Meena: And if we take a = c = 0, say, 7y = 0, that means y = 0, which is the equation of the x-axis that also passes through the origin. So, they conclude: Whenever c = 0, the line ax + by + c = 0 will always pass through the origin, irrespective of the values of a or b. Do you agree with them?
Solution:
Yes, we agree with them. [c = 0]
As per Ranju: When we substitute the value of c = 0 in ax + by + c = 0 equation becomes ax + by = 0 … (1)
Substituting the value of x = 0 in equation (1), we get
a(0) + by = 0
⇒ 0 + by = 0
⇒ y = 0
(0, 0) is the solution of the equation; i.e., the line passes through the origin.
(a ≠ b, b ≠ 0), c = 0
As per Farhan:
2x + 3y = 0
Substituting the value of x = 0 in equation (1), we get
2(0) + 3y = 0
⇒ 0 + 3y = 0
⇒ y = 0
(0, 0) is the solution of the equation, i.e., the line passes through the origin.
a ≠ 0, b ≠ 0, c = 0
As per Toshi:
5x = 0
⇒ x = 0
x = 0 is the equation of the y-axis, which passes through the origin.
As per Meena:
a = 0, c = 0, 7y = 0
⇒ y = 0
which is the equation of the x-axis, which passes through the origin.
So, in all four cases
(i) For any real value of a, b and c = 0
(ii) a ≠ 0, b ≠ 0, c = 0
(iii) a ≠ 0, b = 0, c = 0
(iv) a = 0, c = 0
∴ The line passes through the origin.

Think and Reflect (NCERT Textbook Page No. 97)

Question 1.
Consider the point on the line AB whose x-coordinate is 8. What is the y-coordinate of this point? Express this as an ordered pair.

Solution:
The line in Figure passes through A(2, 0) and B(4, 3), so its slope is
\(\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{4-2}=\frac{3}{2}\)
Starting at A, increasing x from 2 to 8 gives a run of 6.
The rise is \(\frac {3}{2}\) × 6 = 9
Therefore, the ordered pair is (8, 9).
Using the point-slope form (y – y₁) = m(x – x₁) with point A(2, 0)
⇒ y – 0 = \(\frac {3}{2}\)(x – 2) (∵ m = \(\frac {3}{2}\))
⇒ y = \(\frac {3}{2}\)(x – 2)

Question 2.
What is the x-coordinate of the point on the line AB whose y-coordinate is 6? Express this as an ordered pair.
Solution:
Using y = \(\frac {3}{2}\)(x – 2)
⇒ 6 = \(\frac {3}{2}\)(x – 2)
⇒ 4 = x – 2
⇒ x = 6
Therefore, the point is (6, 6).

Think and Reflect (NCERT Textbook Page No. 99)

Question 1.
Can you use the points A, B and C to verify that the slope of this line is indeed -5?

Solution:
Yes. Points are A(0, 3), B(1, -2), C(-1, 8)
Taking points A(0, 3) and B(1, -2)
Slope = \(\frac{-2-3}{1-0}\) = -5
and taking points B(1, -2) and C(-1, 8)
Slope = \(\frac{8-(-2)}{-1-1}=\frac{10}{-2}\) = -5
and taking points A(0, 3) and C(-1, 8)
Slope = \(\frac{8-3}{-1-0}=\frac{5}{-1}\) = -5
Also y = -5x + 3
Comparing it with y = mx + d, we get
m = -5 i.e. slope = -5

Think and Reflect (NCERT Textbook Page No. 104)

Question 1.
The cost of one paratha and 2 bowls of dahi is ₹ 90. Also, the cost of 2 parathas and 7 bowls of dahi is ₹ 260. Can you figure out the cost of one paratha and one bowl of dahi and explain your reasoning?
Solution:
Let the cost of one paratha be ₹ x
and the cost of one bowl of dahi be ₹ y
From the first condition
1 paratha + 2 bowls of dahi = ₹ 90
and from the second condition
2 parathas + 7 bowls of dahi = ₹ 260
⇒ x + 2y =90 ……..(1)
⇒ 2x + 7y = 260 ……….(2)
From (1), x = 90 – 2y ……..(3)
Substituting the value of x into equation (2), we get
⇒ 2(90 – 2y) + 7y = 260
⇒ 180 – 4y + 7y = 260
⇒ 3y = 80
⇒ y = \(\frac {80}{3}\)
⇒ y = 26.67
Substituting the value of y = \(\frac {80}{3}\) in equation (3), we get
x = 90 – \(\frac{2 \times 80}{3}\)
= 90 – \(\frac {160}{3}\)
= \(\frac{270-160}{3}\)
= \(\frac {110}{3}\)
= 36.67
So the cost of one paratha = ₹ 36.67
and the cost of one bowl of dahi = ₹ 26.67
Verification:
Here, x + 2y = 90
LHS = \(\frac{110}{3}+2 \times \frac{80}{3}\)
= \(\frac{110+160}{3}\)
= \(\frac {270}{3}\)
= 90
= RHS

Think and Reflect (NCERT Textbook Page No. 106)

Question 1.
Solve the puzzle and discuss your strategy with your friends. Can you determine the value represented by each shape?

Solution:

Then row 1 gives,
2T + 2R = 28
⇒ T + R = 14 ……(1)
and row 2 gives,
2H + 2R = 30
⇒ H + R = 15 ……(2)
and row 3 gives, 3C + 1T = 18 ……(3)
and row 4 gives, 3C + 1R = 20 ……(4)
Strategy: Subtracting equation (4) from equation (3), we get
1T – 1R = -2 ……..(5)
Now, adding equation (1) and equation (5), we get
1T + 1R + 1T – 1R = 14 – 2
⇒ 2T = 12
⇒ T = 6
Now, substituting the value of T = 6 in equation (1), we get
6 + R = 14
⇒ R = 14 – 6 = 8
⇒ R = 8
Again, substituting the value of R = 8 in equation (2), we get
H + 8 = 15
⇒ H = 15 – 8
⇒ H = 7
Again, substituting the value of T = 6 in equation (3), we get
3C + 6 = 18
⇒ 3C = 12
⇒ C = 4
Yes, we can determine the value represented by each shape.

Think and Reflect (NCERT Textbook Page No. 107)

Question 1.
What if we had expressed y in terms of x, i.e., take Equation (2) and write y = \(\frac {1}{2}\)(3 – x). Would we still get the same solution?
Solution:
Given equations are 7x – 15y = 2 …..(1)
x + 2y = 3 ……(2)
Substituting y = \(\frac {1}{2}\)(3 – x) in equation (1), we get
7x – 15 × \(\frac {1}{2}\) × (3 – x) = 2
⇒ \(7 x-\frac{45}{2}+\frac{15}{2} x=2\)
⇒ 14x – 45 + 15x = 4
⇒ 29x = 49
⇒ x = \(\frac {49}{29}\)
Now, y = \(\frac{1}{2}\left(3-\frac{49}{29}\right)\) = \(\frac{1}{2}\left(\frac{87-49}{29}\right)\) = \(\frac {19}{29}\)

Think and Reflect (NCERT Textbook Page No. 109)

Question 1.
The method used in solving this problem is called the Elimination Method because we eliminate one of the variables to obtain a linear equation in the other variable. In the example above, we eliminated y. Try the same problem by eliminating x instead of y. Also try to solve the two equations using the Substitution Method. Assess the pros and cons of each method.
Solution:
The given equations are
9x – 4y = 2000 ……..(1)
7x – 3y = 2000 ……..(2)
(a) Elimination Method: To eliminate x, we multiply (1) by 7 and (2) by 9
[∵ Coefficient of x is 7 in (2), and it is 9 in (1)]
Now subtracting, -y = -4000
⇒ y = 4000
Substituting the value of y = 4000 in (1)
9x – 4 × 4000 = 2000
⇒ 9x – 16000 = 2000
⇒ 9x = 18000
⇒ x = 2000
Hence, x = 2000, y = 4000

(b) Substitution method:
From (1), x = \(\frac{2000+4 y}{9}\) ……..(3)
Substituting equation (3) into equation (2), we get
\(7\left(\frac{2000+4 y}{9}\right)\) – 3y = 2000
⇒ 14000 + 28y – 27y = 18000
⇒ y = 4000
Substitute y = 4000 into equation (3), we get
x = \(\frac{2000+4 \times 4000}{9}\)
= \(\frac{2000+16000}{9}\)
= \(\frac {18000}{9}\)
= 2000
Hence x = 2000, y = 4000
Pros and Cons: Substitution is natural when one variable already has coefficient 1 (like x + 2y = 3), but here it brings in fractions. Elimination avoids fractions and is quicker when coefficients are easy to match, but may need larger multipliers (63 here). Both always give the same answer.

