Master Workbook: Chapter 13

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

The standard form of a linear equation in two variables is ax + by + c = 0. Substitute the given values into the standard form. 3x + 0y - 1/5 = 0 Final Answer: 3x + 0y - 1/5 = 0

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

Express each equation in the form ax + by + c = 0 to identify the coefficients.

Linear equation	Standard Form	Coefficient of x (a)	Coefficient of y (b)	Constant term (c)
y - 15 = √2x	-√2x + y - 15 = 0	-√2	1	-15
3y - 2x = 0	-2x + 3y + 0 = 0	-2	3	0
5x = 3y	5x - 3y + 0 = 0	5	-3	0
x = 8	x + 0y - 8 = 0	1	0	-8
3y = 1	0x + 3y - 1 = 0	0	3	-1

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, two Indian batsmen together scored 176 runs. Manisha expressed this situation as x + y = 176. Is this a correct representation?

(i) Let pen cost ₹10 (p=10). A notebook costs twice as much, so it costs ₹20. Charlie's equation: t = 2(10) = 20. This matches the logic perfectly. Meera's equation: p = 2(20) = 40. This is incorrect. Final Answer (i): Charlie's representation (t = 2p) is correct. (ii) Let the runs scored by the first batsman be x and the second batsman be y. Their total combined score is 176. Adding their individual scores gives x + y. Final Answer (ii): Yes, Manisha's representation (x + y = 176) is correct.

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

Substitute x = 4 and y = 3 into the left-hand side (LHS) of the equation: LHS = 5(4) - 6(3) LHS = 20 - 18 = 2 Since the LHS equals the right-hand side (RHS = 2), the ordered pair (4, 3) satisfies the equation. Final Answer: Yes, (4, 3) is a valid solution.

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

(i) 7x - 3y = 21

Let x = 0: 7(0) - 3y = 21 => -3y = 21 => y = -7. Solution 1: (0, -7). Let y = 0: 7x - 3(0) = 21 => 7x = 21 => x = 3. Solution 2: (3, 0).

(ii) 2x + 3y = 5

Let x = 1: 2(1) + 3y = 5 => 3y = 3 => y = 1. Solution 1: (1, 1). Let x = -2: 2(-2) + 3y = 5 => -4 + 3y = 5 => 3y = 9 => y = 3. Solution 2: (-2, 3).

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.

Substitute x = 2 and y = -1 into the first equation (2mx + 3y = 7): 2m(2) + 3(-1) = 7 4m - 3 = 7 4m = 10 => m = 10/4 = 5/2 Substitute x = 2 and y = -1 into the second equation (4x + ny = -10): 4(2) + n(-1) = -10 8 - n = -10 -n = -18 => n = 18 Final Answer: m = 5/2, n = 18.

Question 4:
Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants.

(i) 5x + 3y = 7

Let x = -1: -5 + 3y = 7 => 3y = 12 => y = 4. Point (-1, 4) is in Quadrant II. Let y = -1: 5x - 3 = 7 => 5x = 10 => x = 2. Point (2, -1) is in Quadrant IV.

(ii) 5x - 3y = 7

Let x = 2: 10 - 3y = 7 => -3y = -3 => y = 1. Point (2, 1) is in Quadrant I. Let y = -4: 5x + 12 = 7 => 5x = -5 => x = -1. Point (-1, -4) is in Quadrant III.

(iii) -5x + 3y = 7

Let x = 1: -5 + 3y = 7 => 3y = 12 => y = 4. Point (1, 4) is in Quadrant I. Let y = -1: -5x - 3 = 7 => -5x = 10 => x = -2. Point (-2, -1) is in Quadrant III.

(iv) -5x - 3y = 7

Let x = 1: -5 - 3y = 7 => -3y = 12 => y = -4. Point (1, -4) is in Quadrant IV. Let y = 1: -5x - 3 = 7 => -5x = 10 => x = -2. Point (-2, 1) is 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?

Visually from the graph, the point C(2, 3) is situated above the line. It does not lie on the line. Numerically checking the equation: 3(2) - 7(3) = 6 - 21 = -15. Since -15 ≠ 21, the point C(2, 3) does not satisfy the equation. Final Answer: No. A point off the line cannot satisfy its equation, because the line is precisely the graph of all its solutions.

