Ganita Manjari Class 9 Chapter 13 Solutions Two Variables One Line

Students often refer to Ganita Manjari Class 9 Solutions and Part 2 Class 9 Maths Chapter 13 Two Variables One Line Solutions to verify their answers.

Two Variables One Line Class 9 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. 88)

Question 1.
Are (1, \(\frac{9}{2}\)) (0, 6), (4, 1) and \(\frac{1}{2}\) solutions of the
equation 3x + 2y = 12?
Can you find any other solution? How many solutions can you find?
Solution:
Given linear equation: 3x + 2y = 12
For (1, \(\frac{9}{2}\)): Substituting x = 1 and y = \(\frac{9}{2}\) in the given equation, we get
3(1) + 2(\(\frac{9}{2}\)) = 3 + 9 = 12

Since (1, \(\frac{9}{2}\)) satisfies the given equation, it is a solution of the equation.
For (0, 6): Substituting x = 0 and y = 6 in the given equation, we get
3(0)+ 2(6) = 12
Since (0, 6) satisfies the given equation, it is a solution of the equation.
For (4, 1): Substituting x = 4 and y = 1 in the given equation, we get
3(4)+ 2(1) = 12 + 2 = 14 ≠ 12
Since (4. 1) doesn’t satisfy the given equation, it is not a solution of the equation.

For \(\left(\frac{1}{2}, \frac{21}{4}\right)\) : Substituting x = \(\frac{1}{2}\) and y = \(\frac{21}{4}\)
in the given equation, we get
3\(\left(\frac{1}{2}\right)\) + 2\(\left(\frac{21}{4}\right)\) = \(\frac{3}{2}+\frac{21}{2}=\frac{24}{2}\) = 12
Since \(\left(\frac{1}{2}, \frac{21}{4}\right)\) satisfies the given equation, it is a solution.

Think and Reflect (NCERT Textbook Page No. 92)

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 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:
Do it yourself.

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:

From the figure, the point on the line AB whose x-coordinate is 8 is D.
The y-coordinate of D is 9 and the ordered pair is (8, 9).

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:
From the figure, the point on the line AB whose y-coordinate is 6 is C.
The x-coordinate of C is 6 and the ordered pair is (6, 6).

Think and Reflect (NCERT Textbook Page No. 99)

Can you use the points A, B and C to verify that the slope of this line is indeed -5?
Solution:
Using points A(0, 3) and B(1, -2):
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-2-3}{1-0}\) = -5

Using points B(1, -2) and C(-1, 8):
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{8-(-2)}{-1-1}=\frac{10}{-2}\)

Using points A(0, 3) and C(-1, 8):
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{8-3}{-1-0}\) = -5
Therefore, the slope of this line is indeed -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:
We are given,
Cost of 1 paratha and 2 bowls of dahi = ₹ 90
Cost of 2 parathas and 7 bowls of dahi = ₹ 260
Clearly, the difference between 2 parathas, 7 bowls of dahi and 1 paratha, 2 bowls of dahi is 1 paratha, 5 bowls of dahi And the difference in their cost is ₹ 170 (₹ 260 – ₹ 90)
So, the cost of 1 paratha and 5 bowls of dahi = ₹ 170
Now, the difference between 1 paratha, 5 bowls of dahi and 1 paratha, 2 bowls of dahi is 3 bowls of dahi And the difference in their cost is ₹ 80 (₹ 170 – ₹ 90)
So, the cost of 3 bowls of dahi = ₹ 80
the cost of 1 bowl of dahi is ₹ \(\frac{80}{3}\) ≈ ₹ 26.67
Now, the cost of 2 bowls of dahi = ₹ 26.67 × 2 = ₹ 53.33
And the cost of 1 paratha and 2 bowls of dahi = ₹ 90 the cost of 1 paratha is ₹ 36.67 (₹ 90 – ₹ 53.33)
Thus, the cost of one paratha and one bowl of dahi = ₹ 26.67 + ₹ 36.67 = ₹ 63.34

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:
We are given,
7x – 15y = 2 …….. (1)
x + 2y = 3 ……….(2)
In equation (2), 2y = 3 – x ⇒ y = \(\frac{1}{2}\)(3 – x) …….(3)
Substituting y = \(\frac{1}{2}\) (3 – x) in equation (1), we get
7x – 15[\(\frac{1}{2}\)(3 – x)] = 2
⇒ 7x – \(\frac{45}{2}+\frac{15 x}{2}\) = 2
⇒ \(\frac{29 x}{2}=\frac{49}{2}\)
⇒ x = \(\frac{49}{29}\)

Substituting x = \(\frac{49}{29}\) in equation (3), we get
y = \(\frac{1}{2}\)(3 – \(\frac{49}{29}\))
= \(\frac{1}{2}\left(\frac{87-49}{29}\right)=\frac{1}{2}\left(\frac{38}{29}\right)=\frac{19}{29}\)
Thus, we would still get the same solution.

Think and Reflect (NCERT Textbook Page No. 109)

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:
By elimination method:
By eliminating x:
Given equations are: 9x – 4y = 2000 …(1)
7x – 3y – 2000 …(2)
Multiplying equation (1) by 7 and equation (2) by 9, we get
63x – 28y = 14000 …(3)
63x – 27y = 18000 …(4)
Now, subtracting equation (3) from equation (4), we get
(63x – 21 y) – (63x – 28y) = 18000 – 14000
⇒ y = 4000
Now, substituting y = 4000 in equation (1), we get 9x – 4(4000) = 2000
⇒ 9x = 18000
⇒ x = 2000
Hence, the solution of the equations are x = 2000, y = 4000.
By substitution method:
Given equations are: 9x- 4y- 2000 …(1)
7x – 3y = 2000 …(2)
From equation (2):
7x – 3y = 2000
7x = 2000 + 3y
x = \(\frac{1}{2}\)
Substituting x = \(\frac{1}{2}\) in equation (1), we get
9\(\frac{1}{2}\) – 4y = 200
Multiplying both sides by 7, we get
9(2000 + 3y) – 28y = 14000
⇒ 18000 + 21y – 28y = 14000
⇒ 18000 – y = 14000
⇒ 1 – y = 14000 – 18000
⇒ -y = -4000
⇒ y = 4000
Substituting y = 4000 in equation (2), we get
7x – 3(4000) = 2000
⇒ 7x – 12000 = 2000
⇒ 7x = 14000
⇒ x = 2000
Hence, the solution of the equations are x = 2000, y = 4000.