Question 2.
Does a pair of linear equations always have a unique solution? Can you analyse this using the Elimination Method? While subtracting the equations to eliminate a particular variable, if the other variable remains, then clearly there is a unique solution. But what if the second variable gets eliminated too while eliminating the first variable? Can you think of a situation where this happens?
Solution:
No. A pair can have one solution, infinitely many solutions, or no solution.
If elimination leaves an equation such as 3y = 6, it determines y and then x, giving a unique solution.
If both variables disappear, inspect the remaining statement:
1. A true identity, such as 0 = 0, means the equations are equivalent and have infinitely many common solutions.
For example, x + y = 3 and 2x + 2y = 6.
Doubling the first and subtracting gives 0 = 0.

2. A false statement, such as 0 = 1, means no ordered pair can satisfy both.
For example, x + y = 3 and 2x + 2y = 7.
Subtracting twice the first from the second gives 0 = 1.

Think and Reflect (NCERT Textbook Page No. 110)

Question 1.
How many solutions are there to the equations 10x – 5y = 20 and 4x – 2y = 8? Give 3 more examples of pairs of equations that lead to 0 = 0 after subtraction, and therefore have infinitely many solutions. Can you find a simple rule to check when this happens?
Solution:
The given equations are
10x – 5y – 20 = 0 ……(1)
4x – 2y – 8 = 0 …….(2)
Now, comparing the above equations with the general equation of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
∴ a₁ = 10, a₂ = 4, b₁ = -5, b₂ = -2, c₁ = -20, c₂ = -8
\(\frac{a_1}{a_2}=\frac{10}{4}=\frac{5}{2} ; \frac{b_1}{b_2}=\frac{-5}{-2}=\frac{5}{2} ; \frac{c_1}{c_2}=\frac{-20}{-8}=\frac{5}{2}\)
So infinitely many solutions.
Three more examples
1. x + y = 3 and 2x + 2y = 6.
Subtract twice the first from the second to obtain 0 = 0.
2. 2x – 3y = 5 and 6x – 9y = 15.
Subtract three times the first from the second to obtain 0 = 0.
3. x – 4y = -2 and -2x + 8y = 4.
Multiply the first by -2, then subtract it from the second to obtain 0 = 0.
Rule: One complete equation must be a non-zero constant multiple of the other, including its constant term.
For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, this means a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ for the same non-zero k.
When all denominators are non-zero, write this as \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Question 2.
What if one of a₂, b₂, c₂ is zero?
Solution:
We must not divide by zero, so the displayed ratio test may no longer be usable.
Instead, test whether the whole first equation is a non-zero multiple of the second:
a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ for one common non-zero k.
If a₂ = 0, this requires a₁ = 0.
For example, y – 2 = 0 and 3y – 6 = 0 represent the same horizontal line.
If b₂ = 0, this requires b₁ = 0.
For example, x – 3 = 0 and 2x – 6 = 0 represent the same vertical line.
If c₂ = 0, this requires c₁ = 0.
For example, x + 2y = 0 and 3x + 6y = 0 represent the same line through the origin.
Zero coefficients do not by themselves determine the number of solutions.
For example, x = 0 and y = 0 have one common solution, whereas x = 0 and x = 1 have none.
Elimination works in all these cases without dividing by a zero coefficient.

Question 3.
Does a pair of linear equations always have either a unique solution or infinitely many solutions?
Solution:
No, a pair of linear equations may have no solution.
For example
3x – 5y = 7 ……..(1)
and 6x – 10y = 9 ……..(2)
On solving, i.e., multiply (1) by 2 and subtract from (2)

This is wrong, so there is no solution.
So a pair of linear equations always has three possibilities:

• Unique solution
• Infinitely many solutions
• No solution

Think and Reflect (NCERT Textbook Page No. 111)

Question 1.
Give 3 more examples of pairs of equations that have no solution. Can you find a simple rule to check when this will happen?
Solution:
1. x + y = 1 and x + y = 3.
Subtracting gives 0 = 2.
2. 2x – y = 4 and 4x – 2y = 9.
Subtracting twice the first equation from the second gives 0 = 1.
3. 3x + 2y = 5 and 9x + 6y = 16.
Subtracting three times the first equation from the second gives 0 = 1.
Each pair has no solution because elimination produces a false statement.
Rule: The coefficients of x and y are proportional by the same factor, but the constant terms are not proportional by that factor.
In standard form, a₁ = ka₂ and b₁ = kb₂, but c₁ ≠ kc₂.
When the denominators are non-zero, this is \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
The corresponding lines are distinct and parallel.

Think and Reflect (NCERT Textbook Page No. 112)

Question 1.
Are the converses of the above statements true?
Solution:
Yes, under the stated conditions that both equations represent lines and the denominators in the ratios are non-zero.
The three converses are:
1. If a pair has a unique solution, then \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
2. If a pair has infinitely many solutions, then \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
3. If a pair has no solution, then \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Reason: These three coefficient cases are mutually exclusive and cover all possibilities.
Unequal x- and y-coefficient ratios leave a variable after elimination; proportional coefficients eliminate both variables.
In the latter case, equal constant ratios give 0 = 0 and unequal constant ratios give a contradiction.
Therefore, each possible number of solutions forces exactly its stated coefficient case.
If a denominator is zero, use the non-zero-multiple test or elimination instead of the ratios.

Think and Reflect (NCERT Textbook Page No. 113)

Question 1.
How do we find solutions from the graph? What can we say about the number of solutions when the two lines (i) intersect at a point, (ii) are parallel, (iii) are coincident?
Solution:
To find solutions from a graph, look for the points where the two graphed lines overlap or cross each other, as any shared point satisfies both equations.
Number of Solutions for Lines on a Graph:
(i) Intersect at a point: There is one unique solution, which is given by the exact (x, y) coordinates of the single point where the lines cross.
2x + y = 6, x – y = 2

(ii) Are parallel: There are no solutions because the lines never touch or share any common points.
x – 2y = 4; 2x – 4y = 6

(iii) Are coincident: There are infinitely many solutions because the two lines lie directly on top of each other, meaning every single point on the line satisfies both equations.
3x – 2y = 6; 6x – 4y = 12

Think and Reflect (NCERT Textbook Page No. 115)

Question 1.
Can you prove that two lines of equal slope are parallel?
(Hint: Consider the equations of the two lines to be y = mx + d₁ and y = mx + d₂)
Solution:
Let us consider two lines y = mx + d₁ and y = mx + d₂ having the same slope m but different values of the y-intercepts d₁ and d₂.
We are to prove that lines are parallel means they do not have common point.
Let us suppose they have a common point (a, b), then
b = ma + d₁ ……(1)
b = ma + d₂ ………(2)
By Solving (1) & (2)
0 = 0 + (d₁ – d₂)
⇒ d₁ – d₂ = 0
⇒ d₁ = d₂
Hence, when d₁ ≠ d₂, no common point is possible.
The two lines are parallel.

Ex 13.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 13.1 Solutions

Exercise 13.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 13.1 Solutions

Question 1.
Write a linear equation in two variables in which a = 3, b = 0 and c = \(-\frac {1}{5}\).
Solution:
We know that the standard form of a linear equation is ax + by + c = 0
Here a = 3, b = 0, c = \(-\frac {1}{5}\)
Putting the values of a, b, and c, we get
3x + 0y – \(\frac {1}{5}\) = 0
⇒ 3x – \(\frac {1}{5}\) = 0
⇒ \(\frac{15 x-1}{5}\) = 0
⇒ 15x – 1 = 0

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

Solution:

Question 3.
(i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to he 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?
Solution:
(i) Given, the cost of a notebook = ₹ t
and the cost of a pen = ₹ p
So, cost of a notebook = 2 × cost of a pen
t = 2 × p
Then the correct representation is t = 2p
i.e., Charlie’s representation is correct.

(ii) Given the number of runs scored by one batsman = x
Number of runs scored by the other batsman = y
Total score by both batsmen = 176
Runs scored by one batsman and runs scored by the other batsman = Total runs scored by both batsmans
i.e., x + y = 176
So, x + y = 176 is the correct representation.
Yes, Manisha’s representation is correct.

Ex 13.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 13.2 Solutions

Exercise 13.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 13.2 Solutions

Question 1.
Verify if the ordered pair (4, 3) is a solution of 5x – 6y = 2. Explain your reasoning.
Solution:
The given equation is 5x – 6y = 2, and (4, 3) is a solution to the given equation.
Substituting x = 4 and y = 3, we get
5x – 6y = (5 × 4) – (6 × 3)
= 20 – 18
= 2
= R.H.S
Hence, (4, 3) is a solution of 5x – 6y = 2.
An ordered pair is a solution if, on substituting its values, the L.H.S. is equal to the R.H.S.

