Chapter 13: Two Variables, One LineAll 9 Questions on This Page

Class 9 Maths Chapter 13 Exercise Set 13.5 Solutions

Comprehensive step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 13 Exercise Set 13.5 (Questions 1–9). Covers algebraic and graphical methods for pairs of linear equations in two variables, classification using coefficient ratios, and real-world word problems.

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Key Concepts for Exercise Set 13.5

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Algebraic Methods: Elimination and Substitution

A pair of linear equations can be solved algebraically by substitution (expressing one variable in terms of another) or elimination (multiplying by coefficients to cancel a variable upon addition or subtraction).

Substitution: y = f(x) | Elimination: c₁E₁ ± c₂E₂

Classification of Linear Systems by Coefficient Ratios

For a pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0: if a₁/a₂ ≠ b₁/b₂, lines intersect (unique solution); if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, lines are parallel (no solution); if a₁/a₂ = b₁/b₂ = c₁/c₂, lines are coincident (infinitely many solutions).

Intersecting: a₁/a₂ ≠ b₁/b₂ | Parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Coincident: a₁/a₂ = b₁/b₂ = c₁/c₂

Graphical Solution of Linear Systems

Plotting both equations on the Cartesian plane reveals their common solution at the exact coordinates of their point of intersection (x, y). Parallel lines never meet and have no common solution.

Point of intersection (x, y) = Unique common solution

Question 1

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

(i) Two integers

Let the first integer be x and the second integer be y.

Their sum is 5:

x + y = 5

Their difference is −21:

x − y = −21

Adding equations (1) and (2):

2x = −16

x = −8

Substituting x = −8 into equation (1):

−8 + y = 5

y = 13

Therefore, the two integers are −8 and 13.

(ii) Difference between two numbers

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

Their difference is 26:

x − y = 26

The larger number is three times the smaller number:

x = 3y

Substituting equation (2) into equation (1):

3y − y = 26

2y = 26

y = 13

Since x = 3y,

x = 3(13) = 39

Therefore, the two numbers are 39 and 13.

(iii) Cricket bats and balls

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

From the first purchase:

7x + 6y = 8880

From the second purchase:

3x + 5y = 4000

Multiplying equation (1) by 3:

21x + 18y = 26640

Multiplying equation (2) by 7:

21x + 35y = 28000

Subtracting equation (3) from equation (4):

17y = 1360

y = 80

Substituting y = 80 into equation (2):

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) Taxi fare

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

For a journey of 10 km:

x + 10y = 155

For a journey of 15 km:

x + 15y = 220

Subtracting equation (1) from equation (2):

5y = 65

y = 13

Substituting y = 13 into equation (1):

x + 10(13) = 155

x + 130 = 155

x = 25

Therefore, the fixed charge is ₹25 and the charge per kilometre is ₹13.

For a journey of 25 km:

Total fare = x + 25y

= 25 + 25(13)

= 25 + 325

= 350

Therefore, the total fare for 25 km is ₹350.

(v) Finding a fraction

Let the numerator be x and the denominator be y.

The first condition gives:

x + 2y + 2 = 911

Cross-multiplying:

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

11x − 9y = −4

The second condition gives:

x + 3y + 3 = 56

Cross-multiplying:

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

6x − 5y = −3

Multiplying equation (1) by 5:

55x − 45y = −20

Multiplying equation (2) by 9:

54x − 45y = −27

Subtracting equation (4) from equation (3):

x = 7

Substituting x = 7 into equation (2):

6(7) − 5y = −3

42 − 5y = −3

5y = 45

y = 9

Therefore, the fraction is 79.

(vi) Finding another fraction

Let the numerator be x and the denominator be y.

Adding 1 to the numerator and subtracting 1 from the denominator makes the fraction equal to 1:

x + 1y − 1 = 1

x + 1 = y − 1

x − y = −2

Adding 1 only to the denominator makes the fraction equal to 1/2:

x/(y + 1) = 12

Cross-multiplying:

2x = y + 1

2x − y = 1

Subtracting equation (1) from equation (2):

x = 3

Substituting x = 3 into equation (1):

3 − y = −2

y = 5

Therefore, the fraction is 35.

(vii) Ages of Nuri and Sonu

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

Five years ago:

x − 5 = 3(y − 5)

x − 3y = −10

Ten years from now:

x + 10 = 2(y + 10)

x − 2y = 10

Subtracting equation (1) from equation (2):

y = 20

Substituting y = 20 into equation (2):

x − 2(20) = 10

x − 40 = 10

x = 50

Therefore, Nuri's present age is 50 years and Sonu's present age is 20 years.