Elimination Method Pros

• It is useful when the coefficients of one variable are the same or can be made the same easily.
• It involves fewer steps for many equations.
• It avoids dealing with fractions in many cases.
• Both variables can be eliminated by adding or subtracting equations.

Cons

• It may require multiplying equations to make coefficients equal, which can increase calculations.
• There is a greater chance of calculation errors while adding or subtracting equations.

Substitution Method Pros

• It is useful when one variable is already isolated or can be easily expressed in terms of another variable.
• It clearly shows the relationship between the two variables.
• It is helpful for equations where one coefficient is 1.

Cons

• It can become lengthy when substitution creates complex expressions.
• It may involve fractions, making calculations more difficult.
• It requires solving one equation first before substituting into the other.

Think and Reflect (NCERT Textbook Page No. 110)

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 mle to check when this happens?
Solution:
Given equations:
10x – 5y = 20 ……. (1)
4x – 2y = 8 …(2)
Simplifying equation (1) by dividing by 5, we get 2x-y= 4
Simplifying equation (2) by dividing by 2, we get
2x – y = 4
Now, subtracting both equations, we get
(4x + 6y) – (4x + 6y) = 12 – 12 = 5 – 0 = 0
Here, both variables are eliminated and we get a true statement.
Therefore, every solution of one equation is also a solution of the other.
Thus, the equations have infinitely many solutions.
The three examples of pairs of equations that lead to 0 = 0 after subtraction are:
x + 2y = 5 and 2x + 4y = 10
3x + y = 1 and 6x + 2y = 14
5x – 3y = 12 and 10x – 6y = 24

Think and Reflect (NCERT Textbook Page No. 110)

What if one of a₂, b₂, c₂ is zero?
Solution:
If one of a₂, b₂, or c₂ becomes zero, then the corresponding ratio cannot be formed directly because division by zero is not possible.
However, pair of the equations can still have infinitely many solutions if the second equation is a constant multiple of the first equation.
For example: Consider the equations 2x + 3y = 0 and 4x + 6y = 0
The second equation is twice the first equation, so both equations represent the same line and have infinitely many solutions.

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:
Examples of pairs of equations that have no solution are:
(i) x + y = 5 and 2x + 2y = 12
(ii) 2x + 3y = 7 and 4x + 6y = 20
(iii) 3x – 2y = 8 and 6x – 4y = 20
On solving the above pairs of equations, we get that the LHS of the equations is equal, but the RHS is not.

Think and Reflect (NCERT Textbook Page No. 112)

Question 1.
Are the converses of the above statements true?
Solution:
Converse of Statement 1
To prove: If a pair of linear equations has a unique solution, then \(\frac{1}{2}\)
Proof: Suppose \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Then, a₁b₂ = a₂b₂
⇒ a₁b₂ – a₂b₁ = 0
The two equations have the same or parallel coefficients, so, by eliminating one variable, the equations cannot give exactly one value of the other variable. They either have

• infinitely many solutions, or
• no solution.

Therefore, they cannot have a unique solution.
Hence, for a unique solution, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) is necessary.

Converse of Statement 2
To prove: If a pair of linear equations has a infinitely many solutions, then \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Proof:
If a pair has infinitely many solutions, the two lines represent the same line.
Therefore, the second equation must be a multiple of the first equation.
Let, a₂ = ka₁, b₂ = kb₁, c₂ = kc₁ for some constant (k).
Therefore, \(\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}\) = k

Taking reciprocals,
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{k}\) (\(\frac{1}{2}\) also a constent)
Hence proved.

Converse of Statement 3
To prove: If a pair of linear equations has no solution, then
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
proof: If a pair has no solution, the two lines are parallel and distinct.
For parallel lines, the ratios of coefficients of (x) and (y) are equal, that is
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)
But since the lines are different, so the constant terms are not in the same ratio:
\(\frac{c_1}{c_2} \neq \frac{a_1}{a_2}\)
Therefore,
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence proved.
Therefore, the converse of all three statements is true.

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 andy = mx + d₂)
Sol. Consider two lines:
y = mx + d₁ …………(1)
y = mx + d₂ ………..(2)
where both lines have the same slope m and d₁ ≠ d₂.
Comparing equations (1) and (2) with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, respectively, we get
a₁ = m, b₁ = -1, c₁ = d₁, a₂ = m, b₂ = -1, and c₂ = d₂
∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the two equations cannot have no solution. Therefore, the tw o lines do not intersect at any point. Thus, the 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:
A linear equation in two variables can be expressed in the form ax + by + c = 0
We are given a = 3, b = 0, and c = –\(\frac{1}{5}\)
∴ The linear equation is 3x + 0y + \(\left(-\frac{1}{5}\right)\) = 0 ⇒ 3x – \(\frac{1}{5}\) = 0

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

Solution:
Standard form of a linear equation: ax + by + q = 0.

Question 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 isy. Is this a correct representation?
Solution:
(i) Let the cost of a notebook be and the cost of a pen be ₹ p.
According to the question,
Cost of a notebook = 2 x Cost of a pen
⇒ t = 2p
Charlie wrote: t = 2p
Meera wrote: p = 2t
Charlie’s representation (t = 2p) is correct.

(ii) Runs scored by the first batsman = x
Runs scored by the second batsman = y
Total runs scored by the two batsman = x + y = 176
Thus, x + y = 176 is the correct representation.

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:
Given equation: 5x – 6y = 2
Given ordered pair: (x, y) = (4, 3)
Substitutingx = 4 andy = 3, we get 5(4) – 6(3) = 20- 18 = 2
∴ The ordered pair satisfies the equation.
Thus, (4, 3) is a solution of 5x – 6y = 2.

Question 2.
Find any two solutions for each of the following equations:
(i) 7x-3y = 21
(ii) 2x + 3y = 5
Solution:
(i) 7x – 3y = 21
Substituting x = 0, we get
-3y = 21 ⇒ y = -7
So, (0, -7) is a solution of the given equation.
Similarly, substituting y = 0, we get 7x = 21 ⇒ x = 3
So, (3, 0) is a solution of the given equation.
Thus, (0, -7) and (3, 0) are the solutions of 7x – 3y = 21.