Question 2.
Find any two solutions for each of the following equations:
(i) 7x – 3y = 21
(ii) 2x + 3y = 5
Solution:
(i) Given equation is 7x – 3y = 21 …..(1)
Substituting x = 0 into equation (1), we get
7(0) – 3y = 21
⇒ -3y = 21
⇒ y = -7
So, one solution is (0, -7).
Again, substituting y = 0 into equation (1), we get
7x – (3 × 0) = 21
⇒ 7x = 21
⇒ x = 3
So, the other solution is (3, 0).

(ii) Given equation is 2x + 3y = 5 ………(1)
Substituting x = 1 in equation (1), we get
2(1) + 3y = 5
⇒ 2 + 3y = 5
⇒ 3y = 5 – 2
⇒ 3y = 3
⇒ y = 1
So, one solution is (1, 1).
Again, substituting x = 4 into equation (1), we get
2(4) + 3y = 5
⇒ 8 + 3y = 5
⇒ 3y = 5 – 8
⇒ 3y = -3
⇒ y = -1
So, the other solution is (4, -1).

Question 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.
Solution:
Given equation is 2mx + 3y = 7 …..(1)
and (2, -1) is a solution of equation (1)
Substituting x = 2, y = -1 in the given equation
2m(2) + 3(-1) = 7
⇒ 4m – 3 = 7
⇒ 4m = 7 + 3
⇒ m = \(\frac{10}{4}=\frac{5}{2}\)
Also (2, -1) is a solution of equation 4x + ny = -10
∴ Substituting x = 2, y = -1 into the equation (2).
4(2) + n(-1) = -10
⇒ 8 – n = -10
⇒ -n = -10 – 8
⇒ -n = -18
⇒ n = 18
Hence m = \(\frac {5}{2}\) and n = 18.

Question 4.
Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie.
(i) 5x + 3y = 7
(ii) 5x – 3y = 7
(iii) -5x + 3y = 7
(iv) -5x – 3y = 7
Verify your solutions by representing the linear equations on a graph paper.
Solution:
(i) The given equation is 5x + 3y = 7 …..(1)
Substituting x = 2 into equation (1)
5(2) + 3(y) = 7
⇒ 10 + 3y = 7
⇒ 3y = 7 – 10
⇒ 3y = -3
⇒ y = -1
Also, point A(2, -1) lies in the IVth quadrant. (∵ x is positive and y is negative)
Now, substituting x = -1 into equation (1)
5(-1) + 3y = 7
⇒ -5 + 3y = 7
⇒ 3y = 7 + 5
⇒ 3y = 12
⇒ y = 4
Also, point B(-1, 4) lies in the II quadrant. (∵ x -ve and y +ve)
Now mark a point O on the graph paper as the origin.
Draw two lines X’X and YY’ perpendicular to each other and passing through the origin O.
Scale: take 1 cm = 1 unit on x-axis
1 cm = 1 unit on the y-axis
To plot the point (2, -1), start from the origin O, move along OX and reach the point marked 2.
From here, move -1 unit downwards parallel to the y-axis.
Mark the point A and write (2, -1).
To plot point (-1, 4), start from the origin O, move along OX’ and reach the point marked -1.
From here, move 4 units upwards parallel to the y-axis.
Mark the point B and write (-1, 4) there.
Join these points to form a line AB.

(ii) Given equation is 5x – 3y = 7 …..(1)
Substituting x = (-1) in equation (1), we get
5(-1) – 3y = 7
⇒ -5 – 3y = 7
⇒ -3y = 7 + 5
⇒ y = -4
Here both x and y are negative.
Hence, point A(-1, -4) lies in the IIIrd quadrant.
Now, substituting x = 2 in equation (1), we get
5(2) – 3y = 7
⇒ 10 – 3y = 7
⇒ -3y = 7 – 10
⇒ -3y = -3
⇒ y = 1
Here, both x and y are positive.
Hence, point B(2, 1) lies in the first quadrant.
Using points A(-1, -4) and B(2, 1), we can draw the graph of the above equation.

(iii) Given equation is -5x + 3y = 7 …..(1)
Substituting x = 1 in equation (1), we get
-5(1) + 3y = 7
⇒ -5 + 3y = 7
⇒ 3y = 7 + 5
⇒ 3y = 12
⇒ y = 4
Here, both x and y are positive.
Hence, point A(1, 4) lies in the first quadrant.
Now, substituting x = -2 in equation (1), we get
-5(-2) + 3y = 7
⇒ 10 + 3y = 7
⇒ 3y = 7 – 10
⇒ 3y = -3
⇒ y = -1
Here both x and y are negative.
∴ Point B(-2, -1) lies in the IIIrd quadrant.
Using points A(1, 4) and B(-2, -1), we can draw the graph of the above equation.

(iv) Given equation is -5x – 3y = 7 …..(1)
Substituting x = 1 in equation (1), we get
-5(1) – 3y = 7
⇒ -5 – 3y = 7
⇒ -3y = 7 + 5
⇒ y = -4

Here, x is positive, and y is negative.
Hence, point A(1, -4) lies in the IVth quadrant.
Now, substituting x = -2 in given equation
-5(-2) – 3y = 7
⇒ 10 – 3y = 7
⇒ -3y = 7 – 10
⇒ -3y = -3
⇒ y = 1
Here, x is positive, and y is negative.
Hence, point B(2, -1) lies in the IIIrd quadrant.
Using points A(1, -4) and B(-2, 1), we can draw the graph of the above equation.
Now let us prepare a table for the given equations.

Verification: In each case, we observe that when we plotted the points of the solutions, they were found to lie on the lines representing the equation, which confirms that they are solutions.

Question 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?

Solution:
On observing the given graph, we see that point C lies on I quadrant while point A(7, 0) and B(0, -3) lie on x and y axis respectively.
This means point C(2, 3) does not lie on the line.
Substituting x = 2, y = 3 in equation 3x – 7y = 21
L.H.S. = 3(2) – 7(3)
= 6 – 21
= -15 ≠ 21
∴ L.H.S. ≠ R.H.S.
So the point C(2, 3) does not lie on the line.
It does not satisfy the equation.
Every point that satisfies the equation lies on the line, and every point on the line satisfies the equation.

Question 6.
State whether the following sentences are True or False. Justify your answer.
(i) A linear equation in two variables has only one solution.
(ii) The graph of a linear equation in two variables always passes through the origin.
(iii) A linear equation in two variables can never have rational solutions.
(iv) x = 3 is a valid linear equation in two variables.
(v) The equation 2x + 3y = 7 has infinitely many solutions.
(vi) The point (1, 2) is a solution of the equation 2x + 3y = 7.
Solution:
(i) False
A linear equation in two variables has infinitely many solutions because for every value chosen for x, we get a value of y and vice versa; we can choose infinite values of x to get infinite values of y.

(ii) False
The graph of the linear equation y = mx passes through the origin, but the graph of y = mx + c does not pass through the origin.

(iii) False
\(\left(\frac{1}{2}, \frac{1}{2}\right)\) is a rational solution of x + y = 1.

(iv) True
x = 3 can be written as x + 0.y – 3 = 0, which is the standard form of a linear equation in two variables.

(v) True
2x + 3y = 7
⇒ y = \(\frac{7-2 x}{3}\) for any value of x, we can get a corresponding value of y.
This means the given equation has infinitely many solutions.

(vi) False
Substituting (1, 2) i.e., (x = 1 and y = 2) in the given equation 2x + 3y = 1
⇒ 2(1) + 3(2) = 7
⇒ 2 + 6 = 7
⇒ 8 ≠ 7
RHS ≠ LHS.
Hence, the given point is not the solution of the equation 2x + 3y = 7.

Question 7.
(i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other.
(ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.
Solution:
(i) Given, 3x + 4y = 7 …….(1)
6x + 8y = 14 …….(2)
When we multiply both sides of equation (1) by 2, we get
6x + 8y = 14
Which is the same as the given equation (2).
So, both the equations have the same set of solutions.
For example, substitute x = 1 into equation (1)
⇒ 3(1) + 4y = 7
⇒ 4y = 7 – 3 = 4
⇒ y = 1; we get the solution (1, 1).
3x + 4y = 7 as 3(1) + 4(1) = 7
Now, if we substitute the same solution (1, 1) into equation (2), we get
L.H.S. = 6(1) + 8(1)
= 6 + 8
= 14
= R.H.S.
⇒ (1, 1) is also a solution of equation (2).

