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

Class 9 Ganita Manjari Chapter 13 Quick Links:

• Think and Reflect
• Exercise Set 13.1
• Exercise Set 13.2
• Exercise Set 13.3
• Exercise Set 13.4
• Exercise Set 13.5
• End of Chapter Exercises

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Class 9 Maths Ganita Manjari Chapter 13 Think and Reflect Solutions

Answer:
Substitute each ordered pair in the left-hand side.

Another solution is (2, 3), since 3(2) + 2(3) = 12.
Rearranging gives y = (12 − 3x)/2.
Every real value of x gives a corresponding value of y.
Therefore there are infinitely many solutions.

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?
Answer:
Yes, provided a and b are not both zero, as required by the definition of a linear equation in two variables.

When c = 0, the equation is ax + by = 0. Substituting (0, 0) gives a(0) + b(0) = 0.
Hence the origin lies on the line.
This also includes x = 0, the y-axis and y = 0, the x-axis.

If a = b = 0 too, the equation becomes 0 = 0. Every point satisfies it, so it represents the whole plane, not a single line. That case is excluded by the definition.

Answer:
The line in Figure passes through A(2, 0) and B(4, 3), so its slope is (3 − 0)/(4 − 2) = 3/2.

Starting at A, increasing x from 2 to 8 gives a run of 6. The rise is (3/2) × 6 = 9.
Therefore y = 9 and the ordered pair is (8, 9).
Equivalently, the equation is y = (3/2)(x − 2).

Answer:
Using y = (3/2)(x − 2).
6 = (3/2)(x − 2)
⇒ 4 = x − 2, so x = 6.
Therefore the point is (6, 6).

Answer:
The given points are A(0, 3), B(1, −2) and C(−1, 8).

• Slope using A and B = (−2 − 3)/(1 − 0) = −5.
• Slope using A and C = (8 − 3)/(−1 − 0) = 5/(−1) = −5.
• Slope using B and C = (8 − (−2))/(−1 − 1) = 10/(−2) = −5.

Every pair gives the same slope, −5, as required.

Answer:
Let one paratha cost ₹p and one bowl of dahi cost ₹d.
Then p + 2d = 90 and 2p + 7d = 260.
Double the first equation: 2p + 4d = 180.
Subtract it from the second: 3d = 80, so d = 80/3.
Now p = 90 − 2(80/3) = (270 − 160)/3 = 110/3.
One paratha costs ₹110/3 = ₹36⅔ and one bowl of dahi costs ₹80/3 = ₹26⅔.

Checking:
110/3 + 160/3 = 90 and 220/3 + 560/3 = 260.
The stated totals lead to fractional-rupee values. ₹36.67 and ₹26.67 are only rounded approximations.

Answer:
Row 1 gives 2T + 2R = 28, so T + R = 14.
Column 4 gives T + R + 2C = 22.
Hence 14 + 2C = 22, giving C = 4.

Row 3 gives T + 3C = 18, so T + 12 = 18 and T = 6.
Then R = 14 − 6 = 8.
Row 2 gives 2H + 2R = 30, so 2H + 16 = 30 and H = 7.
Therefore triangle = 6, rectangle = 8, hexagon = 7, circle = 4.
The missing first-column total is T + H + C + C = 6 + 7 + 4 + 4 = 21.

Checking:
Row 4 totals 4 + 8 + 4 + 4 = 20
Column 2 totals 8 + 8 + 6 + 8 = 30
Column 3 totals 8 + 7 + 4 + 4 = 23.

Answer:
Yes. The original equations are 7x − 15y = 2 and x + 2y = 3.
Substitute y = (3 − x)/2 into the first equation:
7x − 15(3 − x)/2 = 2.
Multiply throughout by 2:
14x − 45 + 15x = 4
⇒ 29x = 49, so x = 49/29.
Then y = (3 − 49/29)/2 = (38/29)/2 = 19/29.
Thus we obtain the same answer: (x, y) = (49/29, 19/29).

