CBSE Class 9 Maths Ganita Manjari Part 2 Ch 13 Two Variables One Line Short Answer Questions with Answers

Short Answer Questions

Question 1. Write each of the following in standard form and state its a, b and c.
(i) 5y = 2x
(ii) \( \frac{x}{2} \) + \( \frac{y}{3} \) = 1
(iii) 3 = 7 − 4y
Answer:
(i) 2x − 5y = 0 → a = 2, b = −5, c = 0 (passes through the origin).
(ii) Multiply by 6: 3x + 2y = 6 → 3x + 2y − 6 = 0 → a = 3, b = 2, c = −6.
(iii) 4y = 4 → 0·x + 4y − 4 = 0 → a = 0, b = 4, c = −4 (the line y = 1).
In simple words: Bring everything to one side, clear fractions, and read off a, b, c.

Teacher's Note:
1. a = 0 is allowed if b ≠ 0.
2. Clear fractions with the LCM.

Question 2. Find four different solutions of 3x + 4y = 24, of which one lies on the x-axis, one lies on the y-axis, and one has a negative x-coordinate.
Answer:
x-axis (y = 0): 3x = 24 → (8, 0).
y-axis (x = 0): 4y = 24 → (0, 6).
x = −4: −12 + 4y = 24 → y = 9 → (−4, 9).
x = 4: 12 + 4y = 24 → y = 3 → (4, 3).
Each satisfies 3x + 4y = 24.
In simple words: Choose x values and solve for y.

Teacher's Note:
1. Pick x values that make y a whole number.
2. Verify each pair.

Question 3. Show, by comparing slopes, that the points (1, 2), (3, 8) and (5, 14) all lie on one straight line.
Answer:
(1, 2) to (3, 8): \( \frac{8 - 2}{3 - 1} \) = 3.
(3, 8) to (5, 14): \( \frac{14 - 8}{5 - 3} \) = 3.
Equal slopes and a common point (3, 8), so the three points are collinear.
Check: line y = 3x − 1 gives 3(5) − 1 = 14 ✓.
In simple words: Same slope from a shared point means the same line.

Teacher's Note:
1. Mention the shared point.
2. An equation check adds confidence.

Question 4. Find the slope of the line through (0, −4) and (6, 2), and then write the equation of that line in the form y = mx + d and in standard form.
Answer:
m = \( \frac{2 - (-4)}{6 - 0} \) = 1.
(0, −4) is on the y-axis, so d = −4: y = x − 4.
Standard form: x − y − 4 = 0. Both points satisfy it.
In simple words: Slope 1, crossing the y-axis at −4.

Teacher's Note:
1. A point with x = 0 gives d directly.
2. Check both points.

Question 5. Two points (x₁, y₁) and (x₂, y₂) lie on the line y = mx + d. Show that \( \frac{y_2 - y_1}{x_2 - x_1} \) equals m, and say what this tells you about measuring the slope at different places along the line.
Answer:
y₁ = mx₁ + d and y₂ = mx₂ + d.
Subtracting: y₂ − y₁ = m(x₂ − x₁) (d cancels).
Dividing by x₂ − x₁ (≠ 0): \( \frac{y_2 - y_1}{x_2 - x_1} \) = m.
No special points were used, so any two points give the same slope. The steepness is the same everywhere on the line.
In simple words: Wherever you measure on a straight line, you get the same slope.

Teacher's Note:
1. Point out where d cancels.
2. Explain why x₂ ≠ x₁.

Question 6. Show that 4x − 6y = 10 and 6x − 9y = 15 have exactly the same solutions, and write down three ordered pairs that satisfy both.
Answer:
Multiply the first by 3/2: 6x − 9y = 15, the second equation exactly.
Multiplying by a non-zero number does not change the solutions, so the two have the same solutions (both reduce to 2x − 3y = 5).
Three pairs: (4, 1), (1, −1), (−2, −3).
Check (4, 1): 16 − 6 = 10 and 24 − 9 = 15 ✓.
In simple words: Both equations simplify to 2x − 3y = 5.

Teacher's Note:
1. Reduce both to simplest form.
2. Check at least one pair in both.