(ii) We have ax + by = c ………(1)
kax + kby = kc, k ≠ 0 ……. (2)
Let (m, n) be the solution of equation (1)
∴ ax + by = c
⇒ am + bn = c
On multiplying with k both sides, we get kam + kbn = kc
which confirms that (m, n) is a solution of kax + kby = kc.

Ex 13.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 13.3 Solutions

Exercise 13.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 13.3 Solutions

Question 1.
The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest.

Solution:
(i) Here, rise = 60 m and run = 70 m
∴ Slope of hill A = \(\frac{\text { rise }}{\text { run }}=\frac{60}{70}=\frac{6}{7}\) = 0.857 approx.

(ii) Here, rise = 60 m and run = 110 m
∴ Slope of hill B = \(\frac{\text { rise }}{\text { run }}=\frac{60}{110}=\frac{6}{11}\) = 0.545 approx.

(iii) Here, rise = 80 m and run = 100 m
∴ Slope of hill C = \(\frac{\text { rise }}{\text { run }}=\frac{80}{100}=\frac{8}{10}\) = 0.80
∴ 0.545 < 0.80 < 0.857
Therefore, Slope of B < Slope of C < Slope of A.
⇒ Steepness of B < Steepness of C < Steepness of A.
Hence, the order from least to greatest steepness is B, C, A.

Question 2.
The ramp at a loading dock rises 2.5 metres over a run of 4 metres. Find the slope of the ramp.

Solution:
Here, rise = 2.5 m and run = 4 m.
∴ Slope of ramp = \(\frac{\text { rise }}{\text { run }}=\frac{2.5}{4}\) = 0.625

Question 3.
Find the slope of the line l in each of the following diagrams.


Solution:
(i) The line l is horizontal
The rise is 0
∴ Slope = 0 (∵ Slope = \(\frac {rise}{run}\)

(ii) By observing the given graph, we get the points A(-2, 2) and B(2, -2)
Hence, x₁ = -2, y₁ = 2; x₂ = 2, y₂ = -2.
So, slope m = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-2-(2)}{2-(-2)}\)
= \(\frac{-2-2}{4}\)
= \(\frac {-4}{4}\)
= -1

(iii) By observing the given graph, we get the points A(1, 1) and B(-2, -2).
Hence, x₁ = 1; y₁ = 1; x₂ = -2; y₂ = -2.
∴ Slope m = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-2-1}{-2-1}\)
= \(\frac {-3}{-3}\)
= 1

(iv) The line l is vertical.
Here run = 0
∴ Slope = \(\frac{\text { rise }}{\text { run }}=\frac{\text { rise }}{0}\) = ∞
So the slope is undefined.

Question 4.
An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp are 1 cm vertical rise to 12 cm horizontal length. What should be the horizontal length of the ramp if the total height of the stairs is 18 cm?

Solution:
First case (The guideline):
Here, rise = 1 cm
Horizontal length; run = 12 cm
Slope = \(\frac{\text { rise }}{\text { run }}=\frac{1}{12}\) …….(1)
Second case (Required):
Here, rise = 18 cm
Let horizontal length = x
Slope = \(\frac {18}{x}\) ………(2)
From (1) and (2)
\(\frac{1}{12}=\frac{18}{x}\)
⇒ x = 18 × 12
⇒ x = 216 cm

Ex 13.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 13.4 Solutions

Exercise 13.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 13.4 Solutions

Given below are some everyday situations. Construct a pair of linear equations in two variables for each situation:

Question 1.
On two different days, a family buys movie tickets and snack boxes. Let the cost of one movie ticket be ₹ x and the cost of one snack box be ₹ y. Frame a pair of linear equations in x and y to represent the following situations:
(i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹ 850.
(ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹ 1100.
Solution:
Let the cost of 1 movie ticket = ₹ x
and the cost of 1 snack box = ₹ y
(i) Cost of 2 movie tickets = 2 × x = 2x
Cost 3 snack boxes = 3 × y = 3y
Total amount spent = ₹ 850
⇒ 2x + 3y = 850

(ii) Cost of 4 movie tickets = 4 × x = 4x
and cost of 1 snack box = 1 × y = y
Total amount = ₹ 1100
Then the cost of 4 movie tickets and 1 snack box = 1100
⇒ 4x + y = ₹ 1100

Question 2.
Two friends, Sahil and Meena, travelled by taxi. Let the fixed charge for one taxi trip be ₹ x and the additional charge per kilometre be ₹ y. Frame a pair of linear equations in x and y to represent the following situation:
(i) Sahil travelled 6 km and paid a total fare of ₹ 122.
(ii) Meena travelled 8 km and paid a total fare of ₹ 160.
Solution:
(i) The total taxi fare consists of a fixed charge (₹ x) plus the additional charge per kilometer (₹ y) multiplied by the distance travelled.
∴ Total fare = Fixed charge + (distance) × Charge per kilometer
Sahil travelled 6 km, so the additional charge = 6 × y = 6y.
He paid a total fare of ₹ 122.
∴ x + 6y = 122

(ii) Meena travelled 8 km, then the additional charge = 8 × y = 8y.
She paid a total fare of ₹ 160.
∴ x + 8y = 160

Question 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.
Solution:
Let the number of adult tickets sold = x
and the number of children’s tickets sold = y
Now, the cost of 1 adult ticket = ₹ 150
then the cost of x adult tickets = x × 150 = 150x
and the cost of 1 child ticket sold = ₹ 100
Then the cost of y children’s tickets = y × 100 = 100y
Total number of people who attended the meet = 200
⇒ x + y = 200
The total money collected from x adult tickets at ₹ 150 each and y children’s tickets at ₹ 100 each is ₹ 25000.
⇒ 150x + 100y = 25000

Ex 13.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 13.5 Solutions

Exercise 13.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 13.5 Solutions

Question 1.
Form a pair of linear equations for each of the following problems and find their solutions.
(i) The sum of two integers is +5, and their difference is -21. Find the two numbers.
Solution:
Let the two integers be x and y.
According to the question,
The sum of the two integers = 5
⇒ x + y = 5 ……..(1)
and the difference of the two integers = -21
⇒ x – y = -21 …….(2)
Adding equation (1) and equation (2), we get
2x = -16
⇒ x = -8
Putting the value of x = -8 in equation (1), we get
-8 + y = 5
⇒ y = 5 + 8
⇒ y = 13
Hence, the required numbers are -8 and 13.

(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Solution:
Let the larger number = x
Smaller number = y
According to the question.
The difference between the two numbers = 26
⇒ x – y = 26 …….(1)
and the larger number is 3 times the smaller number.
Then x = 3y
⇒ x – 3y = 0 ……(2)
Subtracting equation (2) from equation (1), we get
2y = 26
⇒ y = 13
Substituting the value of y = 13 into equation (2), we get
x – 3 × 13 = 0
⇒ x – 39 = 0
⇒ x = 39
Hence, the numbers are 13 and 39.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 8880. Later, she buys 3 bats and 5 balls for ₹ 4000. Find the cost of each bat and each ball.
Solution:
Let the cost of 1 bat = x
According to the question,
Cost of 1 ball = y
Cost of 7 bats and 6 balls is ₹ 8880
⇒ 7x + 6y = 8880 ……(1)
Cost of 3 bats and 5 balls is ₹ 4000
⇒ 3x + 5y = 4000 ……(2)
Multiply equation (1) by 3 and equation (2) by 7
21x + 18y = 26640 …….(3)
and 21x + 35y = 28000 …….(4)
Now subtract equation (3) from equation (4)
17y = 1360
⇒ y = 80
Substitute the value of y = 80 into equation (1)
7x + 6 × 80 = 8880
⇒ 7x + 480 = 8880
⇒ 7x = 8880 – 480
⇒ 7x = 8400
⇒ x = 1200
Therefore, cost of 1 bat = ₹ 1200 and cost of 1 ball = ₹ 80

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered for every km. For a distance of 10 km, the total amount paid is ₹ 155, and for a journey of 15 km, the total amount paid is ₹ 220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution:
Let fixed charges = ₹ x
Charges per km = ₹ y
According to the question,
Cost of travelling 10 km is ₹ 155
x + 10y = 155 …….(1)
and the cost of travelling 15 km is ₹ 220
x + 15y = 220 ……(2)
Subtract equation (2) from equation (1)
5y = 65
⇒ y = 13
Substitute the value of y = 13 into equation (1)
x + 10 × 13 = 155
⇒ x = 155 – 130
⇒ x = 25
Therefore fix charges = ₹ 25
Charges per km = ₹ 13
Now, for 25 km
Total charges for 25 km = 25 × x
= 25 × 13
= ₹ 325
Hence, a person has to pay for travelling a distance of 25 km = Fixed charge + Charge per km
= 25 + 325
= ₹ 350

