Exercise Set 13.1
Question 1
Write a linear equation in two variables in which a = 3, b = 0, and c = -1/5.
Solution:
The standard form of a linear equation in two variables is:
ax + by + c = 0
Substituting the given values:
a = 3
b = 0
c = -1/5
3x + 0y + (-1/5) = 0
3x – 1/5 = 0
Multiplying the entire equation by 5 to clear the denominator:
15x – 1 = 0 (or written as 15x + 0y – 1 = 0)
Question 2
Complete the following table after expressing the given linear equations in standard form.
| Linear equation in two variables | Standard Form | Coefficient of x | Coefficient of y | Constant term |
|---|---|---|---|---|
| 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 |
Step-by-step Explanation:
- y – 15 = √2x: Rearranging to standard form ax + by + c = 0 gives √2x – y + 15 = 0. Here, a = √2, b = -1, c = 15.
- 3y – 2x = 0: Rearranging gives -2x + 3y + 0 = 0. Here, a = -2, b = 3, c = 0.
- 5x = 3y: Rearranging gives 5x – 3y + 0 = 0. Here, a = 5, b = -3, c = 0.
- x = 8: Rearranging gives 1x + 0y – 8 = 0. Here, a = 1, b = 0, c = -8.
- 3y = 1: Rearranging gives 0x + 3y – 1 = 0. Here, a = 0, b = 3, c = -1.
Question 3
(i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to be Rs t and that of a pen to be Rs p. Charlie wrote t = 2p, whereas Meera wrote p = 2t. Which of these two representations is correct?
Solution:
According to the problem statement:
Cost of a notebook = t
Cost of a pen = p
Given that “Cost of a notebook is twice the cost of a pen”:
Cost of notebook = 2 * (Cost of pen)
t = 2p
Therefore, Charlie’s representation (t = 2p) is correct.
(ii) In a one-day International Cricket match between India and Sri Lanka played in Nagpur, two Indian batsmen together scored 176 runs. Manisha expressed this situation as x + y = 176, where the number of runs scored by one batsman is x, and the number of runs scored by the other is y. Is this a correct representation?
Solution:
Let the runs scored by the first batsman = x
Let the runs scored by the second batsman = y
Total runs scored by both batsmen together = 176
So, x + y = 176.
Therefore, Yes, Manisha’s representation is correct.
Exercise Set 13.2
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
Ordered pair: (x, y) = (4, 3)
Substituting x = 4 and y = 3 into the Left Hand Side (LHS):
LHS = 5(4) – 6(3) = 20 – 18 = 2
Since LHS = RHS = 2, the ordered pair (4, 3) satisfies the equation.
Conclusion: Yes, (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
- Let x = 0: 7(0) – 3y = 21 => -3y = 21 => y = -7. Solution: (0, -7)
- Let y = 0: 7x – 3(0) = 21 => 7x = 21 => x = 3. Solution: (3, 0)
(ii) 2x + 3y = 5
- Let x = 1: 2(1) + 3y = 5 => 3y = 3 => y = 1. Solution: (1, 1)
- Let x = 4: 2(4) + 3y = 5 => 8 + 3y = 5 => 3y = -3 => y = -1. Solution: (4, -1)
Question 3
In the equations given below, m and n are unknown constants:
2mx + 3y = 7
4x + ny = -10
If (2, -1) is the solution of both equations, find the values of m and n.
Solution:
Substitute x = 2 and y = -1 into both equations.
- 2mx + 3y = 7:
2m(2) + 3(-1) = 7 => 4m – 3 = 7 => 4m = 10 => m = 2.5 - 4x + ny = -10:
4(2) + n(-1) = -10 => 8 – n = -10 => -n = -18 => n = 18
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 in which the points lie.
(i) 5x + 3y = 7
- Point (2, -1) lies in Quadrant IV.
- Point (-1, 4) lies in Quadrant II.
(ii) 5x – 3y = 7
- Point (2, 1) lies in Quadrant I.
- Point (-1, -4) lies in Quadrant III.
(iii) -5x + 3y = 7
- Point (1, 4) lies in Quadrant I.
- Point (-2, -1) lies in Quadrant III.
(iv) -5x – 3y = 7
- Point (1, -4) lies in Quadrant IV.
- Point (-2, 1) lies in Quadrant II.
Question 5
Consider the graph of the equation 3x – 7y = 21. 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:
- Does C(2, 3) lie on the line? No, point C(2, 3) is not on the line.
- Does it satisfy the equation? LHS = 3(2) – 7(3) = 6 – 21 = -15. Since -15 ≠ 21, no.
- Can points that do not lie on the line satisfy the equation? No, only points on the graph line satisfy the linear 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. False. It has infinitely many solutions.
- (ii) The graph of a linear equation in two variables always passes through the origin. False. It passes through (0, 0) only if c = 0.
- (iii) A linear equation in two variables can never have rational solutions. False. It can have rational solutions, e.g., (1/2, 1/2) for x + y = 1.
- (iv) x = 3 is a valid linear equation in two variables. True. It can be represented as 1x + 0y = 3.