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. -> FALSE. It has infinitely many solutions forming a straight line. (ii) The graph of a linear equation in two variables always passes through the origin. -> FALSE. It only passes through the origin if the constant term c is zero (e.g., ax + by = 0). (iii) A linear equation in two variables can never have rational solutions. -> FALSE. Coordinates of solutions can easily be fractions (e.g., x = 1/2). (iv) x = 3 is a valid linear equation in two variables. -> TRUE. It can be written in standard form as 1x + 0y - 3 = 0. (v) The equation 2x + 3y = 7 has infinitely many solutions. -> TRUE. For any real value of x, there is a corresponding real value of y. (vi) The point (1, 2) is a solution of the equation 2x + 3y = 7. -> FALSE. Substituting yields 2(1) + 3(2) = 2 + 6 = 8, which is not 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.

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

(i) 3x + 4y = 7 and 6x + 8y = 14

If you multiply the entire first equation (3x + 4y = 7) by a constant factor of 2, you obtain the second equation (6x + 8y = 14). Because one is just a scaled version of the other, they represent the exact same line graphically and share an identical infinite set of solutions.

(ii) ax + by = c and kax + kby = kc (where k ≠ 0)

The second equation can be factored on both sides: k(ax + by) = kc. Since we are explicitly given that k ≠ 0, we can divide both sides of the equation by k. This results in ax + by = c, which perfectly matches the first equation. Thus, any coordinate pair (x, y) that satisfies the first will inherently satisfy the second.

Question 1:
The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest.
• Hill A: Rise = 60 m, Run = 70 m
• Hill B: Rise = 60 m, Run = 110 m
• Hill C: Rise = 80 m, Run = 100 m

Steepness is determined by calculating the slope (Slope = Rise / Run). Slope of Hill A = 60 / 70 = 6/7 ≈ 0.857 Slope of Hill B = 60 / 110 = 6/11 ≈ 0.545 Slope of Hill C = 80 / 100 = 8/10 = 4/5 = 0.800 Compare the decimal values: 0.545 < 0.800 < 0.857 Final Answer: Ranking from least steep to greatest steepness is 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.

Formula for Slope = Rise / Run Substitute the given values = 2.5 / 4 Multiply numerator and denominator by 10 to remove the decimal = 25 / 40 Simplify the fraction by dividing both by 5 = 5 / 8 Final Answer: The slope of the ramp is 5/8 (or 0.625).

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

Diagram (i): The line is perfectly horizontal. The rise is zero for any given run. Slope = 0 / Run = 0. Diagram (ii): The line slants downwards from left to right. It descends one grid unit for each unit to the right. As run is positive, rise is negative. Slope is Negative (-1). Diagram (iii): The line slants upwards from left to right. Marking points (-2, -1) and (1, 1), we get m = (1 - (-1)) / (1 - (-2)) = 2/3. Slope is Positive. Diagram (iv): The line is perfectly vertical. The horizontal run is zero for any given rise. Slope = Rise / 0, which is Undefined.

Question 4:
An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp 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?

Let the required horizontal length (run) be x. The mandated slope ratio (Rise / Run) is given as 1 / 12. Set up the proportion: Rise / Run = 1 / 12 Substitute the known total rise (18 cm): 18 / x = 1 / 12 Cross-multiply to solve for x: x = 18 × 12 x = 216 Final Answer: The horizontal length of the ramp should be 216 cm.

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.

Let x = cost of one movie ticket Let y = cost of one snack box First day purchase: 2 tickets and 3 snack boxes cost 850 Equation 1: 2x + 3y = 850 Second day purchase: 4 tickets and 1 snack box cost 1100 Equation 2: 4x + 1y = 1100

Final Answer: The pair of linear equations is 2x + 3y = 850 and 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.

Let x = fixed charge for the trip Let y = additional charge per kilometre Sahil's fare: Fixed charge + 6 km charge = 122 Equation 1: x + 6y = 122 Meena's fare: Fixed charge + 8 km charge = 160 Equation 2: x + 8y = 160

Final Answer: The pair of linear equations is x + 6y = 122 and x + 8y = 160.

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

Let x = number of adult tickets sold Let y = number of children's tickets sold Total number of attendees is 200 Equation 1: x + y = 200 Total amount collected is calculated by (price × adults) + (price × children) Equation 2: 150x + 100y = 25000 Simplify Equation 2 by dividing all terms by 50 Simplified Equation 2: 3x + 2y = 500

Final Answer: The pair of linear equations is x + y = 200 and 150x + 100y = 25000.