(v) A fraction becomes equal to \(\frac {9}{11}\) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to \(\frac {5}{6}\). Find the fraction.
Solution:
Let the fraction = \(\frac {x}{y}\)
According to the question,
Condition I:
\(\frac{x+2}{y+2}=\frac{9}{11}\)
⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y = -4 …….(1)
Condition II:
\(\frac{x+3}{y+3}=\frac{5}{6}\)
⇒ 6(x + 3) = 5(y + 3)
⇒ 6x + 18 = 5y + 15
We get 6x – 5y = -3 …….(2)
From (1), x = \(\frac{-4+9 y}{11}\) ……(3)
Substituting the value of x from equation (3) in equation (2), we get
\(6\left(\frac{-4+9 y}{11}\right)\) – 5y = -3
⇒ \(\frac{-24+54 y-55 y}{11}\) = -3
⇒ -24 + 54y – 55y = -33
⇒ -y = -9
⇒ y = 9
Now, substituting the value of y = 9 into (3), we get
x = \(\frac{-4+9 \times 9}{11}\)
= \(\frac{-4+81}{11}\)
= \(\frac {77}{11}\)
= 7
Hence, x = 7, y = 9
Therefore, required fraction = \(\frac{x}{y}=\frac{7}{9}\)

(vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to \(\frac {1}{2}\) if we add 1 only to the denominator. What is the fraction?
Solution:
Let the fraction = \(\frac {x}{y}\)
According to the question,
Condition I:
\(\frac{x+1}{y-1}\) = 1
⇒ x + 1 = y – 1
⇒ x – y = -2 ………(1)
Condition II:
\(\frac{x}{y+1}=\frac{1}{2}\)
⇒ 2x = y + 1
⇒ 2x – y = 1 ………(2)
Subtracting equation (2) from equation (1), we get
-x = -3
⇒ x = 3
Now, substituting the value of x = 3 in equation (1), we get
3 – y = -2
⇒ -y = -2 – 3
⇒ y = 5
Thus, x = 3, y = 5
Therefore, required fraction = \(\frac{x}{y}=\frac{3}{5}\)

(vii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution:
Let Sonu’s present age = x years
Nuri’s present age = y years
Then, five years ago
Sonu’s age = x – 5 years
Nuri’s age = y – 5 years
And ten years later
Sonu’s age = x + 10 years
Nuri’s age = y + 10 years
According to the question,
Condition I:
Five years ago, Nuri was thrice as old as Sonu,
⇒ y – 5 = 3(x – 5)
⇒ y – 5 = 3x – 15
⇒ 3x – y = 10 …….(1)
Condition II:
Ten years later, Nuri will be twice as old as Sonu
⇒ y + 10 = 2(x + 10)
⇒ y + 10 = 2x + 20
⇒ 2x – y = -10 …….(2)
Subtracting equation (1) from equation (2), we get
⇒ -x = -20
⇒ x = 20
Substituting the value of x = 20 in equation (1), we get
3 × 20 – y = 10
⇒ -y = 10 – 60
⇒ -y = -50
⇒ y = 50
Hence, Sonu’s age = 20 years
and Nuri’s age = 50 years

(viii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let the digit at the unit place = x
The digit at the tens place = y
Then original number = 10y + x
New number after reversing the order of digits = 10x + y
According to the question,
Condition I:
x + y = 9 ………(1)
Condition II:
9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x + 2y
⇒ 9x – 20x = 2y – 90y
⇒ -11x = -88y
⇒ x = 8y
Substituting the value of x = 8y into equation (1), we get
8y + y = 9
⇒ 9y = 9
⇒ y = 1
Substituting the value of y = 1 into equation (2), we get
x = 8 × 1
⇒ x = 8
Thus, x = 8, y = 1
Now, original number = 10y + x
= 10 × 1 + 8
= 10 + 8
= 18
Hence, the required number is 18.

(ix) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 did she receive?
Solution:
Let number of ₹ 50 notes = x
and Number of ₹ 100 notes = y
According to the question,
Meena got 25 notes in all
⇒ x + y = 25 ……(1)
Total amount she withdrew
⇒ 50x + 100y = 2000 …….(2)
From equation (1)
y = 25 – x ……..(3)
Substituting the value of y in equation (2), we get
50x + 100(25 – x) = 2000
⇒ 50x + 2500 – 100x = 2000
⇒ -50x = -500
⇒ x = 10
Substituting the value of x = 10 into equation (3), we get
y = 25 – 10
⇒ y = 15
Thus, x = 100, y = 15
Therefore, number of ₹ 50 notes = 10 and number of ₹ 100 notes = 15

(x) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
Let the fixed charge for the first 3 days = ₹ x
and Additional charge per day = ₹ y
According to the question,
Saritha paid ₹ 27 for 7 days, then
x + 4y = 27 …….(1)
Susy paid ₹ 21 for 5 days, then
x + 2y = 21 ……..(2)
Subtracting equation (2) from equation (1), we get
2y = 6
⇒ y = 3
Substituting the value of y = 3 in (1), we get
x + 4(3) = 27
⇒ x = 27 – 12
⇒ x = 15
∴ Fixed charge = x = ₹ 15
and additional charge = y = ₹ 3

Question 2.
Form a pair of linear equations and find their common solutions graphically.
10 students of Grade 9 took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution:
Let the number of boys be x
and the number of girls be y
Given the total number of students is 10
⇒ x + y = 10 ……… (1)
and the number of girls is 4 more than the number of boys, then
y = x + 4
⇒ -x + y = 4
⇒ x – y = -4 ……(2)
To graph a straight line, we need to find at least two coordinate points (x, y) for each equation.
For equation x + y = 10
Expressing y in terms of x, we get y = 10 – x
When x = 1, then y = 10 – 1 = 9
When x = 2, then y = 10 – 2 = 8
When x = 3, then y = 10 – 3 = 7
When x = 4, then y = 10 – 4 = 6
Table of values

Hence, (2, 8), (3, 7) and (4, 6) are points on the above line.
For equation x – y = -4
When x = 1, 1 – y = -4
⇒ y = 5
When x = 2, 2 – y = -4
⇒ y = 6
When x = 3, 3 – y = -4
⇒ y = 7
Table of values

Hence, (1, 5), (2, 6) and (3, 7) are points on the line.

Hence, (3, 7) is the common (unique) solution.
Therefore, the number of boys = 3
and the number of girls = 7

Question 3.
Use the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}\), to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel, or are coincident.
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
(i) Given linear equations are 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0
For the first line a₁ = 5, b₁ = -4, c₁ = 8
and for the second line a₂ = 7, b₂ = 6, c₂ = -9
Now, \(\frac{a_1}{a_2}=\frac{5}{7} ; \frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}\)
Since, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the lines intersect at a point.

(ii) Given linear equations are 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
For the first line a₁ = 9, b₁ = 3, c₁ = 12
and for the second line a₂ = 18, b₂ = 6, c₂ = 24
Now \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2} ; \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)
\(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)
Clearly \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the lines are coincident. (Coincident means the same line.)

(iii) Given linear equations are 6x – 3y + 10 = 0 and 2x – y + 9 = 0
For the first line a₁ = 6, b₁ = -3, c₁ = 10
and for the second line a₂ = 2, b₂ = -1, c₂ = 9
Now \(\frac{a_1}{a_2}=\frac{6}{2}=\frac{3}{1} ; \frac{b_1}{b_2}=\frac{-3}{-1} \doteq \frac{3}{1} ; \frac{c_1}{c_2}=\frac{10}{9}\)
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the lines are parallel.

Question 4.
Which of the following pairs of linear equations have solutions? If they have solutions, find them graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:
To determine whether a pair of linear equations has solution, we compare the ratios of their coefficients.
(i) Given linear equations are
x + y = 5, 2x + 2y = 10
Ratios of the corresponding coefficients and constant is
\(\frac{a_1}{a_2}=\frac{1}{2} ; \frac{b_1}{b_2}=\frac{1}{2} ; \frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}\)
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), hence lines are coincident.
So the pair of linear equations has infinite solutions.
x + y = 5
Table of values

2x + 2y = 10
⇒ x + y = 5
Table of values

Because both equations represent the same line, every point on this line is a solution.