- (v) The equation 2x + 3y = 7 has infinitely many solutions. True. Any linear equation in two variables has infinite real solutions.
- (vi) The point (1, 2) is a solution of the equation 2x + 3y = 7. False. 2(1) + 3(2) = 8 ≠ 7.
Exercise Set 13.3
Question 1
Rank the hills in order of their steepness, from least to greatest.
Hill A: Rise = 60 m, Run = 70 m (Slope ≈ 0.857)
Hill B: Rise = 60 m, Run = 110 m (Slope ≈ 0.545)
Hill C: Rise = 80 m, Run = 100 m (Slope = 0.800)
Ranking: 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 = Rise / Run = 2.5 / 4 = 0.625
Answer: Slope = 5/8 or 0.625.
Question 3
Find the slope of the line l in each of the following diagrams.
- (i) Horizontal line: Slope = 0.
- (ii) Sloping downwards: Points (0, 0) and (4, -4) => Slope = -1.
- (iii) Sloping upwards: Points (-3, -1) and (1, 1) => Slope = 1/2.
- (iv) Vertical line: Slope is undefined.
Question 4
Guidelines for a ramp: 1 cm vertical rise to 12 cm horizontal length. What is the horizontal length if height is 18 cm?
Solution:
1/12 = 18/x => x = 18 * 12 = 216 cm.
Answer: 216 cm.
Exercise Set 13.4
Question 1
Construct linear equations for:
- (i) 2 movie tickets and 3 snack boxes for Rs 850: 2x + 3y = 850
- (ii) 4 movie tickets and 1 snack box for Rs 1100: 4x + y = 1100
Question 2
Taxi charges (fixed x, per km y):
- (i) Sahil: 6 km, total Rs 122 => x + 6y = 122
- (ii) Meena: 8 km, total Rs 160 => x + 8y = 160
End-of-Chapter Exercises
Question 1: Water Tank Problem (y = 3x)
- (i) Slope: (6-0)/(2-0) = 3.
- (ii) Y-intercept: 0. Initial water level is 0 cm.
- (iii) 5 minutes: 15 cm. For 21 cm: 7 minutes.
- (iv) (4, 10): No, because 3(4) = 12 ≠ 10.
- (v) Intersection with x = 3: (3, 9).
Question 2: Temperature Conversion (F = 9/5C + 32)
- (ii) 30°C in F: 86°F.
- (iii) 95°F in C: 35°C.
- (iv) 0°C is 32°F; 0°F is -17.78°C.
- (v) Numerically same: -40°.
Question 3: Graphical Solution
Equations: 2x + y = 6 and 2x – y = 2.
Both lines intersect at (2, 2).
Answer: x = 2, y = 2.
Question 12: Jacob’s Watch
- Time 1 = 2:a, Time 2 = 3:b.
- (60 + b) – a = 15 => a – b = 45.
- Given a = 6b => 5b = 45 => b = 9.
Answer: Second time is 3:09.
Question 13: Digit Problem
- x + y = 15
- (10y + x) – (10x + y) = 9 => y – x = 1.
- Solving: x = 7, y = 8.
Answer: The number is 78.
Question 14: Cyclic Quadrilateral
- (x + 7) + (3y + 23) = 180 => x + 3y = 150
- (y + 8) + (4x + 12) = 180 => 4x + y = 160
- Solving: x = 30, y = 40.
Angles: A=37°, B=48°, C=143°, D=132°.
Question 15: Train Distance
- (s + 6)(t – 6) = st => -s + t = 6
- (s – 4)(t + 6) = st => 6s – 4t = 24
- Solving: s = 24 km/h, t = 30 hours.
Answer: Speed = 24 km/h, Distance = 720 km.
Question 16: Father’s Age
- F = C
- (C + 80) = 2(F + 20) => F + 80 = 2F + 40 => F = 40.
Answer: Father is 40 years old.
Exercise Set 13.5
| Problem Type | Scenario | Solution / Equation |
|---|---|---|
| Integers | Sum = 5, Diff = -21 | -8 and 13 |
| Numbers | Diff = 26, x = 3y | 39 and 13 |
| Coach | 7 bats, 6 balls = 8880; 3 bats, 5 balls = 4000 | Bat: Rs 1200, Ball: Rs 80 |
| Taxi | 10km = 155, 15km = 220 | Fixed: Rs 25, Rate: Rs 13/km |
| Fraction | Add 2 to both = 9/11; Add 3 = 5/6 | 7/9 |
| Ages | Nuri & Sonu (5 yrs ago/10 yrs later) | Nuri: 50, Sonu: 20 |
| Digits | Sum = 9, 9 * Original = 2 * Reversed | 18 |
| Currency | Rs 2000 in 50s and 100s (Total 25) | 10 (50s), 15 (100s) |
| Library | Fixed (3 days) + Daily (Extra) | Fixed: Rs 15, Daily: Rs 3 |
Question 9: Height of the Stool
From the equations:
Stool + Cat1 – Cat2 = 85
Stool + Cat2 – Cat1 = 25
Adding them: 2 * Stool = 110 => Stool = 55.
Answer: The height of the stool is 55 cm.