Question 7. Solve 2x + 3y = 13 and 5x − y = 7 by substitution twice over: first by using the second equation to write y in terms of x, and then by using the second equation to write x in terms of y. Confirm that the two routes end at the same solution, and say whether the choice of which variable to isolate can change the answer.
Answer:
Route 1: y = 5x − 7. Then 2x + 3(5x − 7) = 13 → 17x = 34 → x = 2, y = 3.
Route 2: x = (y + 7)/5. Then 2(y + 7)/5 + 3y = 13 → 2y + 14 + 15y = 65 → 17y = 51 → y = 3, x = 2.
Both give (2, 3).
The choice cannot change the answer; it only changes how easy the working is. Isolating y avoided fractions.
In simple words: Different routes, same answer; choose the one with fewer fractions.

Teacher's Note:
1. Coefficient −1 or 1 is easiest to isolate.
2. Check in both equations.

Question 8. Solve 5x + 3y = 27 and 2x − 5y = −14 by elimination. Here neither variable can be removed until both equations have been multiplied by something, so state the two multipliers you use and explain how you chose them.
Answer:
To remove y, make the y-coefficients 15 and −15 (LCM of 3 and 5): multiply the first by 5 and the second by 3.
25x + 15y = 135 and 6x − 15y = −42.
Adding: 31x = 93, x = 3. Then 15 + 3y = 27, y = 4.
Solution (3, 4). Check: 6 − 20 = −14 ✓.
In simple words: Multiply each equation by the other's y-coefficient so the y-terms cancel.

Teacher's Note:
1. Explain the choice of multipliers.
2. Opposite signs → add.

Question 9. A problem in the Gaṇita sāra saṅgraha of Mahāvīrācārya leads to the pair 9x + 7y = 107, 7x + 9y = 101. Notice that the coefficients of x and y have been interchanged between the two equations. Use this feature to solve the pair by adding the two equations and by subtracting them, rather than by the usual elimination.
Answer:
Adding: 16x + 16y = 208 → x + y = 13.
Subtracting: 2x − 2y = 6 → x − y = 3.
So 2x = 16, x = 8, and y = 5.
Check: 72 + 35 = 107 and 56 + 45 = 101 ✓. A citron costs 8 and a wood-apple 5.
In simple words: Adding and subtracting swapped-coefficient equations gives a very simple pair.

Teacher's Note:
1. Look for interchanged coefficients.
2. A lovely historical example.

Question 10. The digits of a two-digit number add up to 11, and the number is 45 more than the number obtained by reversing its digits. Find the number.
Answer:
Let the digits be x (tens) and y (units): number 10x + y, reverse 10y + x.
x + y = 11 and (10x + y) − (10y + x) = 45 → 9x − 9y = 45 → x − y = 5.
x = 8, y = 3. The number is 83.
Check: 8 + 3 = 11 and 83 − 38 = 45 ✓.
In simple words: A number minus its reverse is always 9 times the digit difference.

Teacher's Note:
1. Divide by 9 early.
2. Check against the words.

Question 11. A taxi fare consists of a fixed charge together with a charge for each kilometre travelled. A journey of 12 km costs ₹177 and a journey of 18 km costs ₹255. Find the fixed charge and the charge per kilometre, and then find the fare for a journey of 25 km.
Answer:
x + 12y = 177 and x + 18y = 255.
Subtracting: 6y = 78, y = 13. Then x = 177 − 156 = 21.
Fixed charge ₹21, per km ₹13.
25 km: 21 + 25 × 13 = ₹346.
In simple words: The extra 6 km costs ₹78, so each km costs ₹13.

Teacher's Note:
1. Subtract to remove the fixed charge.
2. Verify both given fares.

Question 12. Five chairs and three tables cost ₹8700, while three chairs and five tables cost ₹10 100. Find the cost of one chair and of one table.
Answer:
5x + 3y = 8700 and 3x + 5y = 10 100.
Adding: 8x + 8y = 18 800 → x + y = 2350. Subtracting: 2x − 2y = −1400 → x − y = −700.
x = 825, y = 1525.
Chair ₹825, table ₹1525.
In simple words: Swapped coefficients: add and subtract to get a simple pair.

Teacher's Note:
1. Check: 4125 + 4575 = 8700.
2. Negative x − y just means a table costs more.

Question 13. Half the perimeter of a rectangular field is 46 m, and its length is 8 m more than its width. Find the length and the width.
Answer:
x + y = 46 and x − y = 8.
2x = 54 → x = 27; y = 19.
Length 27 m, width 19 m.
In simple words: Half the perimeter is length + width.

Teacher's Note:
1. Do not double the 46.
2. Check: 27 − 19 = 8.