(ii) We have x – y = 8, 3x – 3y = 16
Ratios of the corresponding coefficients and constant is
⇒ \(\frac{a_1}{a_2}=\frac{1}{3} ; \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3} ; \quad \frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}\)
⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), hence lines are parallel.
So the pair of linear equations has no solution.

(iii) Given 2x + y – 6 = 0, 4x – 2y – 4 = 0
Ratios of the corresponding coefficients and constant is
\(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2} ; \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2} ; \frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
So the pair of linear equations has a unique solution.
For 2x + y = 6,
If x = 0, y = 6
If y = 0, x = \(\frac{6}{2}\) = 3
If x = 2, y = 2

For 4x – 2y = 4,
if x = 0, y = -2,
if x = 1, y = 0,
if x = 2, y = 2

For the first line plotting (0, 6), (2, 2) and (3, 0)
For the second line plotting (0, -2), (1, 0) and (2, 2).

The graphs intersect at (2, 2).
Hence, x = 2, y = 2 is the unique solution.

(iv) We have 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Ratios of the corresponding coefficients and constant is
\(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2} ; \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2} ; \frac{c_1}{c_2}=\frac{-2}{-5}\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
These lines are parallel and have no solution.

Question 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.
Solution:
Let the length of the garden = x m
and width of the garden = y m
We know that perimeter of a rectangle = 2(length + width) = 2(x + y)
According to question,
Half of the perimeter of the rectangular garden = 36 m
⇒ \(\frac{2(x+y)}{2}\) = 36
⇒ x + y = 36 ………..(1)
Now length is 4 m more than its width, then
x = y + 4 ……..(2)
Substitute the value of x from equation (2) in equation (1)
y + 4 + y = 36
⇒ 2y = 32
⇒ y = 16
Substituting the value of y = 16 in equation (2)
x = 16 + 4 = 20
∴ length = 20 m; width = 16 m

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables so that the graphs of the pair formed represent
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Many answers are possible.
One valid choice for each is:
(i) x + y – 3 = 0
Here the coefficient ratios are \(\frac {2}{1}\) = 2 and \(\frac {3}{1}\) = 3, which differ.
The lines intersect.
In fact, their common point is (1, 2).

(ii) 2x + 3y – 5 = 0
The x- and y-coefficients are unchanged but the constant term differs.
The lines are distinct and parallel.

(iii) 4x + 6y – 16 = 0
This is twice the complete given equation.
Therefore, the lines coincide.

Question 7.
Here is a problem that was posed by Mahaviracharya in Ganita sara sangraha (c. 850 CE).
The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101.
O mathematician, tell me quickly the price of each citron and of each fragrant wood-apple.
(Hint: The given problem can be modelled as
9x + 7y = 107
7x + 9y = 101
Can you figure out a way of solving these equations without directly using elimination or the substitution method? What is special about this pair of equations? The x and y coefficients are interchanged in the 2 equations.
What will happen if we add the pair of equations? What will happen if we subtract the pair of equations? Can the resulting equations be solved to find the values of x and y?)
Solution:
Let the price of 1 citron = ₹ x
and the price of 1 fragrant wood-apple = ₹ y
According to the question,
The price of 9 citrons and 7 fragrant wood apples is 107, then
9x + 7y = 107 ……..(1)
and the price of 7 citrons and 9 fragrant wood apples is 101, then
7x + 9y = 101 ……..(2)
Adding equation (1) and equation (2), we get
(9x + 7y) + (7x + 9y) = 107 + 101
⇒ 16x + 16y = 208
⇒ 16(x + y) = 16(13)
⇒ x + y = 13 …….(3)
Subtracting equation (2) from equation (1), we get
(9x + 7y) – (7x + 9y) = 107 – 101
⇒ 9x – 7x + 7y – 9y = 6
⇒ 2x – 2y = 6
⇒ x – y = 3 ……..(4)
Now, adding equation (3) and equation (4), we get
x + y + x – y = 13 + 3
⇒ 2x = 16
⇒ x = 8
Substitute the value of x = 8 in equation (3)
8 + y = 13
⇒ y = 13 – 8
⇒ y = 5
Therefore, price of 1 citron = ₹ 8
and the price of 1 fragrant wood-apple = ₹ 5.

Question 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.
Solution:
Let the cost of 1 pencil = ₹ x
Cost of 1 pen = ₹ y
According to the question,
The cost of 5 pencils and 7 pens is ₹ 50, then
5x + 7y = 50 ……..(1)
and the cost of 7 pencils and 5 pens is ₹ 46
7x + 5y = 46 ……..(2)
Adding equation (1) and equation (2), we get
(5x + 7y) + (7x + 5y) = 50 + 46
⇒ 12x + 12y = 96
⇒ 12(x + y) = 12 × 8
⇒ x + 7 = 8 ………(3)
Subtracting equation (2) from equation (1), we get
(5x + 7y) – (7x + 5y) = 50 – 46
⇒ 5x – 7x + 7y – 5y = 4
⇒ -2x + 2y = 4
⇒ x – y = -2 ……..(4)
Adding equation (3) and equation (4), we get
(x + y) + (x – y) = 8 – 2
⇒ 2x = 6
⇒ x = 3
Substituting the value of x = 3 into equation (3), we get
3 + y = 8
⇒ y = 5
Therefore cost of 1 pencil = ₹ 3 and cost of 1 pen = ₹ 5

Question 9.
Find the height of the stool.

Solution:
Let the height of the stool = x cm
Height of cat = y cm
According to the question,
When the cat is on the stool, then
x + y = 85 ……..(1)
When the cat is next to the stool, then
x – y = 25 ………(2)
Adding equation (1) and equation (2), we get
2x = 110
Then x = 55
Substituting the value of x = 55 in (2), we get
55 – y = 25
⇒ y = 25 – 55
⇒ y = 30
Therefore, height of stool = 55 cm and height of cat = 30 cm

Ganita Manjari Class 9 Maths Chapter 13 End of Chapter Exercise Solutions

Two Variables One Line End of Chapter Exercise Solutions

Question 1.
The graph of the line y = 3x, passing through (0, 0) and (2, 6), is given below.
(i) Identify the slope of the line from the graph and explain how you calculated it using the two points on the line.
(ii) What is the y-intercept of the line? Explain its significance.
(iii) A water tank is being filled so that the water level y (in cm) after x minutes follows this graph.

(a) How high is the water after 5 minutes?
(b) How many minutes will it take for the water to reach a height of 21 cm?
(iv) Without drawing a new graph, determine whether the point (4, 10) lies on this line. Justify your answer.
(v) Plot the point where this line intersects the line x = 3. Explain how you found the coordinates.
Solution:
(i) The two points given are (0, 0) and (2, 6).
Here, x₁ = 0, y₁ = 0; x₂ = 2, y₂ = 6.
Now, Slope = \(\frac{\text { Rise }}{\text { Run }}=\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{6-0}{2-0}\)
= \(\frac {6}{2}\)
= 3
For each increase of 1 in x, y increases by 3.
In the tank, the water level rises at 3 cm per minute.

(ii) The line y = 3x passes through the origin (0, 0), so the y-intercept is 0.
Moreover comparing y = mx + d with y = 3x
d = 0 i.e. y-intercept = 0
Significance: When d, i.e., the y-intercept, = 0, it means the line must pass through the origin (0, 0).

(iii) (a) Given that x is the time in minutes and y is the water level in cm.
Also given that the water level y = 3x after x minutes.
∴ Water level after 5 min = 3 × 5 = 15 cm
(b) When height = 21 cm
y = 21 cm
⇒ 3x = 21
⇒ x = 7
∴ After 7 minutes, the water will reach a height of 21 cm.

(iv) Substitute x = 4 in line y = 3x
⇒ y = 3 × 4
⇒ y = 12, hence the coordinates are (4, 12).
So, point (4, 10) does not lie on this line.

(v) For line x = 3
Table of values

As the x-coordinate of line x = 3 is always 3, y may have different values: y = 3x.
For line y = 3x
Table of values

Plot the points on the graph.

From the graph, the lines meet at (3, 9).

Question 2.
In countries like the USA, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Celsius to Fahrenheit:
F = \(\frac {9}{5}\)C + 32
(i) Draw the graph of the linear equation above using Celsius on the x-axis and Fahrenheit on the 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.
Solution:
(i) Given equation is F = \(\frac {9}{5}\)C + 32 …….(1)
Putting C = -40°, 0°, 10, 30° in (1), we get the corresponding values of F = 40, 32, 50 and 86, respectively.
e.g.: F = \(\frac {9}{5}\)(10) + 32 = 18 + 32 = 50
Therefore, the table of values is:
Table of values

(ii) When the temperature is 30°C = C (given)
Then F = \(\frac {9}{5}\) × 30 + 32 = 54 + 32 = 86°
⇒ F = 86°
Hence, the temperature in Fahrenheit is 86°F.