(viii) A two-digit number

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

The sum of the digits is 9:

x + y = 9

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

According to the question:

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

90x + 9y = 20y + 2x

88x = 11y

Dividing by 11:

8x = y

Substituting equation (2) into equation (1):

x + 8x = 9

9x = 9

x = 1

Therefore, y = 8(1) = 8.

The original number is 10(1) + 8 = 18.

(ix) ₹50 and ₹100 notes

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

The total number of notes is 25:

x + y = 25

The total value is ₹2000:

50x + 100y = 2000

Dividing equation (2) by 50:

x + 2y = 40

Subtracting equation (1) from equation (3):

y = 15

Substituting y = 15 into equation (1):

x + 15 = 25

x = 10

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

(x) Lending library charges

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

Saritha kept the book for seven days (3 fixed days + 4 extra days):

x + 4y = 27

Susy kept the book for five days (3 fixed days + 2 extra days):

x + 2y = 21

Subtracting equation (2) from equation (1):

2y = 6

y = 3

Substituting y = 3 into equation (2):

x + 2(3) = 21

x + 6 = 21

x = 15

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

Question 2

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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.

Graphical Solution for Question 2

Graph showing the intersection of x + y = 10 and y − x = 4 at point (3, 7).

Solution

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

There are 10 students in total:

x + y = 10

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

y = x + 4

Rearranging:

y − x = 4

We will plot both equations on the same coordinate graph.

For equation (1): x + y = 10

•If x = 0, then y = 10, giving the point (0, 10).

•If y = 0, then x = 10, giving the point (10, 0).

Draw a straight line passing through (0, 10) and (10, 0).

For equation (2): y − x = 4

•If x = 0, then y = 4, giving the point (0, 4).

•If x = 6, then y = 10, giving the point (6, 10).

Draw a straight line passing through (0, 4) and (6, 10).

The two lines intersect at (3, 7).

Therefore,

x = 3, y = 7

Answer:Boys = 3, Girls = 7 (Common solution from graph: (3, 7))

Question 3

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Use the ratios a₁/a₂, b₁/b₂, c₁/c₂, 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

For equations in the standard form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:

•If a₁/a₂ ≠ b₁/b₂, the lines intersect at one point.

•If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.

•If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident.

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

The coefficient ratios are:

a₁/a₂ = 57

b₁/b₂ = −46 = −23

c₁/c₂ = 8/(−9) = −89

Since 57 ≠ −23, the first two ratios are unequal:

a₁/a₂ ≠ b₁/b₂

Therefore, the lines intersect at one point.

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

The coefficient ratios are:

a₁/a₂ = 918 = 12

b₁/b₂ = 36 = 12

c₁/c₂ = 1224 = 12

All three ratios are equal:

a₁/a₂ = b₁/b₂ = c₁/c₂ = 12

Therefore, the lines are coincident (they represent the same line).

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

The coefficient ratios are:

a₁/a₂ = 62 = 3

b₁/b₂ = −3/(−1) = 3

c₁/c₂ = 109

The first two ratios are equal, but the third ratio is different:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (since 3 ≠ 109)

Therefore, the lines are parallel (they do not intersect).

Answer:(i) Intersect at a point | (ii) Coincident lines | (iii) Parallel lines

Question 4

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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

Graphical Solutions for Question 4

Graphs for parts (i), (ii), (iii), and (iv) showing coincident lines in (i), parallel lines in (ii) and (iv), and intersecting lines in (iii).

Solution

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

Dividing the second equation 2x + 2y = 10 by 2 gives:

x + y = 5

The second equation is twice the first. Both equations represent the same straight line.

Table of values for x + y = 5:

•If x = 0, then y = 5: point (0, 5).

•If x = 5, then y = 0: point (5, 0).

Plotting both equations results in one coincident line.

Therefore, the pair of linear equations has infinitely many solutions (coincident lines).

(ii) x − y = 8 and 3x − 3y = 16

Expressing each equation in slope-intercept form (y = mx + c):

From x − y = 8:

y = x − 8

From 3x − 3y = 16:

3y = 3x − 16

y = x − 163

The slopes are equal (m = 1), but the intercepts differ (−8 and −163). Therefore, the lines are parallel and never intersect.

Plotting the two lines:

•The first line (x − y = 8) passes through (8, 0) and (0, −8).

•The second line (3x − 3y = 16) passes through (163, 0) and (0, −163).

Therefore, the pair of linear equations has no solution (parallel lines).