(ii) 2x + 3y = 5
Substituting x = 0, we get
3y = 5 ⇒ y = \(\frac{5}{3}\)
So, (0, \(\frac{5}{3}\)) is a solution of the given equation. Similarly, substituting y = 0, we get
2x = 5 ⇒ x = \(\frac{-3}{5}\)
So, (\(\frac{5}{3}\), 0) is a solution of the given equation.
Thus, (0, \(\frac{5}{3}\)) and (\(\frac{5}{3}\), 0) are the solutions of 2x + 3y = 5.

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:
Substituting (2, -1) in the first equation, (i.e. x = 2 and y = -1) we get
2m(2) + 3(-1) = 7
⇒ 4m – 3 = 7
⇒ 4m = 10
⇒ m = \(\frac{5}{2}\)
Now, substituting (2, -1) in the second equation, (i. e. x = 2 andy = -1), we get
4(2) + n(-1) = -10
⇒ 8 – n = -10
⇒ n = 18
Therefore, 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
Solution:
5x + 3y = 7
Taking x = 1:
5(1) + 3y = 7
⇒ 5 + 3y = 7
⇒ 3y = 2
⇒ y = \(\frac{2}{3}\)
Thus, (1, \(\frac{2}{3}\)) is a solution of 5x + 3y = 7, which lies in Quadrant I.
Now, taking x = -1:
5(-1) + 3y = 7
⇒ -5 + 3y = 7
⇒ 3y = 12
⇒ y = 4
Thus, (-1, 4) is a solution of 5x + 3y = 7, which lies in Quadrant II.

(ii) 5x – 3y = 7
Solution:
5x – 3y = 7
Taking x = 1:
5(1) – 3y = 7
⇒ 5 – 3y = 1
⇒ -3y = 2
⇒ y = –\(\frac{2}{3}\)
Thus, (1, \(\frac{-2}{3}\)) is a solution of 5x – 3y = 7, which lies in Quadrant IV.
Now, taking x = -1:
⇒ 5(-1) – 3y = 1
⇒ -5 – 3y = 7
⇒ -3y = 12
⇒ y = -4
Thus, (-1, -4) is a solution of 5x – 3y = 7, which lies in Quadrant III.

(iii) -5x + 3y = 7
Solution:
-5x + 3 v = 7
Taking x = 1:
⇒ -5(1) + 3y = 7
⇒ -5 + 3y = l
⇒ 3y = 12
⇒ y = 4
Thus, (1,4) is a solution of-5x + 3y = 7, which lies in Quadrant I.
Now, taking y = – 1:
-5x + 3(-1) = 7
⇒ -5x – 3 = 7
⇒ -5x = 10
⇒ x = -2
Thus, (-2, -1) is a solution of -5x + 3y = 7, which lies in Quadrant III.

(iv) -5x – 3y = 7
Solution:
—5x – 3y = 7
Taking x = 1:
-5(1) – 3y = 7
⇒ -5 – 3y = 7
⇒ -3y = 12
⇒ y = -4
Thus, (1, -4) is a solution of -5x – 3y = 1, which lies in Quadrant IV.
Now, taking y = 1:
-5x – 3(1) = 7
⇒ -5x – 3 = 7
⇒ -5x = 10
⇒ x = -2
Thus, (-2, 1) is a solution of -5x – 3y = 1, which lies in Quadrant II.

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:
In the given graph, the point C(2, 3) does not lie on the line.
In equation 3x – 7y = 21, taking x = 2 and y = 3, we get
3(2) – 7(3) = 6 – 21 = -15 ≠ 21
⇒ Point C(2, 3) does not satisfy the equation.
Thus, the points that do not lie on the line do not satisfy 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.
Solution:
False. A linear equation in two variables has infinitely many solutions because different values of one variable can give different values of the other variable.

(ii) The graph of a linear equation in two variables always passes through the origin,
Solution:
False. A graph passes through the origin only if substituting x = 0 and y = 0 makes the equation true. When an equation has a nonzero constant term, its graph does not pass through the origin.

(iii) A linear equation in two variables can never have rational solutions.
Solution:
False. A linear equation in two variables can have rational solutions. For example, In linear equation 2x + 3y = 5, if x = 1, then 2(1) + 3y = 5 ⇒ 3y – 3 ⇒ y = 1. So, the solutiori of the linear equation is x = 1 andy = 1, which are both rational numbers.

(iv) x = 3 is a valid linear equation in two variables.
Solution:
True, x = 3 is a valid linear equation in two variables because it can be expressed in the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero.
For x = 3, we can write x + 0y – 3 = 0
Here a = 1, b = 0 and c = 3.

(v) The equation 2x + 3y = 7 has infinitely many solutions.
Solution:
True. The equation 2x + 3y = 7 has infinitely many solutions because choosing any arbitrary value for x yields a corresponding unique value fory, giving an infinite set of solution pairs.

(vi) The point (1, 2) is a solution of the equation 2x + 3y = 7.
Solution:
False. Substituting x = 1 andy = 2 in the equation 2x + 3y = 7, we get 2(1) + 3(2) = 2 + 6 = 8 ≠ 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) 3x + 4y = 7 …(1)
6x + 8y = 14 …(2)
For equation (1):
Let x = 1,
3(1) + 4y = 7 ⇒ 4y = 4 ⇒ y = 1
So, (1, 1) is the solution of equation (1).
Now, letx = 3,
3(3) + 4y = 7 ⇒ 4y = -2 ⇒ y = –\(\frac{1}{2}\)
So, (3, –\(\frac{1}{2}\)) is the solution of equation (1).

Now, checking these solutions for equation (2):
For (1, 1),
6(1) + 8(1) = 6 + 8 = 14
So, (1, 1) is also the solution of equation (2).
For (3, –\(\frac{1}{2}\)),
6(3) + 8(-\(\frac{1}{2}\)) = 14
So, (3, –\(\frac{1}{2}\)) is also the solution of equation (2).

Therefore, every solution of 3x + 4y = 7 is also a solution of 6x + 8y = 14.
Thus, both equations have the same solution sets.