(iii) When the temperature is 95°F
Then F = \(\frac {9}{5}\)C + 32
Now, F = \(\frac {9}{5}\)C + 32
⇒ 95° = \(\frac {9}{5}\)C + 32
⇒ 95 – 32 = \(\frac {9}{5}\)C
⇒ 63 = \(\frac {9}{5}\)C
⇒ \(\frac{63 \times 5}{9}\) = C
⇒ 35 = C
⇒ C = 35°
Therefore, the temperature in Celsius = 35°C

(iv) When the temperature in Celsius = 0°C = C
Then, F = \(\frac {9}{5}\)C + 32
= \(\frac {9}{5}\) × 0 + 32
= 0 + 32
= 32
Therefore, the temperature in Fahrenheit is 32°F.
When the temperature in Fahrenheit = 0°F
∴ F = 0
Then F = \(\frac {9}{5}\)C + 32
⇒ \(\frac {9}{5}\)C + 32 = 0
⇒ \(\frac {9}{5}\)C = -32
⇒ C = \(\frac{-32 \times 5}{9}\)
⇒ C = \(\frac {-160}{9}\) or -17.78
Therefore, the temperature in Celsius is- 17.78°C.

(v) Let us check
Given equation is F = \(\frac {9}{5}\)C + 32 ……….(1)
If the temperature is the same number on both scales, then F = C
C = \(\frac {9}{5}\)C + 32
⇒ C – \(\frac {9}{5}\)C = 32
⇒ \(\frac {-4}{5}\)C = 32
⇒ C = \(\frac{-32 \times 5}{4}\)
⇒ C = -40 F
Hence, Yes. -40° is the only temperature that has the same numerical value in both Celsius and Fahrenheit.

Question 3.
Solve the following system of equations graphically:
2x + y = 6, 2x – y – 2 = 0
Solution:
Given equation is 2x + y = 6 ………(1)
Putting x = 0, 1, 2 in (1), we get
The corresponding values of y are 6, 4, and 2, respectively.
Therefore, the table of values is

Now in equation 2x – y – 2 = 0
⇒ y = 2x – 2 ……..(2)
Putting x = 0, 1, 2 in equation (2), we get the corresponding values of y = -2, 0 and 2, respectively.
Therefore, the table of values is

Now let us draw the points on the graph.

As the two lines intersect at (2, 2), the required solution is (2, 2).

Question 4.
Find the point of intersection of the lines shown on the cover page.
Solution:
On the cover page, the given equations are
15x – 10y – 115 = 0 …….(1)
15x + 100y = 335 ……….(2)
Using the Substitution Method
We can also write equation (1) as
15x = 10y + 115
⇒ x = \(\frac{5(2 y+23)}{5 \times 3}\)
⇒ x = \(\frac{2 y+23}{3}\) …….(3)
Now substitute the value of x in equation (2)
\(15\left(\frac{2 y+23}{3}\right)\) + 100y = 335
⇒ 10y + 115 + 100y = 335
⇒ 110y = 335 – 115
⇒ 110y = 220
⇒ y = 2
Now substitute the value of y = 2 into equation (3); we get
x = \(\frac{2(2)+23}{3}\)
⇒ x = 9
Hence point of intersection of gives lines is x = 9, y = 2.

Question 5.
Give a formula to find the x-intercept of the line y = mx + c.
Solution:
x-intercept is where the line intersects the x-axis
i.e., y-coordinate = 0
We know y = mx + c
Putting y = 0, we get
0 = mx + c
⇒ x = \(\frac {-c}{m}\)
So x-intercept = (\(\frac {-c}{m}\), o)

Question 6.
A person is choosing between two mobile plans.
Plan A: ₹ 50 monthly fee + ₹ 0.20 per minute of call time.
Plan B: ₹ 30 monthly fee + ₹ 0.30 per minute of call time.
For how many minutes of calling per month is Plan A cheaper than Plan B? For how many minutes is Plan B cheaper? Also find the number of minutes at which both plans cost the same.
Solution:
Let the call time per month = t
Monthly cost = P
For Plan A
P = 50 + 0.20t ………(1)
For Plan B
P = 30 + 0.30t ……….(2)
For equal plan costs
50 + 0.20t = 30 + 0.30t
0. 20t – 0.30t = 30 – 50
⇒ -0.10t = -20
⇒ t = 200 min
For t = 200 m
Cost for Plan A = 50 + 0.20 × 200
= 50 + 40.00
= ₹ 90
Cost for Plan B = 30 + 0.30 × 200
= 30 + 60
= ₹ 90
Now, cost of plan B – Cost of plan A = 30 + 0.30t – (50 + 0.20t)
= 30 + 0.30t – 50 – 0.20t
= -20 + 0.10t
= 0.10t – 20
(i) If t < 200
Say 190
Then 0.10t – 20 = 0.10 × 190 – 20
= 19 – 20
= -1
⇒ Cost of plan B – Cost of plan A = negative
⇒ Plan B is cheaper.

(ii) If t > 200
Say 210
Then 0.10t – 20
= 0.10 × 210 – 20
= 21 – 20
= 1
⇒ Cost of plan B – Cost of plan A = +ve
⇒ Plan A is cheaper
So plan A is cheaper than plan B when time is more than 200 minutes.
Plan B is cheaper than plan A when time is less than 200 minutes.

(iii) For 200 minutes
Cost of plan A = Cost of plan B

Question 7.
How many lines exist that
(i) have a given slope?
(ii) have a given slope and pass through a given point?
Solution:
(i) There are infinitely many lines which have a given slope.
Keep m fixed in y = mx + c and vary c.
This gives different parallel lines.

(ii) If the given point is (x0, y0) then c = y0 – mx0 is fixed.
For an undefined slope, only x = x0 passes through the point.

Question 8.
For what values of p does the pair of equations given below have a unique solution?
4x + py + 8 = 0; 2x + 2y + 2 = 0.
Solution:
We have 4x + py + 8 = 0 ……(1)
2x + 2y + 2 = 0 ……..(2)
For unique solution
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
∵ If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\), that means the lines are parallel, and there is no solution.
∴ \(\frac{a_1}{a_2}=\frac{4}{2}=2 ; \quad \frac{b_1}{b_2}=\frac{p}{2}\)
⇒ \(\frac {p}{2}\) ≠ 2
⇒ p ≠ 4
Therefore, the pair of equations has a unique solution for all real values of p except 4.

Question 9.
Find the values of a and b for which the following system of equations has infinitely many solutions:
(a + b)x – 2by = 5a + 2b + 1; 3x – y = 14.
Solution:
(a + b)x – 2by = 5a + 2b + 1; 3x – y = 14
For infinitely many solutions
Condition: \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) (∵ lines are coincident)
Hence \(\frac{a+b}{3}=\frac{-2 b}{-1}=\frac{5 a+2 b+1}{14}\)
Now, \(\frac{a+b}{3}=\frac{-2 b}{-1}\)
⇒ a + b = 6b
⇒ a = 6b – b
⇒ a = 5b ………..(1)
and \(\frac{-2 b}{-1}=\frac{5 a+2 b+1}{14}\)
⇒ 28b = 5a + 2b + 1
⇒ 28b – 2b = 5a + 1
⇒ 26b = 5a + 1
Substitute the value of a from equation (1); we get
26b = 5(5b) + 1
⇒ 26b – 25b = 1
⇒ b = 1
Substitute the value of b = 1 into equation (1), we get
a = 5(1)
⇒ a = 5
So the values of a and b are 5 and 1, respectively.

Question 10.
Find the value of ‘k’ for which the following system of equations represents a pair of coincident lines:
x + 2y = 3; (k – 1)x + (k + 1)y = k + 3.
Solution:
Given equations are x + 2y = 3
(k – 1)x + (k + 1)y = k + 3
For coincident lines
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
⇒ \(\frac{1}{k-1}=\frac{2}{k+1}=\frac{3}{k+3}\)
Now, \(\frac{1}{k-1}=\frac{2}{k+1}\)
⇒ 2k – 2 = k + 1
⇒ 2k – k = 1 + 2
⇒ k = 3
and \(\frac{2}{k+1}=\frac{3}{k+3}\)
⇒ 2k + 6 = 3k + 3
⇒ 6 – 3 = 3k – 2k
⇒ 3 = k or k = 3
So, for k = 3, the pair of linear equations represents coincident lines.