Answer:
The equations in Example 9 are 9x − 4y = 2000 and 7x − 3y = 2000. The incomes are 9x and 7x; the expenditures are 4y and 3y.

Eliminating x:
Multiply the first equation by 7: 63x − 28y = 14,000.
Multiply the second equation by 9: 63x − 27y = 18,000.
Subtract the first new equation from the second: y = 4000.
Substitute into 7x − 3y = 2000:
7x − 12,000 = 2000, so x = 2000.

Using substitution:
From 7x − 3y = 2000, x = (2000 + 3y)/7.
Substitute into 9x − 4y = 2000:
9(2000 + 3y)/7 − 4y = 2000.
Multiply by 7: 18,000 + 27y − 28y = 14,000.
Thus y = 4000 and x = 2000.
The monthly incomes are 9x = ₹18,000 and 7x = ₹14,000.

Comparison:
Elimination is convenient here because multiplication makes the selected coefficients equal while keeping all calculations in integers. Eliminating y, as the textbook does, uses smaller multipliers 3 and 4. Substitution is equally valid but introduces a fraction in this example. It is especially convenient when an equation already has a variable with coefficient 1 or −1.

Answer:
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.

Answer:
Divide 10x − 5y = 20 by 5 and 4x − 2y = 8 by 2. Both become 2x − y = 4.
Therefore they have infinitely many common solutions, namely (t, 2t − 4) for every real t. Examples are (0, −4), (2, 0) and (3, 2).

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 a₁/a₂ = b₁/b₂ = c₁/c₂.

Answer:
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.

Answer:
No. A pair can also have no solution.
For example, x + y = 2 and x + y = 5 cannot both hold for the same x and y.
Subtracting gives 0 = 3, a contradiction.
Their graphs are distinct parallel lines and have no common point.

Answer:

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 a₁/a₂ = b₁/b₂ ≠ c₁/c₂. The corresponding lines are distinct and parallel.

Answer:
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 a₁/a₂ ≠ b₁/b₂.
2. If a pair has infinitely many solutions, then a₁/a₂ = b₁/b₂ = c₁/c₂.
3. If a pair has no solution, then a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

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.

Answer:
Yes. One useful presentation is the comparison method:
Express the same variable from both equations and equate the two expressions.
For example, take x + y = 7 and 2x − y = 2.
The first gives y = 7 − x, while the second gives y = 2x − 2.
Since both expressions equal y, write 7 − x = 2x − 2.
This is a linear equation in x alone. It gives 3x = 9, so x = 3 and y = 4.
Comparison is closely related to substitution; it is a different way to organise the work, rather than a completely unrelated principle.
For equations with interchanged coefficients, using the sum x + y and difference x − y can also simplify the work.

Answer:
A common solution is an ordered pair belonging to both graphs. Look for the points shared by the two lines.

• (i) If they intersect at one point, its coordinates give the unique solution.
• (ii) If they are distinct parallel lines, they have no common point, so there is no solution.
• (iii) If they coincide, every point on the common line satisfies both equations. There are infinitely many solutions.

The four pairs immediately above this box illustrate these cases. The following table also supplies the plotting points requested in that activity.

For Pair 1, the graph gives approximately (2.67, 0.67).
Substituting y = x − 2 in 2x + y = 6 gives 3x = 8, confirming the exact intersection (8/3, 2/3).

(Hint: Consider the equations of the two lines to be y = mx + d₁ and y = mx + d₂.)
Answer:
Two distinct lines with the same defined slope are parallel.
Suppose y = mx + d₁ and y = mx + d₂ had a common point (u, v).
Then v = mu + d₁ and v = mu + d₂.
Subtracting gives d₁ = d₂.
Therefore, when d₁ ≠ d₂, no common point is possible.
The two lines are parallel.

If d₁ = d₂, the equations describe the same line, so they coincide.
This distinction is necessary: equal slopes alone allow either parallel distinct lines or coincident lines.

Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.1 Solutions

Answer:
The standard form is ax + by + c = 0.
Substituting the given values gives 3x + 0y − 1/5 = 0.
Thus the required equation is 3x − 1/5 = 0.
The equivalent equation 15x − 1 = 0 represents the same line, but 3x + 0y − 1/5 = 0 displays the requested coefficients exactly.

Answer:
Moving all terms to the left-hand side and compare with ax + by + c = 0. Here √2x means (√2) × x.

Multiplying every term of an equation by −1 gives another valid standard form. Its coefficients and constant must all change sign together.

Answer:
(i) Charlie is correct.
The notebook costs twice as much as the pen, so t = 2p. For example, if a pen costs ₹10, the notebook costs ₹20. Meera’s equation reverses the relationship.

(ii) Yes.
The two batsmen’s scores add to 176, so x + y = 176 correctly represents the situation. In this context x and y are non-negative integers.

Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.2 Solutions

Answer:
Substituting x = 4 and y = 3 gives
5x − 6y = 5(4) − 6(3) = 20 − 18 = 2.
The left-hand side equals the right-hand side. Therefore (4, 3) is a solution.

Answer:
(i) For 7x − 3y = 21:
Put y = 0: 7x = 21, so x = 3.
Put x = 0: −3y = 21, so y = −7.
Two solutions are (3, 0) and (0, −7).

(ii) For 2x + 3y = 5:
Put x = 1: 2 + 3y = 5, so y = 1.
Put x = 4: 8 + 3y = 5, so y = −1.
Two solutions are (1, 1) and (4, −1).

Answer:
Substitute x = 2 and y = −1 into the first equation:
2m(2) + 3(−1) = 7
⇒ 4m − 3 = 7
⇒ 4m = 10, so m = 5/2.
Substitute into the second equation:
4(2) + n(−1) = −10
⇒ 8 − n = −10, so n = 18.
Therefore m = 5/2 and n = 18.

(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.
Answer:
Choosing a value of x and calculate y, keeping both coordinates non-zero.

Drawing the coordinate axes using equal unit lengths on the two axes. Plotting the two points for each equation and draw the straight line through them. Each graph shows the relevant points in their stated quadrants.

Answer:
No, C(2, 3) does not lie on the line.
At C, 3x − 7y = 3(2) − 7(3) = 6 − 21 = −15, which is not 21. Thus C does not satisfy the equation.
The line is exactly the set of points whose coordinates satisfy the equation. Therefore a point that does not lie on this line cannot satisfy its equation.
For drawing the line, we use its intercepts A(7, 0) and B(0, −3).

(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.
Answer:
(i) False.
It has infinitely many real solutions. For example, x + y = 2 has (0, 2), (1, 1), (2, 0), and infinitely many others.

(ii) False.
For ax + by + c = 0, substituting the origin gives c = 0. Only equations with c = 0 represent lines through the origin. For example, x + y = 2 does not pass through it.

(iii) False.
For example, (1/2, 1/2) is a rational solution of x + y = 1.

(iv) True.
Write it as x + 0y − 3 = 0. Here a = 1, b = 0, c = −3; a and b are not both zero. Its solutions are (3, y) for every real y.

(v) True.
Rearranging gives y = (7 − 2x)/3. Every real value of x gives a solution.

(vi) False.
At (1, 2), 2x + 3y = 2 + 6 = 8 ≠ 7.

Answer:
(i) Multiplying both sides of 3x + 4y = 7 by 2 gives 6x + 8y = 14. Thus every solution of the first equation satisfies the second.
Conversely, dividing both sides of the second equation by 2 gives the first.
Hence their solution sets are identical.

(ii) If (u, v) satisfies au + bv = c, multiplying by k gives kau + kbv = kc. So it satisfies the second equation.
Conversely, divide kau + kbv = kc by the non-zero number k to get au + bv = c.
Thus it satisfies the first equation.
The restriction k ≠ 0 is essential. Multiplying by zero gives 0 = 0, which is satisfied by every point.

Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.3 Solutions

Answer:
Comparing rise ÷ horizontal run, rather than height alone.

• Slope of A = 60/70 = 6/7 ≈ 0.857.
• Slope of B = 60/110 = 6/11 ≈ 0.545.
• Slope of C = 80/100 = 4/5 = 0.8.

Since 6/11 < 4/5 < 6/7, the order from least to greatest steepness is B, C, A.

Answer:
Slope = rise/run
= 2.5/4
= 5/8
= 0.625.
Thus the ramp’s slope is 5/8.

Answer:
(i) The line is horizontal. Between two different points, rise = 0 and run ≠ 0.
Therefore slope = 0.

(ii) Moving right along the line, the vertical fall equals the horizontal run. Counting the large grid intervals between the marked points gives a rise of −4 units and a run of 4 units.
Therefore slope = −4/4 = −1.

(iii) Between the marked points, move 3 large grid intervals to the right and 2 upward.
Therefore slope = 2/3.

(iv) The line is vertical. Its run is 0, while its rise is non-zero. Since division by zero is undefined, its slope is undefined. It is not zero.

Answer:
The given rise-to-run ratio is 1:12.
For a vertical rise of 18 cm, the horizontal length is 18 × 12 = 216 cm.
Therefore the required horizontal length is 216 cm or 2.16 m.

Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.4 Solutions

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

(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.
Answer:
(i) The movie tickets cost ₹2x and the snack boxes cost ₹3y.
Therefore 2x + 3y = 850.

(ii) The movie tickets cost ₹4x and the snack box costs ₹y.
Therefore 4x + y = 1100.

Hence, the required pair is 2x + 3y = 850 and 4x + y = 1100.

(i) Sahil travelled 6 km and paid a total fare of ₹122.
(ii) Meena travelled 8 km and paid a total fare of ₹160.
Answer:
Total fare = fixed charge + (distance × charge per kilometre).
(i) Sahil’s fare gives x + 6y = 122.
(ii) Meena’s fare gives x + 8y = 160.
These are the required equations.

Answer:
The total number of tickets gives x + y = 200.
The total collection gives 150x + 100y = 25,000.
Thus the required pair is x + y = 200 and 150x + 100y = 25,000.
The second equation can also be simplified to 3x + 2y = 500.

Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.5 Solutions

(i) The sum of two integers is +5 and their difference is −21. Find the two numbers.
Answer:
Let the integers, in the stated subtraction order, be x and y.
x + y = 5 and x − y = −21.
Adding gives 2x = −16, so x = −8.
Then y = 5 − (−8) = 13.
The integers are −8 and 13.

(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Answer:
For the usual positive-number interpretation, let the larger number be x and the smaller be y.
x − y = 26 and x = 3y.
Solving the equations, we get 3y − y = 26
⇒ 2y = 26
⇒ y = 13.
Hence x = 39.
The positive 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.
Answer:
Let a bat cost ₹x and a ball cost ₹y.
7x + 6y = 8880 …(1)
and
3x + 5y = 4000 …(2)
Multiply the equation (1) by 5: 35x + 30y = 44,400.
Multiply the equation (2) by 6: 18x + 30y = 24,000.
By subtraction, we get: 17x = 20,400
⇒ x = 1200.
Then from equation (2), we have 3(1200) + 5y = 4000
⇒ 5y = 400
⇒ y = 80.
Hence, each bat costs ₹1200 and each ball costs ₹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?
Answer:
Let the fixed charge be ₹x and the rate per km be ₹y.
x + 10y = 155 and x + 15y = 220.
By subtraction, we get: 5y = 65
⇒ y = 13.
Then x = 155 − 10(13) = 25.
Hence, the fixed charge = ₹25; rate = ₹13 per km.
Now for 25 km, fare = 25 + 25(13) = ₹350.