(ii) Given equations:
ax + by = c …(1)
kax + kby = kc …(2)
Let (x₁, y₁) be a solution of equation (1).
Therefore, ax₁ + by₁ = c
Multiplying both sides by k:
k(ax₁ + by₁) = kc
⇒ kax₁ + kby₁ = kc
Hence, (x₁, y₁) also satisfies equation (2).
Now, let (xh y,) be a solution of equation (2). Therefore, kax₁ + kby₁ = kc Since k ± 0, dividing both sides by k: ax₁ + by₁ = c
Hence, (x₁, y₁) also satisfies equation (1). Therefore, every solution of equation (1) is also a solution of equation (2), and vice versa.
Hence, both equations have the same solution sets.

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:
Slope of hill A = \(\frac{\text { Rise }}{\text { Run }}=\frac{60}{70}=\frac{6}{7}\)

Slope of hill B = \(\frac{\text { Rise }}{\text { Run }}=\frac{60}{110}=\frac{6}{11}\)

Slope of hill C = \(\frac{\text { Rise }}{\text { Run }}=\frac{80}{100}=\frac{4}{5}\)

Slopes, in ascending order, are \(\frac{6}{11}<\frac{4}{5}<\frac{6}{7}\)
The hills in order of their steepness (from least to greatest), are hill B < hill C < hill 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:
Slope of the ramp = \(\frac{\text { Rise }}{\text { Run }}=\frac{2.5}{4}=\frac{5}{8}\)
Therefore, the slope of the ramp is \(\frac{5}{8}\).

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

Solution:
Let A(0, 1.5) and B(1, 1.5) be two points on line l.

Then, slope = \(\frac{\text { Rise }}{\text { Run }}=\frac{y_2-y_1}{x_2-x_1}=\frac{1.5-1.5}{1-0}\) = 0

Solution:

Let the marked points on line l be C(-2, 2) and D(2, -2)
Then, Slope = \(\frac{\text { Rise }}{\text { Run }}=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-2}{2-(-2)}=\frac{-4}{4}\) = -1

Solution:

Let the marked points on line l be P(1, 1) and Q(-2, -1).
Then, slope = \(\frac{\text { Rise }}{\text { Run }}=\frac{y_2-y_1}{x_2-x_1}=\frac{-1-1}{-2-1}=\frac{-2}{-3}=\frac{2}{3}\)

Solution:

Let A(-1, 1) and B(-1, 0) be two points on line l.
Then, slope = \(\frac{\text { Rise }}{\text { Run }}=\frac{y_2-y_1}{x_2-x_1}=\frac{0-1}{-1-(-1)}=\frac{-1}{-1+1}\)
= undefined

Question 4.
An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp is 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:
Slope of the ramp according to the guidelines = \(\frac{\text { Rise }}{\text { Run }}\) = \(\frac{1}{12}\)
Given, the total height of the stairs is 18 cm
Let the horizontal length of the ramp be x.
Then, \(\frac{18}{x}=\frac{1}{12}\) ⇒ x = 18 × 12 = 216 cm
Therefore, the horizontal length of the ramp should be 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 one movie ticket be tx and the cost of one snack box be ₹ y.
(i) Cost of 2 movie tickets = ₹ x
Cost of 3 snack boxes = ₹ 3y
Total cost of 2 movie tickets and 3 snacks boxes is ₹ 850.
Thus, the required equation is 2x + 3y = 850.

(ii) Cost of 4 movie tickets = ₹ 4x
Cost of 1 snack box = ₹ y
Total cost of 4 movie tickets and 1 snacks box is ₹ 1100.
Thus, the required equation is 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:
Let the fixed charge for one taxi trip be ₹ x and the additional charge per kilometre be ₹ y. ,
(i) Fixed charge paid by Sahil = ₹ x
Additional charge paid by Sahil for 6 km = ₹ 6y
Thus, the required equation is x + 6y = 122.

(ii) Fixed charge paid by Meena = ₹ x
Additional charge paid by Meena for 8 km = ₹ 8y
Thus, the required equation is x + 8y = 160.

Question 3.
In a sports meet, tickets for adults cost ₹ 150 each and tickets for children cost 1100 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 andy to represent this situation.
Solution:
Let the number of adult tickets sold be x and the number of children’s tickets sold bey.
So, the number of adults who attended the meet is x and the number of children who attended the meet is y
∴ the number of people attended the meet is x + y = 200
Now, cost of one adult ticket = ₹ 150
And cost of one child’s ticket = ₹ 100
∴ the total amount collected from tickets sales is 150x + 100y = 25,000
Thus, the required pair of linear equations is x + y = 200 and 150x + 100y = 25,000.

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.
ATQ:
x + y = 5 …(1)
x – y = -21 …(2)
Adding both equations, we get
x + y + x – y = 5 – 21
⇒ 2x = – 16
⇒ x = – 8
Substituting x = -8 in equation (1),
we get -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. Find the numbers.
Solution:
Let the two numbers be x and y.
ATQ:
x – y = 26 ….(1)
One number is three times the other:
x = 3y
Substituting x = 3y in equation (1), we get
3y – y = 26
2y = 26 ⇒ y = 13
Now,
x = 3(13) ⇒ x = 39
Therefore, the two numbers are 39 and 13.

(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 one bat be ₹ x and one ball be ₹ y.
ATQ:
7x + 6y = 8880 …(1)
3x + 5y = 4000 …(2)
Multiplying equation (1) by 3, we get
21x+18y = 26640 …(3)
Multiplying equation (2) by 7, we get
21x + 35y = 28000 …(4)
Subtracting equation (3) from equation (4), we get
(21x + 35y) – (21x + 18y) = 28000 – 26640
⇒ 17y = 1360
⇒ y = 80
Substituting y = 80 in equation (2), we get
3x + 5(80) = 4000
⇒ 3x + 400 = 4000
⇒ 3x = 3600
⇒ x = 1200
Therefore, the cost of one bat is ₹ 1200 and the cost of one ball is ₹ 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 the fixed charge be ₹ x.
Let the charge per kilometre be ty.
ATQ:
For 10 km:
x+ 10y = 155
For 15 km:
x + 15y = 220 …(2)
Subtracting equation (1) from equation (2), we get
(x + 15y) – (x + 10y) = 220 – 155
⇒ 5y = 65
⇒ y = 13
Substituting y = 13 in equation (1), we get
x + 10(13) = 155
⇒ x + 130 = 155
⇒ x = 25
Therefore, the fixed charge is ₹ 25 and the charge per kilometre is ₹ 13.
Now, for 25 km:
Total charge = x + 25y
= 25 + 25(13)
= 25 + 325 = 350
Therefore, the amount paid for 25 km is ₹ 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 numerator be x and the denominator be y.
Then, the fraction is \(\frac{x}{y}\).
ATQ:
When 2 is added to both the numerator and denominator:
\(\frac{x+2}{y+2}=\frac{9}{11}\)
Cross-multiplying, we get
11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y = -4 …(1)
When 3 is added to both the numerator and denominator:
\(\frac{x+3}{y+3}=\frac{5}{6}\)
Cross-multiplying, we get
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x – 5y = -3 ….(2)
Multiplying equation (1) by 5, we get
55x – 45y = -20 ….(3)
Multiplying equation (2) by 9, we get
54x – 45y = -27 ……(4)
Now subtracting equation (4) from equation (3), we get
(55x – 45y) – (54x – 45y) = -20 – (-27)
⇒ x = 7
Substituting x = 7 in equation (2), we get
6(7) – 5y = – 3
⇒ 5y = 45
⇒ y = 9
Therefore, the fraction is \(\frac{7}{9}\).