Question 14. For each pair below, use the ratios of the coefficients to say whether the two lines intersect at a point, are parallel, or are coincident.
(i) 3x − y + 5 = 0 and 6x − 2y + 10 = 0
(ii) x + 2y − 4 = 0 and 2x + 4y − 3 = 0
(iii) 4x + y − 7 = 0 and 3x − 2y + 1 = 0
Answer:
(i) 3/6 = −1/−2 = 5/10 = 1/2: all equal → coincident.
(ii) 1/2 = 2/4 but −4/−3 = 4/3 ≠ 1/2 → parallel.
(iii) 4/3 ≠ 1/(−2) → intersect at one point (at (13/11, 25/11)).
In simple words: Compare a-ratio and b-ratio first; use the c-ratio only if they are equal.

Teacher's Note:
1. Write ratios neatly in a row.
2. Unequal first two ratios settle it at once.

Question 15. Find the value of p for which the pair 4x + py + 8 = 0 and 2x + 2y + 2 = 0 fails to have a unique solution, and state the nature of the solutions for that value of p.
Answer:
Not unique when 4/2 = p/2, so p = 4.
Then the ratios are 2, 2 and 8/2 = 4. First two equal, third different → no solution (parallel lines).
Directly: x + y = −2 and x + y = −1 cannot both hold.
In simple words: At p = 4 the lines are parallel, so there is no solution.

Teacher's Note:
1. Find p from the first two ratios.
2. Then use the third to decide.

Question 16. Find the value of k for which the lines x + 2y = 3 and (k − 1)x + (k + 1)y = k + 3 are coincident.
Answer:
Need \( \frac{1}{k-1} \) = \( \frac{2}{k+1} \) = \( \frac{3}{k+3} \).
From the first two: k + 1 = 2(k − 1) → k = 3.
Check the third: 1/2 = 2/4 = 3/6 ✓. At k = 3 the second equation is 2x + 4y = 6, twice the first.
In simple words: k = 3 makes the second equation exactly double the first.

Teacher's Note:
1. Always test the third ratio.
2. Cross-multiply carefully.

Question 17. Five years ago a father was six times as old as his son. In four years from now he will be three times as old as his son. Find their present ages.
Answer:
Let the father be x and the son y years.
x − 5 = 6(y − 5) → x − 6y = −25; x + 4 = 3(y + 4) → x − 3y = 8.
Subtracting: 3y = 33, y = 11; x = 41.
Father 41 years, son 11 years.
In simple words: Write each age condition as an equation, then subtract.

Teacher's Note:
1. Check: 36 = 6 × 6 and 45 = 3 × 15.
2. Past: subtract; future: add.

Question 18. When 1 is subtracted from the numerator of a certain fraction, the fraction becomes \( \frac{1}{2} \). When 8 is added to its denominator instead, the fraction becomes \( \frac{3}{8} \). Find the fraction.
Answer:
Let the fraction be x/y.
(x − 1)/y = 1/2 → 2x − y = 2; x/(y + 8) = 3/8 → 8x − 3y = 24.
3 × first: 6x − 3y = 6. Subtract: 2x = 18, x = 9, y = 16.
The fraction is \( \frac{9}{16} \).
In simple words: Turn each condition into an equation by cross-multiplying.

Teacher's Note:
1. Check: 8/16 = 1/2 and 9/24 = 3/8.
2. Use the original fraction in the second condition.

Question 19. A customer withdrew ₹3000 from a bank and asked for the money in ₹100 notes and ₹500 notes only. She received 14 notes altogether. Find how many notes of each kind she received.
Answer:
x + y = 14 and 100x + 500y = 3000 → x + 5y = 30.
Subtracting: 4y = 16, y = 4; x = 10.
10 notes of ₹100 and 4 notes of ₹500.
In simple words: Count equation and value equation, then subtract.

Teacher's Note:
1. Divide the value equation by 100.
2. Check: 1000 + 2000 = 3000.

Question 20. Two electricity tariffs are on offer. Tariff A charges a fixed ₹40 a month together with ₹0.50 for each unit used; tariff B charges a fixed ₹100 a month together with ₹0.20 for each unit used. Find the number of units for which the two tariffs cost the same, and state which tariff is cheaper for a household that uses fewer units than that.
Answer:
Equal cost: 40 + 0.50u = 100 + 0.20u → 0.30u = 60 → u = 200 units (₹140 each).
Below 200 units, tariff A is cheaper (e.g. 100 units: A ₹90, B ₹120).
In simple words: A has a lower fixed charge, so it wins for small usage.

Teacher's Note:
1. Test one value on each side.
2. Above 200 units B is cheaper.