(v) A fraction becomes equal to 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 5/6. Find the fraction.
Answer:
Let the numerator be x and the denominator be y.
So, (x + 2)/(y + 2) = 9/11 gives 11x + 22 = 9y + 18 ⇒ 11x − 9y = −4 …(1)
and (x + 3)/(y + 3) = 5/6 gives 6x + 18 = 5y + 15 ⇒ 6x − 5y = −3 …(2)
Multiply these equations (1) and (2) by 5 and 9, respectively, we get
55x − 45y = −20 and 54x − 45y = −27.
By subtracting, we have x = 7.
Then 42 − 5y = −3 ⇒ y = 9.
Hence, the fraction is 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 1/2 if we add 1 only to the denominator. What is the fraction?
Answer:
Let the numerator be x and the denominator be y.
So, (x + 1)/(y − 1) = 1 gives x + 1 = y − 1 ⇒ y = x + 2.
and x/(y + 1) = 1/2 gives 2x = y + 1.
Substitute y = x + 2, we get: 2x = x + 3 ⇒ x = 3 and y = 5.
Hence, the fraction is 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?
Answer:
Let Nuri’s present age be x years and Sonu’s present age be y years.
Five years ago: x − 5 = 3(y − 5) ⇒ x − 3y = −10 …(1)
Ten years from now: x + 10 = 2(y + 10) ⇒ x − 2y = 10 …(2)
Subtracting equation (1) from (2), we get: y = 20.
Then from (2), x − 40 = 10 ⇒ x = 50.
Hence, Nuri is 50 years old and Sonu is 20 years old.

(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.
Answer:
Let the tens digit be x and the units digit be y.
Therefore the number = 10x + y
Therefore, sum of digits: x + y = 9.
The number is 10x + y and the reversed number is 10y + x.
Thus 9(10x + y) = 2(10y + x).
⇒ 90x + 9y = 20y + 2x
⇒ 88x = 11y
⇒ y = 8x.
Substitute into x + y = 9, we get: 9x = 9
⇒ x = 1 and y = 8.
Hence, 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?
Answer:
Let x be the number of ₹50 notes and y the number of ₹100 notes.
So, x + y = 25 …(1)
and 50x + 100y = 2000 …(2)
Divide the equation (2) by 50, we get: x + 2y = 40 …(3)
Subtracting equation (1) from (3), we get
y = 15. Then x = 10.
Hence, Meena received 10 notes of ₹50 and 15 notes of ₹100.

(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.
Answer:
Let the fixed charge for the first three days be ₹x and the charge per extra day be ₹y.
Seven days include 7 − 3 = 4 extra days, so x + 4y = 27 …(1)
Five days include 5 − 3 = 2 extra days, so x + 2y = 21 …(2)
Subtracting equation (2) from (1), we get
2y = 6
⇒ y = 3. Then from the equation (2), x = 21 − 6 = 15.
Hence, the fixed charge is ₹15 and each extra day costs ₹3.

Answer:
Let x be the number of boys and y the number of girls.
Then x + y = 10 and y = x + 4.
Points to draw the lines:

Plot the points on common axes, with boys on the x-axis and girls on the y-axis. The lines intersect at (3, 7).
Therefore, 3 boys and 7 girls took part.

(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
Answer:

Thus the pairs have one, infinitely many and no common solutions, respectively.

(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
Answer:
(i) Divide the second equation by 2 to get x + y = 5. Both graphs are the same line.
Plot (0, 5) and (5, 0) for the first equation, and (1, 4) and (4, 1) for the second. The lines coincide.
There are infinitely many solutions: (t, 5 − t), for every real t. For example, (0, 5), (2, 3) and (5, 0).

(ii) The equations become y = x − 8 and y = x − 16/3.
For the first line, plotting (4, −4) and (8, 0).
For the second, plotting (4, −4/3) and (8, 8/3).
They have the same slope 1 and different y-intercepts. They are distinct parallel lines, so there is no solution.