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.

Let the two integers be x and y. x + y = 5 ... (Equation 1) x - y = -21 ... (Equation 2) Adding Equation 1 and Equation 2: 2x = -16 => x = -8 Substitute x = -8 in Equation 1: -8 + y = 5 => y = 13 Final Answer: 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.

Let the larger number be x and the smaller number be y. x - y = 26 ... (Equation 1) x = 3y ... (Equation 2) Substitute Equation 2 into Equation 1: 3y - y = 26 2y = 26 => y = 13 Substitute y = 13 in Equation 2: x = 3(13) => x = 39 Final Answer: The 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.

Let the cost of one bat be ₹x and one ball be ₹y. 7x + 6y = 8880 ... (Equation 1) 3x + 5y = 4000 ... (Equation 2) From Equation 2: 3x = 4000 - 5y => x = (4000 - 5y)/3 Substitute x in Equation 1: 7((4000 - 5y)/3) + 6y = 8880 Multiply equation by 3: 7(4000 - 5y) + 18y = 26640 28000 - 35y + 18y = 26640 -17y = -1360 => y = 80 Substitute y = 80 into x: x = (4000 - 5(80))/3 = (4000 - 400)/3 = 3600/3 = 1200 Final Answer: Cost of a bat is ₹1200 and a 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?

Let the fixed charge be ₹x and charge per km be ₹y. x + 10y = 155 ... (Equation 1) x + 15y = 220 ... (Equation 2) Subtracting Equation 1 from Equation 2: 5y = 65 => y = 13 Substitute y = 13 in Equation 1: x + 10(13) = 155 => x + 130 = 155 => x = 25 Fare for 25 km = x + 25y = 25 + 25(13) = 25 + 325 = 350 Final Answer: Fixed charge is ₹25, charge per km is ₹13. Fare for 25 km is ₹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.

Let the fraction be x/y. (x+2)/(y+2) = 9/11 => 11(x + 2) = 9(y + 2) => 11x - 9y = -4 ... (Equation 1) (x+3)/(y+3) = 5/6 => 6(x + 3) = 5(y + 3) => 6x - 5y = -3 ... (Equation 2) From Equation 2: 6x = 5y - 3 => x = (5y - 3)/6 Substitute x in Equation 1: 11((5y - 3)/6) - 9y = -4 Multiply by 6: 11(5y - 3) - 54y = -24 55y - 33 - 54y = -24 => y = 9 Substitute y = 9 into x: x = (5(9) - 3)/6 = (45 - 3)/6 = 42/6 = 7 Final Answer: 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?

Let the fraction be x/y. (x+1)/(y-1) = 1 => x + 1 = y - 1 => x - y = -2 ... (Equation 1) x/(y+1) = 1/2 => 2x = y + 1 => 2x - y = 1 ... (Equation 2) Subtracting Equation 1 from Equation 2: (2x - x) - (y - y) = 1 - (-2) => x = 3 Substitute x = 3 in Equation 1: 3 - y = -2 => y = 5 Final Answer: 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?

Let Nuri's current age be x and Sonu's current age be y. Five years ago: (x - 5) = 3(y - 5) => x - 3y = -10 ... (Equation 1) Ten years later: (x + 10) = 2(y + 10) => x - 2y = 10 ... (Equation 2) Subtracting Equation 1 from Equation 2: (-2y) - (-3y) = 10 - (-10) => y = 20 Substitute y = 20 in Equation 2: x - 2(20) = 10 => x - 40 = 10 => x = 50 Final Answer: 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.

Let the tens digit be x and the units digit be y. Original Number = 10x + y. x + y = 9 ... (Equation 1) Reversed Number = 10y + x. 9(10x + y) = 2(10y + x) => 90x + 9y = 20y + 2x 88x - 11y = 0 => Dividing by 11: 8x - y = 0 => y = 8x ... (Equation 2) Substitute Equation 2 into Equation 1: x + 8x = 9 => 9x = 9 => x = 1 Substitute x = 1 into y = 8x: y = 8(1) = 8 Final Answer: 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?