(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 be \(\frac{x}{y}\)
Then, according to the question,
\(\frac{x+1}{y-1}\) = 1
⇒ x + 1 = y – 1
⇒ x – y = -2 ….(1)
And \(\frac{x}{y+1}=\frac{1}{2}\)
⇒ 2x = y + 1
⇒ 2x – y = 1 ……… (2)
Subtracting equation (1) from equation (2), we get
(2x – y) – (x – y) = 1 – (-2)
⇒ x = 3
Substituting x = 3 in equation (1), we get
3 – y = – 2
⇒ y = 5
Therefore, the fraction is \(\frac{3}{5}\)

(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 Nuri’s present age be x years and Sonu’s present age be y years.
Five years ago:
Nuri’s age = x – 5
Sonu’s age = y – 5
ATQ:
x – 5 = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = – 10 …………(1)

Ten years later:
Nuri’s age = x + 10
Sonu’s age = y + 1-
ATQ:
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 10 ………..(2)

Subtracting equation (2) from equation (1), we get
(x – 3y) – (x – 2y) = -10 – 10
⇒ -y = —20
⇒ y = 20
Substituting y = 20 in equation (2), we get
x – 40 = 10
⇒ x = 50
Therefore, the present ages of Nuri and Sonu are 50 years and 20 years, respectively.

(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 tens digit of the number be x and the ones digit be y.
∴ The number is 10x +y.
and the number obtained by reversing the digits is 10y + x
Now, according to the question,
Sum of digits is 9:
x + y = 9 …(1)
Nine times the number is twice the reversed number:
9(10x + y) = 2(10y + x)
⇒ 90x + 9y = 20y + 2x
⇒ 88x – 11y = 0 .
⇒ 8x – y = 0 [Dividing by 11]
⇒ y = 8x …(2)
Substituting equation (2) in equation (1), we get
x + 8x = 9
⇒ 9x = 9
⇒ x = 1
∴ y = 8(1) = 8
Therefore, the 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 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
⇒ x + 2y = 40 [Dividing by 50] …(2)
Subtracting equation (1) from equation (2), we get
(x + 2y) – (x + y) = 40 – 25
⇒ y = 15
Substituting y = 15 in equation (1), we get
x + 15 = 25
⇒ x = 10
Therefore, she received ten ₹ 50 notes and fifteen ₹ 100 notes.

(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 be ₹ x and the additional charge per extra day be ₹ y.
For 7 days:
Extra days are 7 – 3 = 4
x + 4y = 27 …(1)
For 5 days:
Extra days are 5 – 3 = 2
x + 2y = 21 ……… (2)
Subtracting equation (2) from equation (1), we get
(x + 4y) – (x + 2y) = 27 – 21
⇒ 2y = 6
⇒ y = 3
Substituting y = 3 in equation (2), we get x + 2(3) = 21
⇒ x + 6 = 21
⇒ x = 15
Therefore, the fixed charge is 115 and the charge for each extra day is ₹ 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.
Total number of students:
x + y = 10 …(1)

The number of girls is 4 more than the number of boys:
y = x + 4 …(2)
Now, in equation (1), considering x to be first 0 and then 10, we obtain the corresponding values ofy as 10 and 0, respectively. Hence, (0, 10) and (10, 0) are points on the line.

Now, in equation (2), considering x to be first 0 and then -4, we obtain the corresponding values of y as 4 and 0, respectively. Hence, (0, 4) and (-4, 0) are points on the line.

Now, by plotting the points and joining them to form a line, we get

Here, both the lines intersect at point (3, 7).
Therefore, the solution of the pair of equations is x = 3 and y = 7.
Thus, 3 boys and 7 girls took part in the quiz,

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
Solution:
5x – 4y + 8 = 0
7x + 6y – 9 = 0
Comparing with:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
We have:
a₁ = 5, b₁ = -4, c₁ = 8,
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 \(\frac{5}{7} \neq-\frac{2}{3}\)
∴ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Clearly, the pair of equations has a unique solution. Thus, the pair of equations intersect at a point.

(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
Solution:
9x + 3y + 12 = 0
18x + 6y + 24 = 0

Comparing with:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
We have:
a₁ = 9, b₁ = 3, c₁ = 12,
a₂ = 18, b₂ = 6, c₂ = 24

Now, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}\)

\(\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\) \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)

∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Clearly, the pair of equations has infinitely many solutions.
Thus, the pair of equations are coincident.

(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
6x – 3y + 10 = 0
2x-y + 9 = 0
Comparing with:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
We have:
a₁ = 6, b₁ = -3, c₁ = 10,
a₂ = 2, b₂ = -1, c₁ = 9

Now, \(\frac{a_1}{a_2}=\frac{6}{2}=\) = 3

\(\frac{b_1}{b_2}=\frac{-3}{-1}\) = 3

\(\frac{c_1}{c_2}=\frac{10}{9}\)

∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Clearly, the pair of equations has no solution.
Thus, the pair of equations 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
Solution:
x + y = 5,
2x + 2y = 10

Writing in standard form:
x + y – 5 = 0,
2x + 2y – 10 = 0

Here, \(\frac{a_1}{a_2}=\frac{1}{2}\)

\(\frac{b_1}{b_2}=\frac{1}{2}\) \(\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)

∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Hence, the pair of equations has infinitely many solutions.
Now, x + y = 5

2x + 2y = 10

Now, plotting the points and joining them to form a line, we get

We find that both equations are represented by the same straight line.
Therefore, all the points lying on the formed line are the solutions of the pair of equations, such as (0, 5) and (5,0).

