NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 13 Two Variables One Line End of Chapter Exercises

End-of-Chapter Exercises

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.
(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) From (0, 0) to (2, 6): run = 2 − 0 = 2, rise = 6 − 0 = 6.
Slope = \( \frac{\text{rise}}{\text{run}} \) = \( \frac{6}{2} \) = 3. This matches m = 3 in y = 3x: every step of 1 to the right lifts the line by 3.
(ii) The line cuts the y-axis at (0, 0), so the y-intercept is 0 (y = 3x + 0).
Meaning: the line passes through the origin. For the tank, at 0 minutes the water level is 0 cm, i.e. the tank was empty when filling started.
(iii) The level rises 3 cm per minute: y = 3x.
(a) x = 5: y = 15. The water is 15 cm high after 5 minutes.
(b) y = 21: 3x = 21, x = 7. It takes 7 minutes.
(iv) At x = 4 the line has y = 3 × 4 = 12, not 10. Since 10 ≠ 12, (4, 10) does not satisfy y = 3x, so it does not lie on the line.
(v) Every point of x = 3 has x-coordinate 3. The common point must also satisfy y = 3x, so y = 3 × 3 = 9.
The lines meet at (3, 9), shown below.

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.
(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) Table (C on the x-axis, F on the y-axis): C = −40, F = −40; C = 0, F = 32; C = 30, F = 86; C = 100, F = 212.
(e.g. C = 100: F = \( \frac{9}{5} \) × 100 + 32 = 180 + 32 = 212.) The graph is a straight line with slope \( \frac{9}{5} \) and y-intercept 32.

(ii) C = 30: F = \( \frac{9}{5} \) × 30 + 32 = 54 + 32 = 86. So 30 °C = 86 °F.
(iii) F = 95: 95 = \( \frac{9}{5} \)C + 32 → \( \frac{9}{5} \)C = 63 → C = 63 × \( \frac{5}{9} \) = 35. So 95 °F = 35 °C.
(iv) C = 0: F = 32, so 0 °C = 32 °F.
F = 0: \( \frac{9}{5} \)C = −32 → C = −32 × \( \frac{5}{9} \) = \( -\frac{160}{9} \) ≈ −17.8. So 0 °F = \( -\frac{160}{9} \) °C ≈ −17.8 °C.
(v) Put F = C: C = \( \frac{9}{5} \)C + 32 → 5C = 9C + 160 → −4C = 160 → C = −40.
Yes: −40 °C = −40 °F. On the graph, this is where the line meets the dashed line F = C.

Teacher's Note:
1. To go from F back to C, undo the steps in reverse order: subtract 32, then multiply by 5/9.
2. Part (v) is the intersection of two lines, F = (9/5)C + 32 and F = C.

Question 3. Solve the following system of equations graphically: 2x + y = 6, 2x − y − 2 = 0.
Answer:
For 2x + y = 6: (0, 6), (3, 0), (1, 4).
For 2x − y = 2: (0, −2), (1, 0), (3, 4).
Plot the points and draw both lines.

The lines cross at (2, 2). Check: 4 + 2 = 6 ✓; 4 − 2 − 2 = 0 ✓.
Solution: x = 2, y = 2 (a unique solution).

Question 4. Find the point of intersection of the lines shown on the cover page.
Answer:
The cover page is not part of the chapter, so the exact lines are not reproduced here. Use the following method with the lines on your book's cover:
1. Read two clear points on each line (for example where it crosses an axis or passes through a grid corner).
2. Find each slope using m = \( \frac{y_2 - y_1}{x_2 - x_1} \) and the y-intercept d, and write each line as y = mx + d (or in standard form).
3. Solve the two equations by substitution or elimination; the solution (x, y) is the point of intersection.
4. Check that this point lies on both lines on the cover.
Illustration: if one line passed through (0, 1) and (1, 3), and the other through (0, 4) and (4, 0), the equations would be y = 2x + 1 and x + y = 4. Then x + 2x + 1 = 4 gives x = 1, y = 3, so they would meet at (1, 3).