Let the number of ₹50 notes be x and the number of ₹100 notes be y. Total number of notes: x + y = 25 ... (Equation 1) Total value of notes: 50x + 100y = 2000 Dividing by 50: x + 2y = 40 ... (Equation 2) Subtracting Equation 1 from Equation 2: y = 15 Substitute y = 15 in Equation 1: x + 15 = 25 => x = 10 Final Answer: She 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.

Let the fixed charge (for the first 3 days) be ₹x and the extra daily charge be ₹y. Saritha kept it for 7 days (3 fixed + 4 extra): x + 4y = 27 ... (Equation 1) Susy kept it for 5 days (3 fixed + 2 extra): x + 2y = 21 ... (Equation 2) Subtracting Equation 2 from Equation 1: 2y = 6 => y = 3 Substitute y = 3 in Equation 2: x + 2(3) = 21 => x + 6 = 21 => x = 15 Final Answer: The fixed charge is ₹15 and the extra day charge 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.

Let the number of boys be x and the number of girls be y. Total number of students: x + y = 10 The number of girls is 4 more than boys: y = x + 4 => -x + y = 4

The point of intersection on the graph is (3, 7). Final Answer: There are 3 boys and 7 girls.

Question 3:
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

a1/a2 = 5/7, b1/b2 = -4/6 = -2/3. Since a1/a2 ≠ b1/b2, the lines Intersect at a point.

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

a1/a2 = 9/18 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = 12/24 = 1/2. Since a1/a2 = b1/b2 = c1/c2, the lines are Coincident.

(iii) 6x - 3y + 10 = 0 ; 2x - y + 9 = 0

a1/a2 = 6/2 = 3, b1/b2 = -3/-1 = 3, c1/c2 = 10/9. Since a1/a2 = b1/b2 ≠ c1/c2, the lines are Parallel.

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

a1/a2 = 1/2, b1/b2 = 1/2, c1/c2 = 5/10 = 1/2. Coincident lines. They have infinitely many solutions.

(ii) x - y = 8, 3x - 3y = 16

a1/a2 = 1/3, b1/b2 = -1/-3 = 1/3, c1/c2 = 8/16 = 1/2. Parallel lines. They have no solution.

(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0

a1/a2 = 2/4 = 1/2, b1/b2 = 1/-2. Since a1/a2 ≠ b1/b2, they have a unique solution.

Graphical intersection is at x = 2, y = 2.

(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0

a1/a2 = 2/4 = 1/2, b1/b2 = -2/-4 = 1/2, c1/c2 = -2/-5 = 2/5. Parallel lines. They have no solution.

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

Let the length be x and the width be y. The length is 4 m more than width: x = y + 4 => x - y = 4 ... (Equation 1) Half the perimeter is 36 m: x + y = 36 ... (Equation 2) Adding Equation 1 and Equation 2: 2x = 40 => x = 20 Substitute x = 20 into Equation 2: 20 + y = 36 => y = 16 Final Answer: The length is 20 m and the width is 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.

(i) For intersecting lines, we need a1/a2 ≠ b1/b2. An example is: 3x + 2y - 5 = 0. (ii) For parallel lines, we need a1/a2 = b1/b2 ≠ c1/c2. An example is: 4x + 6y - 10 = 0. (iii) For coincident lines, we need a1/a2 = b1/b2 = c1/c2. An example is: 4x + 6y - 16 = 0.

Question 7:
Here is a problem that was posed by Mahaviracharya. 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. Tell me quickly the price of each citron and of each fragrant wood-apple.

Let the price of a citron be x and a wood-apple be y. 9x + 7y = 107 ... (Equation 1) 7x + 9y = 101 ... (Equation 2) Adding Equation 1 and Equation 2: 16x + 16y = 208. Dividing by 16: x + y = 13 ... (Equation 3) Subtracting Equation 2 from Equation 1: 2x - 2y = 6. Dividing by 2: x - y = 3 ... (Equation 4) Adding Equation 3 and Equation 4: 2x = 16 => x = 8 Substitute x = 8 into Equation 3: 8 + y = 13 => y = 5 Final Answer: A citron costs 8 and a fragrant wood-apple costs 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.