(ii) x – y = 8, 3x – 3y = 16
Solution:
x – y = 8,
3x – 3y = 16
Writing in standard form:
x – y – 8 = 0
3x – 3y – 16 = 0
Here, \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{1}{3}, \frac{c_1}{c_2}=\frac{1}{2}\)

Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the pair of equations has no solution.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
Solution:
2x + y – 6 = 0,
4x – 2y – 4 = 0
Here, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{1}{-2}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the pair of equations has a unique solution.
Now, 2x + y – 6 = 0

4x – 2y – 4 = 0

Now, plotting the points and joining them to form a line, we get

Here, both the lines intersect at point (2, 2)
Therefore, the solution of the pair of equations is x = 2 and y = 2.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:
2x – 2y – 2 = 0,
4x – 4y – 5 = 0
Here, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{2}{5}\)

Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the pair of equations has 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 length = x m and width = y m.
Half perimeter = 36 m ⇒ Perimeter = 72 m
ATQ: 2(x + y) = 72
x + y = 36……..(1)
x = y + 4
x – y = 4 ……….(2)
Adding both equations, we get
2x = 40
⇒ x = 20
Substituting x = 20 in equation (1), we get
20 + y = 36
y = 16
Therefore, length = 20 m and 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:
Given equation:
2x + 3y – 8 = 0
(i) For intersecting lines:
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Let the another linear equation be x + y = 5
Here, \(\frac{2}{1} \neq \frac{3}{1}\)
Therefore, the lines intersect.
∴ x + y = 5 is the another linear equation.

(ii) For parallel lines:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Let the another linear equation 4x + 6y = 10
Here, \(\frac{2}{4}=\frac{3}{6} \neq \frac{-8}{-10}\)
Therefore, the lines are parallel.
4x + 6y = 10 is the another linear equation.

(iii) For coincident lines:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Multiply the given equation by 2:
4x + 6y – 16 = 0
Here,
\(\frac{2}{4}=\frac{3}{6}=\frac{-8}{-16}=\frac{1}{2}\)
Therefore, the lines coincide.
4x + 6y – 16 = 0 is the another linear equation.
(Answer may vary)

Question 7.
Here is a probleni that was posed by Mahavlracharya 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 in 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 one citron be ₹ x and the price of one fragrant wood-apple be ₹ y.
ATQ:
9x + 7y = 107 …(1)
7x + 9y = 101 …(2)
Adding both equations, we get
16x+ 16y= 208
⇒ x + y = 13
[Dividing by 16] …(3)
Subtracting the second equation from the first equation, we get
2x – 2y = 6
⇒ x – y = 3 [Dividing by 2] …(4)
Adding equations (3) and (4), we get
2x = 16 ⇒ x = 8
Substituting x = 8 in equation (1), we get
8 + y = 13 ⇒ y = 5
Therefore, the price of one citron is ₹ 8 and the price of one fragrant wood-apple is ₹ 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 one pencil be ₹ x and one pen be ty.
Cost of 5 pencils and 7 pens = ₹ 50
5x + 7y = 50 ………. (1)
Cost of 7 pencils and 5 pens = ?46
7x + 5y= 46 …(2)
Multiplying equation (1) by 7, we get
35x + 49y = 350 ………(3)
Multiply equation (2) by 5, we get
35x + 25y = 230 ……(4)
Subtracting equation (4) from (3), we get
24y = 120
⇒ y = 5
Substituting y = 5 in equation (1), we get
5x + 35 = 50
⇒ 5x = 15
⇒ x = 3
Therefore, the cost of one pencil is ₹ 3 and the cost of one pen is ₹ 5.

Question 9.
Find the height of the stool.

Solution:
Let the height of the stool be x cm and the height of the cat be y cm.
According to the question,
x + y = 85
x – y = 25 …(2)
Adding equation (1) and (2), we get
2x = 110
⇒ x = 55
Therefore, the height of the stool is 55 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.
Solution:
Let the two points on the line be (x1,^1) = (0,0) and 02,72) = (2, 6).
Slope (m) = \(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)
= \(\frac{6-0}{2-0}=\frac{6}{2}\) = 3
Hence, the slope of the line is 3.

(ii) What is the y-intercept of the line? Explain its signifiance.
Solution:
Comparing v = 3x with the standard slope-intercept form = mx + d, we get d= 0. Thus, the y-intercept is 0 (the line crosses the y-axis at (0, 0)).
Significance: The y-intercept represents the Value ofy when x = 0. Since they-intercept is 0, the value of the quantity is zero at the starting point. In a real- life situation, it means that when the measurement begins (x = 0), there is no initial amount present. As x increases, the quantity increases according to the slope of the line.

(iii) A water tank is being filed 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?
Solution:
The relation is modelled by y = 3x.
(a) For x = 5 minutes: y = 3(5) = 15 cm
Thus, the water is 15 cm high after 5 minutes.

(b) For y = 21 cm: 21 = 3x ⇒ x = \(\frac{21}{3}\) = 7 minutes
Thus, it will take 7 minutes to reach 21 cm.

(iv) Without drawing a new graph, determine whether the point (4, 10) lies on this line. Justify your answer.
Solution:
A point (x, y) lies on a line if and only if its coordinates satisfy the equation y = 3x. Substituting x = 4 in the equation, we get y = 3(4) = 12. However, the given y-coordinate is 10, which is not equal to 12. Since LHS ^ RHS, the point (4,10) does not lie on this line.