Question 5. Give a formula to find the x-intercept of the line y = mx + c.
Answer:
On the x-axis y = 0, so 0 = mx + c → mx = −c → x = \( -\frac{c}{m} \) (m ≠ 0).
The x-intercept is \( -\frac{c}{m} \), and the line cuts the x-axis at (\( -\frac{c}{m} \), 0), provided m ≠ 0.
If m = 0 and c ≠ 0, the line y = c is parallel to the x-axis and has no x-intercept; if m = 0 and c = 0, the line is the x-axis itself.
Example: y = 2x − 6 has x-intercept \( -\frac{-6}{2} \) = 3, and indeed 2(3) − 6 = 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.
Answer:
Let t be the minutes of calls per month and ₹C the monthly cost.
Plan A: C = 50 + 0.20t. Plan B: C = 30 + 0.30t.
Equal cost: 50 + 0.20t = 30 + 0.30t → 20 = 0.10t → t = 200. Both plans then cost 50 + 40 = 30 + 60 = ₹90.
Difference: cost of B − cost of A = 0.10t − 20.
If t > 200, this is positive, so B costs more: Plan A is cheaper for more than 200 minutes (e.g. t = 300: A ₹110, B ₹120).
If t < 200, this is negative, so A costs more: Plan B is cheaper for less than 200 minutes (e.g. t = 100: A ₹70, B ₹60).
At exactly 200 minutes both plans cost ₹90.

Question 7. How many lines exist that (i) have a given slope? (ii) have a given slope and pass through a given point?
Answer:
(i) Infinitely many. Every line with slope m is y = mx + d, and each value of d gives a different line; all of them are parallel. E.g. y = 2x, y = 2x + 1, y = 2x − 5 all have slope 2.
(ii) Exactly one. If y = mx + d must pass through (x₁, y₁), then y₁ = mx₁ + d, so d = y₁ − mx₁. This fixes a single d, hence a single line.
(If the slope is undefined, the only such line is the vertical line x = x₁.)

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.
Answer:
a₁ = 4, b₁ = p, c₁ = 8; a₂ = 2, b₂ = 2, c₂ = 2.
Unique solution needs \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \): \( \frac{4}{2} \) ≠ \( \frac{p}{2} \) → p ≠ 4.
The pair has a unique solution for all real values of p except p = 4.
(At p = 4: \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) = 2 but \( \frac{c_1}{c_2} \) = 4, so the lines are parallel and there is no solution.)

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.
Answer:
Standard form: (a + b)x − 2by − (5a + 2b + 1) = 0 and 3x − y − 14 = 0.
Infinitely many solutions need \( \frac{a+b}{3} \) = \( \frac{-2b}{-1} \) = \( \frac{5a+2b+1}{14} \).
First two ratios: \( \frac{a+b}{3} \) = 2b → a + b = 6b → a = 5b ... (1).
Second and third: 2b = \( \frac{5a+2b+1}{14} \) → 28b = 5a + 2b + 1 → 5a − 26b = −1 ... (2).
Put a = 5b in (2): 25b − 26b = −1 → b = 1, and a = 5.
Check: the first equation becomes 6x − 2y = 28, which is 2 × (3x − y = 14) ✓.
a = 5 and b = 1.

Teacher's Note:
1. Split the three-way equality into two separate equations.
2. Checking that one equation is a multiple of the other confirms the answer.

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.
Answer:
Coincident lines need \( \frac{1}{k-1} \) = \( \frac{2}{k+1} \) = \( \frac{3}{k+3} \).
From the first two: k + 1 = 2(k − 1) → k + 1 = 2k − 2 → k = 3.
Third ratio check at k = 3: \( \frac{1}{2} \), \( \frac{2}{4} \) = \( \frac{1}{2} \), \( \frac{3}{6} \) = \( \frac{1}{2} \), all equal ✓.
(With k = 3 the second equation is 2x + 4y = 6 = 2 × (x + 2y = 3).)
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?
Answer:
After each full move a robot has gone a fixed run and a fixed rise, so its turning points lie on a straight line through its starting point with slope \( \frac{\text{rise}}{\text{run}} \). We model each path by that line.
(i) Robot 1: through (0, 0), slope \( \frac{4}{3} \): y = \( \frac{4}{3} \)x ... (1).
Robot 2: through (10, 0), slope 2: y = 2(x − 10) = 2x − 20 ... (2).
The slopes differ, so the lines meet. \( \frac{4}{3} \)x = 2x − 20 → 4x = 6x − 60 → x = 30, y = 40.
Both robots actually reach (30, 40): Robot 1 is at (3n, 4n) after n moves (n = 10), and Robot 2 at (10 + n, 2n) (n = 20).
Yes, the paths meet at (30, 40).

