Ganita Manjari Class 9 Maths Chapter 13 Two Variables, One Line End-of-Chapter Exercises Solutions

We take the two points (0, 0) and (2, 6): the rise is 6 and the run is 2.

The line crosses the y-axis at (0, 0), so the y-intercept is 0. For the tank, the water level is 0 cm at time 0: the tank starts empty.

• After 5 minutes: y = 3 × 5 = 15 cm.
• For 21 cm: 21 = 3x, so x = 7 minutes.

Every point on x = 3 has x-coordinate 3. Put x = 3 in y = 3x to get y = 9. The lines meet at (3, 9).

Take points: C = 0 gives F = 32 and C = 100 gives F = 212. The line through (0, 32) and (100, 212) has slope 95.

0 °C: F = 0 + 32 = 32 °F.

We need F = C:

At −40 the two scales agree: the line meets the line F = C at (−40, −40).

The lines meet at (2, 2). Check: 4 + 2 = 6 ✓ and 4 − 2 − 2 = 0 ✓.

From the book

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One line is labelled 15x − 10y − 115 = 0. Dividing by 5 gives 3x − 2y = 23, that is, y = 32x − 232. It has slope 32, crosses the x-axis at (233, 0) and the y-axis at (0, −11.5). On the cover this is the steep line that goes up to the right.

The second line (nearly horizontal and going slightly down to the right) carries no equation, and the picture has no scale on its axes, so its equation cannot be read exactly from the cover. I have therefore not invented a number for the intersection point.

1. Write the second line as a₂x + b₂y + c₂ = 0.
2. Solve it together with 15x − 10y − 115 = 0 by elimination or substitution (they intersect if 15a₂ ≠ −10b₂).
3. Check the point in both equations.

From the picture, the crossing point is above the x-axis, so on the line 3x − 2y = 23 it has y > 0 and therefore x > 233 ≈ 7.7.

The x-intercept is the point (−cm, 0).

• If m = 0 and c ≠ 0, the line is horizontal and never meets the x-axis: there is no x-intercept.
• If m = 0 and c = 0, the line is the x-axis itself.

At 200 minutes both cost 50 + 40 = ₹90.

Check t = 100: A costs 70, B costs 60 (B cheaper). t = 300: A costs 110, B costs 120 (A cheaper).

Lines of the same slope m are y = mx + d for any d. Every value of d gives a different line, and they are all parallel. So there are infinitely many lines.

If the line must also pass through (x₁, y₁), then y₁ = mx₁ + d fixes d = y₁ − mx₁. So there is exactly one such line, y − y₁ = m(x − x₁).

A unique solution needs a₁a₂ ≠ b₁b₂.

Then the equations are 4x + 4y + 8 = 0 and 2x + 2y + 2 = 0, which gives 4x + 4y + 4 = 0 after doubling the second: parallel lines with no solution.

Infinitely many solutions need a + b3 = −2b−1 = 5a + 2b + 114.

• From a + b3 = 2b: a + b = 6b, so a = 5b.
• From 2b = 5a + 2b + 114: 28b = 5a + 2b + 1, so 26b = 5a + 1.
• Put a = 5b: 26b = 25b + 1, so b = 1 and a = 5.

With a = 5, b = 1 the first equation is 6x − 2y = 28, which is twice 3x − y = 14. ✓

From the first two: k + 1 = 2(k − 1), so k + 1 = 2k − 2, so k = 3.

Check with the third ratio: 3k + 3 = 36 = 12, and 1k − 1 = 12 ✓.

For k = 3 the second equation is 2x + 4y = 6, which is 2 times x + 2y = 3. ✓

The first time is 2 h and a minutes. The second time is 3 h and b minutes. The gap is 15 minutes, which is 60 + b − a.

So 6b = b + 45, so b = 9 and a = 54.

The first time is 2:54, and 15 minutes later it is 3:09. And 54 = 6 × 9 ✓.

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

The second gives 9y − 9x = 9, so y − x = 1. Adding to x + y = 15: 2y = 16, so y = 8 and x = 7.

The number is 78, and reversed it is 87, which is 9 more. 7 + 8 = 15 ✓.

From the second, y = 160 − 4x. Put in the first: x + 480 − 12x = 150, so 11x = 330, so x = 30 and y = 40.

Check: 37 + 143 = 180 ✓, 48 + 132 = 180 ✓ and the total is 360° ✓.

Let the speed be v km/h and the time t hours, so the distance is vt.

• Faster by 6, time 6 less: (v + 6)(t − 6) = vt, which gives −6v + 6t − 36 = 0, so t − v = 6.
• Slower by 4, time 6 more: (v − 4)(t + 6) = vt, which gives 6v − 4t − 24 = 0, so 3v − 2t = 12.

Put t = v + 6 into 3v − 2t = 12: 3v − 2v − 12 = 12, so v = 24 and t = 30.

Check: (24 + 6)(30 − 6) = 30 × 24 = 720 ✓ and (24 − 4)(30 + 6) = 20 × 36 = 720 ✓.

Let the father's age be F and the sum of the children's ages be S.

After 20 years each of the 4 children is 20 years older, so their sum is S + 80, and the father is F + 20.

Replace S by F: F + 80 = 2F + 40, so F = 40.

The children's ages add up to 40. In 20 years: 40 + 80 = 120 and the father is 60, and 120 = 2 × 60 ✓.