(v) Plot the point where this line intersects the line x = 3. Explain how you found the coordinates.
Solution:
To find where y = 3x intersects x = 3, substituting x = 3 in the equation, we get y = 3(3) = 9.
Hence, the point of intersection is (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.
Solution:
F = \(\frac{9}{5}\)C + 32
When C = 0,
F = \(\frac{9}{5}\)(0) + 32
⇒ F = 32

When C = 5,
F = \(\frac{9}{5}\) (5) + 32
⇒ F = 9 + 32 = 41

When C = -5,
F = \(\frac{9}{5}\)(-5) + 32
⇒ F = -9 + 32 = 23
Now, plotting points on the graph where C is along the x-axis and F is along they-axis:

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
Solution:
Temperature in Celsius = 30°C
Substituting C = 30° in F = \(\frac{9}{5}\)C + 32, we get
F = \(\frac{9}{5}\)(30) + 32 = 9(6) + 32
= 54 + 32 = 86°F

(iii) If the temperature is 95°F, what is the temperature in Celsius?
Solution:
Temperature in Fahrenheit = 95°F
Substititing F = 95° in F = \(\frac{9}{5}\)C + 32, we get
95 = \(\frac{9}{5}\)C + 32
⇒ 63 = \(\frac{9}{5}\)C
C = \(\frac{63 \times 5}{9}\) = 35°F

(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?
Solution:
At C = 0°C; F = \(\frac{9}{5}\)(0) + 32 = 32°F
At F = 0°C; 0 = \(\frac{9}{5}\)C + 32
⇒ \(\frac{9}{5}\)C = -32
⇒ C = \(\frac{1}{2}\) ≈ -17.78°F

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it
Solution:
Let the common temperature be x, such that F = C = x
⇒ x = \(\frac{9}{5}\)x + 32
⇒ x – \(\frac{9}{5}\)x = 32
⇒ \(\frac{-4}{5}\)x = 32
⇒ x = \(=\frac{-32 \times 5}{4}\) = -40
Thus, -40° is numerically identical in both scales (-40°C = -40°F).

Question 3.
Solve the following system of equations graphically:
2x + y = 6,
2x – y – 2 = 0.
Solution:
2x + y = 6
Expressing y in terms of x,
y = 6 – 2x

For equation:
2z – y – 2 = 0
Expressing y in terms of x, we get
y = 2x – 2

Now, by plotting the points and joining them to form a line, we get

Here, both the lines intersect at point (2, 2).
Therefore, the solution of the pair of equations is x = 2 and y = 2

Question 4.
Find the point of intersection of the lines shown on the cover page.
Solution:
The equations of the lines given on the cover page are 15x- 10y – 115 = 0 and 15x + 100y = 335.
For equation, 15x – 10y – 115 =0
For equation: 15x + 100y = 335
Expressing x in terms of y,
y = \(\frac{15 x-115}{10}\)

For equation: 15x + 100y = 335
Expressing x in terms of y,
x = \(\frac{335-100 y}{15}\)

Now, by plotting the points and joining them to form a line, we get

Here, both the lines intersect at point (9, 2).
Therefore, the point of intersection of the pair of equations is at x = 9 and y = 2.

Question 5.
Give a formula to find the x-intercept of the line y = mx + d.
Solution:
The x-intercept is the x-coordinate of the point where the line crosses the x-axis. At this point, the v-coordinate is zero.
Substituting y = 0 in the given equation, we get
0 = mx + d
mx = -d
x = –\(\frac{d}{m}\) (provided m ≠ 0).

Therefore, the formula for the x-intercept is –\(\frac{d}{m}\), and the point on thexaxis is (-\(\frac{d}{m}\), 0)

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 m denote the total number of calling minutes in a month.
Total monthly cost for Plan A: C_A = 50 + 0.20m
Total monthly cost for Plan B: C_B = 30 + 0.30m

For C_A < C_B:
50 + 0.20m < 30 + 0.30m
⇒ 20 < 0.10m ⇒ m > 200 minutes
Plan A is cheaper when calling exceeds 200 minutes per month.

For C_B < C_A:
30 + 0.30 m < 50 + 0.20w
⇒ 0.10 m < 20
⇒ m < 200 minutes
Plan B is cheaper when the calling is less than 200 minutes per month.
For C_A = C_B:
⇒ 50 + 0.20m = 30 + 0.30m
⇒ 50 – 30 = 0.30m – 0.20m
⇒ 20 = 0.10m
⇒ m = \(\frac{20}{0.10}\)
= 200 minutes
Both plans cost exactly the same (?90) at 200 minutes.

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) In the slope-intercept form y = mx + d, if the slope m is fixed, they-intercept dcan be any arbitrary real number. Each choice of d produces a distinct parallel line. Therefore, infinitely many lines exist that have a given slope.

(ii) A line with a given slope m passing through a fixed point (x₁, y₁) is uniquely determined by the point- slope equation y – y₁ = m(x – x₁). No other line satisfies both conditions simultaneously. Therefore, exactly one unique line exists that have a given slope and pass through a given 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:
4x +py + 8=0
2x + 2y + 2= 0
Comparing the given equations with standard forms
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
a₁ = 4, b₁ = p, c₁ = 8, a₂ = 2, b₂ = 2, c₂ = 2
Condition for a unique solution:
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{4}{2} \neq \frac{p}{2}\)

⇒ 2 ≠ \(\frac{p}{2}\)
⇒ 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
Writing both equations in the standard form, we get
(a + b)x – 2by – (5a + 2b + 1) = 0
⇒ a₁ = a + b, b₁ = -2b, c₁ = -(5a + 2b + 1)
3x – y – 14 = 0
⇒ a₂ = 3, b₂ = -1, c₂ = -14
For the system of equations to possess infinitely many solutions:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{a+b}{3}=\frac{-2 b}{-1}=\frac{-(5 a+2 b+1)}{-14}\)

⇒ \(\frac{a+b}{3}=2 b=\frac{(5 a+2 b+1)}{14}\)

Now, \(\frac{a+b}{3}\) = 2b
⇒ a + b = 6b
⇒ a = 5b
⇒ 2b = \(\frac{1}{2}\)
⇒ 28b = 5a + 2b + 1
⇒ 5a – 26b + 1 = 0

Substituting a = 5b in equation (2), we get
5(5b) – 26b + 1 = 0
⇒ 25b – 26b + 1 = 0
⇒ -b + 1 = 0
⇒ b = 1

Substititing b = 1 in equation (1), we get
a = 5(1) = 5
Therefore, a = 5 and b = 1

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:
For two lines to be coincident, the system of equations must have infinitely many solutions
⇒ \(\frac{1}{2}\)
Here a₁ = 1, b₂ = 2, c₁ = -3
a₂ = k – 1, b₂ = k + 1, c₂ = -(k + 3)
⇒ \(\frac{1}{k-1}=\frac{2}{k+1}=\frac{-3}{-(k+3)}\)
Equating the first two terms:
⇒ \(\frac{1}{k-1}=\frac{2}{k+1}\)
⇒ k + 1 = 2(k – 1)
k+ 1 = 2k – 2
Verifying the three terms at k = 3
\(\frac{1}{3-1}=\frac{2}{3+1}=\frac{-3}{-(3+3)}\)
⇒ \(\frac{1}{2}=\frac{2}{4}=\frac{3}{6}\)
⇒ \(\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\)
Since all three ratios are equal to \(\frac{1}{2}\), the system represents coincident lines when k = 3.