(iii) Write the equations as y = 6 − 2x and y = 2x − 2.
For the first line, plotting (0, 6), (2, 2) and (3, 0).
For the second, plotting (0, −2), (1, 0), and (2, 2).
The graphs intersect at (2, 2). Therefore x = 2 and y = 2 is the unique solution.
Check: 2(2) + 2 − 6 = 0 and 4(2) − 2(2) − 4 = 0.

(iv) The equations become y = x − 1 and y = x − 5/4.
For the first line, plotting (0, −1) and (2, 1).
For the second, plotting (0, −5/4) and (2, 3/4).
Both have slope 1 and different y-intercepts. Their graphs are distinct parallel lines, so there is no solution. A fine grid helps show their small separation.
Therefore only pairs (i) and (iii) have common solutions.

Answer:
Let the length be l metres and width be w metres.
Half the perimeter is l + w, so l + w = 36 …(1)
Also l − w = 4 …(2)
Adding (1) and (2) gives 2l = 40, so l = 20.
Then w = 36 − 20 = 16.
The garden is 20 m long and 16 m wide.

Answer:
Many answers are possible. One valid choice for each is:
(i) x + y − 3 = 0.
Here the coefficient ratios are 2/1 = 2 and 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.

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?)
Answer:
Let one citron cost x units and one fragrant wood-apple cost y units.
The equations are 9x + 7y = 107 and 7x + 9y = 101.
The coefficients 9 and 7 exchange positions, so adding and subtracting gives the sum and difference of the prices directly.
By adding: 16x + 16y = 208, giving x + y = 13.
Subtract the second equation from the first: 2x − 2y = 6, giving x − y = 3.
The two numbers with sum 13 and difference 3 are
x = (13 + 3)/2 = 8 and y = (13 − 3)/2 = 5.
Each citron costs 8 units and each fragrant wood-apple costs 5 units.

Answer:
Let a pencil cost ₹x and a pen cost ₹y.
Then 5x + 7y = 50 and 7x + 5y = 46.
By adding: 12x + 12y = 96, so x + y = 8.
Subtract the first equation from the second: 2x − 2y = −4, so x − y = −2.
Adding these simpler equations, we get
2x = 6, so x = 3 and y = 5.
A pencil costs ₹3 and a pen costs ₹5.

Answer:
Let the stool height be s cm and the cat’s height in the shown posture be c cm.
The first arrangement gives s + c = 85.
The second gives s − c = 25.

Adding the two equations give 2s = 110, so s = 55.
Therefore the stool is 55 cm high.
Hence, the cat’s height is 85 − 55 = 30 cm.

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

(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.

Answer:
(i) Slope = (change in y)/(change in x) = (6 − 0)/(2 − 0) = 6/2 = 3.
For each increase of 1 in x, y increases by 3. In the tank context, the water level rises at 3 cm per minute.

(ii) The y-intercept is 0. The line crosses the y-axis at the origin (0, 0). In the tank model, the water level is 0 cm at time 0.

(iii)(a) At x = 5, y = 3(5) = 15. The water is 15 cm high.
(b) Putting y = 21: 21 = 3x, gives x = 7. It takes 7 minutes.

(iv) No.
For x = 4, the equation requires y = 12, whereas the given y-coordinate is 10. Since 10 ≠ 3(4), (4, 10) is not on the line.

(v) At the intersection with x = 3, substitute x = 3 into y = 3x to obtain y = 9. Plotting (3, 9).
It lies on both lines, so this is their intersection.

(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.
Answer:
(i) Using the equation to find points: (C, F) = (-40, -40), (0, 32), (10, 50), (30, 86) and (35, 95)
Now, plotting these ordered pairs (C, F), with C on the horizontal axis and F on the vertical axis. Draw the straight line through them and label it F = (9/5)C + 32. Using clearly marked scales, for example 10 degrees per major grid interval on each axis.