Let the cost of a pencil be ₹x and a pen be ₹y. 5x + 7y = 50 ... (Equation 1) 7x + 5y = 46 ... (Equation 2) Adding both equations: 12x + 12y = 96 => x + y = 8 ... (Equation 3) Subtracting Equation 2 from Equation 1: -2x + 2y = 4 => -x + y = 2 ... (Equation 4) Adding Equation 3 and Equation 4: 2y = 10 => y = 5 Substitute y = 5 into Equation 3: x + 5 = 8 => x = 3 Final Answer: The cost of one pencil is ₹3 and one pen is ₹5.

Question 9:
Find the height of the stool.

Based on the diagram, let the height of the Stool be S and the height of the Cat be C. When the cat is on the stool: S + C = 85 ... (Equation 1) When the stool is above the cat: S - C = 25 ... (Equation 2) Adding Equation 1 and Equation 2: 2S = 110 => S = 55 Final Answer: The height of the stool is 55 cm.

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.

m = (y2 - y1) / (x2 - x1) m = (6 - 0) / (2 - 0) = 6 / 2 = 3 Ans: Slope is 3.

(ii) What is the y-intercept of the line? Explain its significance.

Line passes through origin (0,0). y-intercept = 0. Ans: Tank starts empty at time 0.

(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?

(a) Put x = 5 in y = 3x: y = 3(5) = 15 cm. (b) Put y = 21 in y = 3x: 21 = 3x => x = 7 mins.

(iv) Without drawing a new graph, determine whether the point (4, 10) lies on this line. Justify your answer.

Put x = 4 in y = 3x: 3(4) = 12. 12 ≠ 10. Ans: Point (4, 10) does not lie on the line.

(v) Plot the point where this line intersects the line x = 3. Explain how you found the coordinates.

Put x = 3 in y = 3x: y = 3(3) = 9. Ans: Coordinates are (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 = 9/5 C + 32

(i) Draw the graph of the linear equation above using Celsius on the x-axis and Fahrenheit on the y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

F = 9/5(30) + 32 = 9(6) + 32 = 54 + 32 = 86°F.

(iii) If the temperature is 95°F, what is the temperature in Celsius?

95 = 9/5 C + 32 => 63 = 9/5 C => C = (63*5)/9 = 35°C.

(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?

If C = 0: F = 9/5(0) + 32 = 32°F. If F = 0: 0 = 9/5 C + 32 => 9/5 C = -32 => C = -160/9 ≈ -17.8°C.

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Let F = C = x. x = 9/5 x + 32 x - 9/5 x = 32 => -4/5 x = 32 x = (32 * 5) / -4 = -40 Ans: Yes, -40°.

Question 3: Solve the following system of equations: 2x + y = 6, 2x - y - 2 = 0

2x + y = 6 ...(i) 2x - y = 2 ...(ii) Add (i) & (ii): 4x = 8 => x = 2 Put x=2 in (i): 2(2) + y = 6 => 4 + y = 6 => y = 2 Ans: x = 2, y = 2.

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

Cover page equation: 60x + 50y = 280. It is a single linear equation, so it represents one line with infinitely many points. Ans: No unique point of intersection exists without a second line.

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

For x-intercept, y = 0. 0 = mx + c => mx = -c => x = -c/m Ans: x = -c/m.

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.

Let minutes = x. Cost A = 50 + 0.20x. Cost B = 30 + 0.30x. Cost A = Cost B: 50 + 0.20x = 30 + 0.30x 20 = 0.10x => x = 200 mins (Costs are same). If x > 200, Plan A is cheaper. If x < 200, Plan B is cheaper.

Question 7: How many lines exist that (i) have a given slope? (ii) have a given slope and pass through a given point?

(i) Infinitely many lines (parallel lines with different y-intercepts). (ii) Exactly one unique line.

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

Unique solution condition: a1/a2 ≠ b1/b2 4/2 ≠ p/2 2 ≠ p/2 => p ≠ 4 Ans: All real values 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

Infinite solutions condition: a1/a2 = b1/b2 = c1/c2 (a+b)/3 = -2b/-1 = (5a+2b+1)/14 From first two: (a+b)/3 = 2b => a+b = 6b => a = 5b From last two: 2b = (5a+2b+1)/14 => 28b = 5a + 2b + 1 Substitute a=5b: 28b = 5(5b) + 2b + 1 => 28b = 27b + 1 => b = 1 a = 5(1) = 5 Ans: a = 5, 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

Coincident lines condition: a1/a2 = b1/b2 = c1/c2 1/(k-1) = 2/(k+1) = 3/(k+3) Take first two: 1/(k-1) = 2/(k+1) => k+1 = 2(k-1) k+1 = 2k - 2 => k = 3 Verify with third ratio: 3/(3+3) = 3/6 = 1/2. Matches 1/(3-1) = 1/2. Ans: k = 3.