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 starts from (0, 0); moves 3 units right, 4 units up.
⇒ Slope m₁ = \(\frac{1}{2}\)
⇒ Equation of the path i s y = \(\frac{4}{3}\)x.
4x – 3y = 0
Robot 2 starts from (10, 0); moves 1 unit right, 2 units up.
⇒ Slope m₂ = \(\frac{\text { Change in } y}{\text { Change in } x}=\frac{2}{1}\) = 2
⇒ Equation of the path is y – 0 = 2(x – 10)
⇒ y = 2x – 20
Now, to find the point of intersection:
⇒ \(\frac{4}{3}\)x = 2x – 20
⇒ 4x = 6x – 60
⇒ 2x = 60
⇒ x = 30
⇒ y= 2(30) – 20 = 40
Since x = 30 andy = 40, this point lies on the paths traced by both robots.
⇒ Yes, the paths traced by two robots will intersect at (30, 40).

(ii) Robot 1 starts from (3, 0); moves 5 units right, 5 units up.
⇒ Slope m₁ = \(\frac{5}{5}\) = 1
⇒ Equation of the path is y = x – 3
(defined for x > 3, y > 0)
Robot 2 starts from (7, 0); moves 5 units right, 3 units down.
⇒ Slope m₂ = \(\frac{-3}{5}\)
⇒ Equation of the path is y = \(\frac{-3}{5}\) (x – 7)
(defined for x > 7, y < 0).
5y = -3x + 21
3x + 5y = 21 …(2)
Now, to find the point of intersection, substitute the value of y (from equation (1) to equation (2), we get,
x – 3 = \(\frac{-3}{5}\) (x – 7)
⇒ 5x – 15 = -3x + 21
⇒ 8x = 36
⇒ x = 4.5
Since Robot 2 begins at x = 7 and only moves to the right (x ≥ 7), the value x = 4.5 is behind Robot 2’s starting location. Therefore, although the extended lines meet at (4.5, 1.5), the actual paths traced by the robots do not intersect.

Question 12.
At a certain time, Jacob notices that his digital watch reads ‘a’ minutes after two o’clock. Fifteen minutes later, it reads 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:
Let the time be measured in total minutes elapsed after 12 o’clock
First reading (a minutes after 2:00)
= 2 × 60 + a = 120 + a minutes

Second reading (b minutes after 3:00)
= 3 × 60 + 6 = 180 + b minutes

The second reading occurs exactly 15 minutes after the first reading:
(180 + b) – (120 + a) = 15
60 + b – a = 15
a – b = 45 …(1)
We are given that a is six times b ⇒ a = 6b …(2)
Substituting a = 6b in equation (1), we get
6b – b = 45
⇒ 5b = 45
⇒ b = 9
Therefore, when Jacob looked at his watch for the second time, the watch read 9 minutes after 3 O’clock.
Thus, the time was 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 tens digit be x and the units digit be y.
Original number = 10x + y
Reversed number = 10y + x
According to the question,
x + y = 15 …(1)
And (10y + x) – (10x + y) = 9
⇒ 9y – 9x = 9
⇒ y – x = 1 [Dividing by 9] …(2)
Adding equations (1) and (2), we get
(x + y) + (y – x) = 15 + 1
⇒ 2y = 16
⇒ y = 8
Substituting y = 8 in equation (1), we get
x + 8 = 15
⇒ x = 7
Therefore, the number is 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, the sum of each pair of opposite angles is 180°.
⇒ ∠A + ∠C = 180°,
And ∠B + ∠D = 180°
⇒ (x + 7) + (3y + 23) = 180
⇒ x + 3y + 30 = 180
⇒ x + 3y = 150 …(1)
And (y + 8) + (4x + 12) = 180
4x + y + 20 = 180
⇒ 4x + y = 160
⇒ y = 160 – 4x ……….(2)
Substituting y = 160 – 4x in equation (1), we get
x + 3(160 -4x)= 150
⇒ x + 480- 12x = 150
⇒ -11x = —330
⇒ x = 30
Substituting x = 30 in equation (2), we get
y = 160 – 4(30)
⇒ 160 – 120 = 40
Now, calculating each angle:
∠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 of the train be v km/h and the usual time taken be t hours.
Distance travelled D = v x t
When speed = (v + 6), time = (t — 6):
(v + 6) (t – 6) = vt
⇒ vt – 6v + 6/ – 36 = vt
⇒ —6v + 61 = 36
⇒ —v + t= 6
⇒ t = v + 6 ……….(1)
When speed = (v – 4), time = (t + 6):
(v – 4)(t + 6) = vt
⇒ vt + 6v – 4t – 24 = vt
⇒ 6v – 4t = 24
⇒ 3v – 2t = 12 …(2)
Substituting equation (1) in equation (2), we get
3v – 2(v + 6) = 12
⇒ 3v – 2v – 12 = 12
⇒ v = 24 km/h
Substituting v = 24 km/h in equation (1), we get
t – 24 + 6 = 30 hours
Now, D = v × t = 24 × 30 = 720 km
Therefore, the speed of the train is 24 km/h, and the distance travelled = 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 present age of the father be F years and the sum of the present ages of the four children be S years. According to the question,
F = S …(1)
After 20 years, the father’s age will be (F + 20) years and since there are 4 children, each child grows 20 years older, so the total increase in their combined age is 4 × 20 = 80 years. Therefore, the combined age of children after years will be (S + 80) years.
ATQ:
S + 80 = 2(F + 20)
⇒ S + 80 = 2F + 40 …(2)
Substituting S = F from equation (1) in equation (2), we get
F + 80 = 2F + 40
⇒ 80 – 40 = 2F – F
⇒ F = 40 years
Therefore, the present age of the father is 40 years.