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

Rewrite the equations as:

2x + y = 6

4x − 2y = 4

For the first line, use the points (0, 6) and (3, 0).

For the second line, use the points (0, −2) and (1, 0).

Plotting both lines shows they intersect at (2, 2).

x = 2, y = 2

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

Rewrite the equations:

2x − 2y = 2 ⇒ x − y = 1

4x − 4y = 5 ⇒ x − y = 54

These equations have the same coefficients of x and y after simplification, but different constant terms.

Their graphs are parallel lines with slope m = 1, so they do not intersect.

Therefore, the pair has no solution.

Answer:(i) Infinitely many solutions (coincident lines) | (ii) No solution (parallel lines) | (iii) x = 2, y = 2 | (iv) No solution (parallel lines)

Question 5

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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 be x metres and the width be y metres.

Half the perimeter is 36 m. For a rectangle, half the perimeter equals length plus width:

x + y = 36

The length is 4 m more than the width:

x − y = 4

Adding equations (1) and (2):

2x = 40

x = 20

Substituting x = 20 into equation (1):

20 + y = 36

y = 16

Therefore, the length is 20 m and the width is 16 m.

Answer:Length = 20 m, Width = 16 m

Question 6

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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.

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

Solution

The given equation is:

2x + 3y − 8 = 0

(i) Intersecting lines

For intersecting lines, we require a₁/a₂ ≠ b₁/b₂.

Choose another equation:

x + y − 1 = 0

The coefficient ratios are:

a₁/a₂ = 21 = 2, b₁/b₂ = 31 = 3

Since 2 ≠ 3, the lines intersect at one point.

(ii) Parallel lines

For parallel lines, we require a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Choose another equation:

4x + 6y − 10 = 0

The coefficient ratios are:

a₁/a₂ = 24 = 12, b₁/b₂ = 36 = 12, c₁/c₂ = −8−10 = 45

Since 12 = 12 ≠ 45, the lines are parallel.

(iii) Coincident lines

For coincident lines, we require a₁/a₂ = b₁/b₂ = c₁/c₂.

Multiplying the original equation by 2:

2(2x + 3y − 8) = 0

4x + 6y − 16 = 0

All three coefficient ratios are equal to 12, so both equations represent the same straight line.

Answer:(i) x + y − 1 = 0 | (ii) 4x + 6y − 10 = 0 | (iii) 4x + 6y − 16 = 0

Question 7

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Here is a problem that was posed by Mahāvīrāchārya in Gaṇita sāra saṅgraha (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, in the same units as the question.

From the first condition:

9x + 7y = 107

From the second condition:

7x + 9y = 101

Adding equations (1) and (2):

16x + 16y = 208

Dividing by 16:

x + y = 13

Subtracting equation (2) from equation (1):

2x − 2y = 6

Dividing by 2:

x − y = 3

Adding equations (3) and (4):

2x = 16

x = 8

Substituting x = 8 into equation (3):

8 + y = 13

y = 5

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

Answer:Price of one citron = 8 units, Price of one fragrant wood-apple = 5 units

Question 8

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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 the cost of one pen be ₹y.

From the first condition:

5x + 7y = 50

From the second condition:

7x + 5y = 46

Adding equations (1) and (2):

12x + 12y = 96

Dividing by 12:

x + y = 8

Subtracting equation (1) from equation (2):

2x − 2y = −4

Dividing by 2:

x − y = −2

Adding equations (3) and (4):

2x = 6

x = 3

Substituting x = 3 into equation (3):

3 + y = 8

y = 5

Therefore, the cost of one pencil is ₹3 and the cost of one pen is ₹5.

Answer:Cost of one pencil = ₹3, Cost of one pen = ₹5

Question 9

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Find the height of the stool.

Stool and Cat Height Puzzle Diagram

Illustration showing total height (85 cm) with cat on stool, and height difference (25 cm) with cat on floor.

Solution

Let the height of the stool be x cm and the height of the cat be y cm.

When the cat sits on the stool, the total height from the floor to the top of the cat is 85 cm:

x + y = 85

When the cat is on the floor, the stool is 25 cm taller than the cat:

x − y = 25

Adding equations (1) and (2):

2x = 110

x = 55

Substituting x = 55 into equation (1):

55 + y = 85

y = 30

Therefore, the height of the stool is 55 cm (and the height of the cat is 30 cm).

Answer:Height of the stool = 55 cm

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💡Key Concepts for Exercise Set 13.5

Algebraic Methods: Elimination and Substitution

Classification of Linear Systems by Coefficient Ratios

Graphical Solution of Linear Systems

Key Concepts for Exercise Set 13.5