Question 11.
(i) Robot 1 starts from the origin and traces a path by repeatedly moving 3 units to the right and then 4 units upward. Robot 2 starts from the point (10, 0) and repeatedly moves 1 unit to the right and then 2 units upward. Will the paths traced by these two robots intersect, and if so, where?
(ii) Robot 1 starts from (3, 0) and traces a path by repeatedly moving 5 units to the right and then 5 units upward. Robot 2 starts from the point (7, 0) and repeatedly moves 5 units to the right and then 3 units downwards. Will the paths traced by these two robots intersect, and if so, where?
Solution:
(i) Robot 1
For each step, the robot moves +3 units to the right and +4 units upwards.
The overarching straight-line trajectory connecting these points passes through the origin (0, 0).

Slope, m = \(\frac {4}{3}\)
then y = \(\frac {4}{3}\)x ……(1)
Robot 2
It starts at (10, 0) and moves +1 unit horizontally and +2 units vertically per step.
The overarching straight-line trajectory passes through (10, 0).
Slope m2 = \(\frac {2}{1}\), starting point = (10, 0)
⇒ y – 0 = 2(x – 10)
⇒ y = 2x – 20 …….(2)
Robot 1 and Robot 2 will intersect when equation (1) = equation (2)
\(\frac {4}{3}\)x = 2x – 20
⇒ \(\frac {4}{3}\)x – 2x = -20
⇒ \(\frac{4 x-6 x}{3}\) = -20
⇒ 4x – 6x = -20
⇒ -2x = -60
⇒ x = 30
When x = 30, then y = \(\frac {4}{3}\) × 30 = 40
So the path traced by these two robots will intersect at point (30, 40).

(ii) Robot 1
Starting point (3, 0)
Slope m = \(\frac {5}{5}\) = 1
∴ y – 0 = 1(x – 3)
⇒ y = x – 3 …….(1)
Robot 2
Starting point (7, 0)

Slope m2 = \(-\frac {3}{5}\) as rise = -3, run = 5
y – 0 = \(-\frac {3}{5}\)(x – 7)
⇒ y = \(\frac{-3(x-7)}{5}\)
⇒ 5y = -3x + 21
⇒ 5(x – 3) = -3x + 21 [From (1)]
⇒ 5x – 15 = -3x + 21
⇒ 8x = 36
⇒ x = \(\frac{36}{8}=\frac{9}{2}\) = 4.5
⇒ y = 4.5 – 3 = 1.5
So the path traced by these two robots will intersect at point (4.5, 1.5).

Question 12.
At a certain time, Jacob notices that his digital watch reads ‘a’ minutes after two o’clock. Fifteen minutes later, it reads ‘b’ minutes after three o’clock. He noticed that a is six times greater than b. What time was it when he looked at his watch for the second time?
Solution:
The first watch shows a minutes after 2:00 pm
After 15 minutes, i.e., a + 15 minutes after 2:00 pm and b minutes after three o’clock.
According to the question,
2:00 + a min + 15 min = 3:00 + b min
⇒ 2:00 + a min + 15 min = 2:00 + 60 min + b min
⇒ a + 15 = 60 + b
⇒ a – b = 60 – 15
⇒ a – b = 45 ……(1)
Also given a = 6b ……..(2)
Substitute the value of a from (2) into equation (1)
6b – b = 45
⇒ 5b = 45
⇒ b = 9 minutes
Substitute the value of b = 9 into equation (2)
a = 6 × 9 = 54 minutes
Time when he sees watch first = 2:00 + a = 2:54
After 15 minutes time when he sees watch second time = 2 : 54 + 15 min = 3 : 09

Question 13.
The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.
Solution:
Let the number be \(\frac{x}{\text { ten’s digit }} \frac{y}{\text { one’s digit }}\)
Hence, the tens digit = x
One’s digit = y
Original number = 10x +y
On reversing the digits
New number = 10y + x
According to question: x + y = 15
And on interchanging the digits, the new no. exceeds by 9
∴ 10y + x = 10x + y + 9
⇒ 10y – y = 10x – x + 9
⇒ 9y = 9x + 9
⇒ x – y = -1 ……(1)
Also, x + y = 15 ….. (2)
Adding (1) and (2), we get
2x = 14
⇒ x = 7
Substitute the value of x = 7 in equation (2)
7 + y = 15
⇒ y = 15 – 7 = 8
∴ Required number = 10x + y
= 10 × 7 + 8
= 70 + 8
= 78

Question 14.
In a cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles of the cyclic quadrilateral.
Solution:
In a cyclic quadrilateral ABCD

∠A + ∠C = 180° (Opp angles)
⇒ (x + 7) + (3y + 23) = 180°
⇒ x + 3y = 180 – 30
⇒ x + 3y = 150 ……..(1)
And ∠B + ∠D = 180°
⇒ (y + 8) + (4x + 12) = 180°
⇒ 4x + y = 180 – 20
⇒ 4x + y = 160 ……….(2)
⇒ 12x + 3y = 480 ……….(3) (Multiplying by 3 to equalise)
Subtract equation (1) from equation (3)
11x = 330
⇒ x = 30
Substitute the value of x = 30 in (1)
30 + 3y = 150
⇒ 3y = 120
⇒ y = 40
Hence ∠A = x + 7 = 30 + 7 = 37°
∠B = y + 8 = 40 + 8 = 48°
∠C = 3y + 23 = 3 × 40 + 23 = 143°
∠D = 4x + 12 = 4 × 30 + 12 = 132°

Question 15.
A train moving with uniform speed for a certain distance takes 6 hours less if its speed is increased by 6 km/hour. It would have taken 6 hours more had its speed been decreased by 4 km/hour. Find the distance travelled and the speed of the train.
Solution:
Let the uniform speed = x km/h
Time taken = t hours
Distance = xt (∵ Speed = \(\frac{\text { Distance }}{\text { Time }}\))
1st case: speed = x + 6, time = (t – 6),
distance = xt
⇒ (x + 6) (t – 6) = xt
⇒ xt – 6x + 6t – 36 = xt
⇒ -6x + 6t = 36
⇒ -x + t = 6
⇒ t = 6 + x ……….(1)
2nd case: speed = x – 4, time = (t + 6),
distance = xt
⇒ (x – 4) (t + 6) = xt
⇒ xt + 6x – 4t – 24 = xt
⇒ 3x – 2t = 12 …….(2)
⇒ 3x – 2(6 + x) = 12 [from eq. (1)]
⇒ 3x – 12 – 2x = 12
⇒ x = 24
and t = 6 + x = 6 + 24 = 30
∴ Speed of the train = 24 km/hr
Time = 30 hrs
Distance travelled = 24 × 30 = 720 km

Question 16.
The age of a father is equal to the sum of the ages of his four children. After 20 years, the sum of the ages of the children will be twice the age of the father. Find the age of the father.
Solution:
Let the father’s present age = x years
and the sum of the four children’s present ages = y
According to the question
x = y ……(1)
After 20 years
Father’s age = x + 20
Sum of four children’s ages after 20 year = y + (20 × 4)
⇒ y + (4 × 20) = 2(x + 20)
⇒ y + 80 = 2(x + 20)
⇒ y + 80 = 2x + 40
⇒ 80 – 40 = 2x – x [Using (1)]
⇒ 40 = x
⇒ y = 40
Hence, father’s age = 40 years.

The post Two Variables One Line Class 9 Solutions Maths Ganita Manjari Chapter 13 appeared first on Learn CBSE.

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Empowering Students with AI-Driven Learning Solutions

Welcome to NCsolve — your trusted educational platform designed to support students worldwide. Whether you're preparing for Class 10, Class 11, or Class 12, NCsolve offers a wide range of learning resources powered by AI Education.

Our platform is committed to providing detailed solutions, effective study techniques, and reliable content to help you achieve academic success. With our AI-driven tools, you can now access personalized study guides, practice tests, and interactive learning experiences from anywhere in the world.

🔎 Why Choose NCsolve?

At NCsolve, we believe in smart learning. Our platform offers:

• ✅ AI-powered solutions for faster and accurate learning.
• ✅ Step-by-step NCERT Solutions for all subjects.
• ✅ Access to Sample Papers and Previous Year Questions.
• ✅ Detailed explanations to strengthen your concepts.
• ✅ Regular updates on exams, syllabus changes, and study tips.
• ✅ Support for students worldwide with multi-language content.

🌐 Explore Our Websites:

🔹 ncsolve.blogspot.com
🔹 ncsolve-global.blogspot.com
🔹 edu-ai.blogspot.com

📲 Connect With Us:

👍 Facebook: NCsolve
📧 Email: ncsolve@yopmail.com

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