(Thin lines show the step-by-step zig-zag tracks. Followed exactly, the tracks also touch a few times near this point, the first being (24, 28); the common turning point given by the straight-line model is (30, 40).)
(ii) Robot 1: through (3, 0), slope \( \frac{5}{5} \) = 1: y = x − 3, for x ≥ 3.
Robot 2: through (7, 0), moving right and down, slope \( -\frac{3}{5} \): y = \( -\frac{3}{5} \)(x − 7), for x ≥ 7.
Solving: x − 3 = \( -\frac{3}{5} \)(x − 7) → 5x − 15 = −3x + 21 → 8x = 36 → x = 4.5, y = 1.5.
But Robot 2 starts at x = 7 and only moves right, so it never reaches x = 4.5. Also Robot 1 only goes up (y ≥ 0) while Robot 2 only goes down (y ≤ 0).

So the two paths do not cross: the meeting point (4.5, 1.5) of the lines lies behind Robot 2's start.
(Followed move by move, both robots begin with a horizontal step on the x-axis, Robot 1 from (3, 0) to (8, 0) and Robot 2 from (7, 0) to (12, 0), so these first steps overlap between (7, 0) and (8, 0); after that the paths separate and never meet.)

Teacher's Note:
1. Turn each repeated move into a slope, and each starting point into a point on the line.
2. Always check that the intersection lies ahead of both starting points; in (ii) it does not.
3. Drawing the zig-zag paths shows why the straight-line model is only an approximation.

Question 12. At a certain time, Jacob notices that his digital watch reads 'a' minutes after two o'clock. Fifteen minutes later, it reads 'b' 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?
Answer:
Count all times in minutes after 2:00. First reading: a. Fifteen minutes later: a + 15.
That second reading is b minutes after 3:00, i.e. 60 + b minutes after 2:00. So a + 15 = 60 + b → a − b = 45 ... (1).
Also a = 6b ... (2).
Substituting: 6b − b = 45 → b = 9, a = 54.
Check: 2:54 + 15 minutes = 3:09 ✓ and 54 = 6 × 9 ✓.
The second time he looked, the watch showed 3:09 (the first time was 2:54).

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.
Answer:
Tens digit x, units digit y. Number = 10x + y; reversed = 10y + x.
x + y = 15 ... (1).
(10y + x) − (10x + y) = 9 → 9y − 9x = 9 → y − x = 1 ... (2).
Adding: 2y = 16, y = 8, x = 7. Number = 78.
Check: 7 + 8 = 15 ✓; 87 − 78 = 9 ✓.
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.

Answer:
Opposite angles of a cyclic quadrilateral add up to 180°.
∠A + ∠C = 180: (x + 7) + (3y + 23) = 180 → x + 3y = 150 ... (1).
∠B + ∠D = 180: (y + 8) + (4x + 12) = 180 → 4x + y = 160 ... (2).
(1) × 4: 4x + 12y = 600 ... (3). (3) − (2): 11y = 440, y = 40. From (1): x = 150 − 120 = 30.
∠A = 37°, ∠B = 48°, ∠C = 3 × 40 + 23 = 143°, ∠D = 4 × 30 + 12 = 132°.
Check: 37 + 143 = 180 ✓; 48 + 132 = 180 ✓; total 360° ✓.
∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 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.
Answer:
Let the speed be x km/h and the time t hours; distance = xt km.
Condition 1: (x + 6)(t − 6) = xt → xt − 6x + 6t − 36 = xt → −6x + 6t = 36 → −x + t = 6 ... (1).
Condition 2: (x − 4)(t + 6) = xt → xt + 6x − 4t − 24 = xt → 6x − 4t = 24 → 3x − 2t = 12 ... (2).
From (1): t = x + 6. In (2): 3x − 2x − 12 = 12 → x = 24, t = 30.
Distance = 24 × 30 = 720 km.
Check: at 30 km/h, 720 ÷ 30 = 24 h (6 h less) ✓; at 20 km/h, 720 ÷ 20 = 36 h (6 h more) ✓.
The distance is 720 km and the speed is 24 km/h.

Teacher's Note:
1. The xt terms cancel on both sides, which is what makes the equations linear.
2. Check the answer against the story, not only the equations.

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.
Answer:
Father's present age x years; present sum of the four children's ages y years.
Now: x = y, i.e. x − y = 0 ... (1).
After 20 years the father is x + 20, and each of the 4 children gains 20 years, so their total becomes y + 80.
y + 80 = 2(x + 20) → y + 80 = 2x + 40 → 2x − y = 40 ... (2).
(2) − (1): x = 40, so y = 40.
Check: after 20 years, father 60, children's total 40 + 80 = 120 = 2 × 60 ✓.
The father's age is 40 years.

Teacher's Note:
1. The children's total grows by 4 × 20 = 80, not by 20; this is the usual trap.
2. Check the condition 'after 20 years' with the final numbers.