Question 11: Robot Paths.

(i) Robot 1 starts from origin, traces path moving 3 right, 4 up. Robot 2 starts from (10, 0), moves 1 right, 2 up. Will paths intersect?

Robot 1: Point (0,0), m = 4/3. Eq: y = 4/3 x => 4x - 3y = 0. Robot 2: Point (10,0), m = 2/1 = 2. Eq: y - 0 = 2(x - 10) => y = 2x - 20. Substitute y: 4x - 3(2x - 20) = 0 => 4x - 6x + 60 = 0 => 2x = 60 => x = 30. y = 2(30) - 20 = 40. Ans: Yes, at (30, 40).

(ii) Robot 1 starts from (3, 0), moves 5 right, 5 up. Robot 2 starts from (7, 0), moves 5 right, 3 downwards. Will they intersect?

Robot 1: Point (3,0), m = 5/5 = 1. Eq: y = 1(x - 3) => y = x - 3. Robot 2: Point (7,0), m = -3/5. Eq: y = -3/5(x - 7) => 5y = -3x + 21. Substitute y: 5(x - 3) = -3x + 21 => 5x - 15 = -3x + 21 => 8x = 36 => x = 4.5. y = 4.5 - 3 = 1.5. Ans: Yes, at (4.5, 1.5).

Question 12: At a certain time, Jacob notices his watch reads 'a' minutes after two o'clock. 15 minutes later, it reads 'b' minutes after three o'clock. a is six times greater than b. What time was it when he looked at his watch for the second time?

Time 1: 2:00 + a minutes. Time 2: 3:00 + b minutes (or 2:00 + 60 + b minutes). Time difference is 15 mins: (60 + b) - a = 15 => a - b = 45. Given: a = 6b. Substitute a: 6b - b = 45 => 5b = 45 => b = 9. Time 2 is b minutes after 3:00. Ans: 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.

Let tens = x, units = y. Orig = 10x + y. x + y = 15 ...(i) Reversed (10y + x) - Orig (10x + y) = 9 => 9y - 9x = 9 => y - x = 1 ...(ii) Add (i) & (ii): 2y = 16 => y = 8 x + 8 = 15 => x = 7 Ans: 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.

Opposite angles sum to 180°: A + C = 180 and B + D = 180. (x+7) + (3y+23) = 180 => x + 3y = 150 ...(i) (y+8) + (4x+12) = 180 => 4x + y = 160 ...(ii) From (ii): y = 160 - 4x. Substitute in (i): x + 3(160 - 4x) = 150 => x + 480 - 12x = 150 => 11x = 330 => x = 30. y = 160 - 4(30) = 40. ∠A = 30+7 = 37°. ∠B = 40+8 = 48°. ∠C = 3(40)+23 = 143°. ∠D = 4(30)+12 = 132°. Ans: 37°, 48°, 143°, 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.

Let speed = v, time = t. Distance D = vt. Case 1: (v+6)(t-6) = vt => vt - 6v + 6t - 36 = vt => t - v = 6 => t = v + 6 ...(i) Case 2: (v-4)(t+6) = vt => vt + 6v - 4t - 24 = vt => 3v - 2t = 12 ...(ii) Substitute (i) in (ii): 3v - 2(v+6) = 12 => 3v - 2v - 12 = 12 => v = 24 km/h. t = 24 + 6 = 30 hours. D = 24 * 30 = 720 km. Ans: Speed = 24 km/h, Distance = 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.

Let Father's age = F. Sum of 4 children's ages = C. F = C ...(i) After 20 years: Father = F + 20. Sum of children = C + 4(20) = C + 80. C + 80 = 2(F + 20) ...(ii) Substitute C=F: F + 80 = 2F + 40 => F = 40. Ans: Father's age is 40.

Master Workbook: Chapter 13

Exercise Set 13.1

Exercise Set 13.2

Exercise Set 13.3

Exercise Set 13.4

Exercise Set 13.5

End-of-Chapter Exercises