(ii) At C = 30, F = (9/5)(30) + 32 = 54 + 32 = 86 °F.

(iii) Put F = 95:
95 = (9/5)C + 32
63 = (9/5)C
C = 63 × 5/9 = 35 °C.

(iv) When C = 0, F = 32 °F.
When F = 0, 0 = (9/5)C + 32, so C = −32 × 5/9 = −160/9 °C ≈ −17.78 °C.

(v) Let the common numerical reading be t, so F = C = t.
t = (9/5)t + 32.
Multiply by 5: 5t = 9t + 160.
Thus −4t = 160 and t = −40.
Therefore −40 °C and −40 °F represent the same temperature.

Answer:
Write the equations as y = 6 − 2x and y = 2x − 2.

Plotting the points on common coordinate axes and drawing both straight lines.
They intersect at (2, 2), giving the unique answer x = 2, y = 2.

Answer:
It is an activity. Do yourself.

Answer:
At the x-axis, y = 0. Therefore 0 = mx + c.
For m ≠ 0, x = −c/m.
Thus the x-intercept is −c/m and the intercept point is (−c/m, 0).
Special cases: If m = 0 and c ≠ 0, the horizontal line y = c never meets the x-axis, so it has no x-intercept. If m = 0 and c = 0, the line is the x-axis itself; every point on it is common, so there is no single intercept point.

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.
Answer:
Let x ≥ 0 be the number of calling minutes in a month.
Cost of Plan A = ₹(50 + 0.20x).
Cost of Plan B = ₹(30 + 0.30x).
For equal costs, 50 + 0.20x = 30 + 0.30x.
Hence 20 = 0.10x, so x = 200.
At 200 minutes, each plan costs ₹90.

To compare, subtract the costs:
Cost A − Cost B = 20 − 0.10x.
If x > 200, this is negative, so Plan A is cheaper.
If 0 ≤ x < 200, this is positive, so Plan B is cheaper.
At x = 200, neither is cheaper; they cost the same.
If only whole minutes are considered, Plan A is cheaper from 201 minutes onward and Plan B is cheaper for 0–199 minutes.

(i) have a given slope?
(ii) have a given slope and pass through a given point?
Answer:
(i) Infinitely many. For a fixed real slope m, every equation y = mx + c has that slope.
Choosing different real values of c gives different parallel lines.

(ii) Exactly one. If the point is (a, b), substituting into y = mx + c gives c = b − ma.
Thus the unique equation is y = mx + (b − ma) or y − b = m(x − a).
For completeness, if the intended direction is vertical, there are infinitely many vertical lines x = k, but exactly one through (a, b): x = a.
Their slope is undefined rather than a real number.

4x + py + 8 = 0; 2x + 2y + 2 = 0.
Answer:
A unique solution occurs when the x- and y-coefficient ratios differ:
4/2 ≠ p/2.
Thus 2 ≠ p/2, giving p ≠ 4.
Therefore every real value of p except 4 gives a unique solution.

(a + b)x − 2by = 5a + 2b + 1; 3x − y = 14.
Answer:
For infinitely many solutions, the complete first equation must be a non-zero multiple k of 3x − y = 14.
Comparing the y-coefficients gives −2b = −k, so k = 2b.
Comparing x-coefficients gives a + b = 3k = 6b, so a = 5b.

The right-hand constants must also match:
5a + 2b + 1 = 14k = 28b.
Substitute a = 5b:
25b + 2b + 1 = 28b, so b = 1.
Therefore a = 5 and b = 1.

x + 2y = 3; (k − 1)x + (k + 1)y = k + 3.
Answer:
For coincidence, the second equation must be a non-zero multiple λ of the first.
Comparing x-coefficients gives λ = k − 1.
Comparing y-coefficients gives k + 1 = 2λ = 2(k − 1).
Thus k + 1 = 2k − 2, so k = 3.