NCERT Solutions for Class 9 Maths Chapter 13 — End-of-Chapter Exercises#

From $(0, 0)$ to $(2, 6)$, the line rises 6 while running 2:

$$\text{Slope} = \dfrac{6 - 0}{2 - 0}$$

$$\text{Slope} = 3$$

The line crosses the y-axis at $(0, 0)$, so the y-intercept is 0. It is the value of $y$ when $x = 0$ — here, the line passes through the origin.

(a) Putting $x = 5$:

$$y = 3 \times 5 = 15 \text{ cm}$$

(b) Putting $y = 21$:

$$3x = 21$$

$$\Rightarrow x = 7 \text{ minutes}$$

(The y-intercept 0 means the tank was empty at the start, and the slope 3 means the level rises 3 cm every minute.)

On the line, $x = 4$ gives

$$y = 3 \times 4 = 12$$$$\Rightarrow 12 \neq 10$$

∴ $(4, 10)$ does not satisfy $y = 3x$, so it does not lie on the line.

Every point on $x = 3$ has $x$-coordinate 3. Substituting $x = 3$ in $y = 3x$:

$$y = 9$$

∴ The lines meet at $(3, 9)$.

Some points: $(0, 32)$, $(100, 212)$, $(-40, -40)$, $(30, 86)$.

$$F = \dfrac{9}{5} \times 30 + 32$$

$$F = 54 + 32$$

$$F = 86$$

∴ 86 °F.

$$95 = \dfrac{9}{5}C + 32$$

$$\Rightarrow \dfrac{9}{5}C = 63$$

$$\Rightarrow C = 35$$

∴ 35 °C.

$$C = 0: \quad F = 32$$$$F = 0: \quad 0 = \dfrac{9}{5}C + 32$$

$$\Rightarrow C = -\dfrac{160}{9} \approx -17.8$$

∴ 0 °C is 32 °F, and 0 °F is about −17.8 °C.

We need $F = C$:

$$C = \dfrac{9}{5}C + 32$$

$$\Rightarrow -\dfrac{4}{5}C = 32$$

$$\Rightarrow C = -40$$

∴ Yes: −40 °C = −40 °F. On the graph, it is where the line meets the line $F = C$.

Line	Points
$2x + y = 6$	$(0, 6)$, $(3, 0)$, $(1, 4)$
$2x - y - 2 = 0$	$(0, -2)$, $(1, 0)$, $(3, 4)$

The lines meet at $(2, 2)$.

∴ $x = 2$, $y = 2$. (Check: $4 + 2 = 6$ ✓ and $4 - 2 - 2 = 0$ ✓)

The cover image is not included in NCERT’s online chapter PDFs, so here is the method to use with your printed book:

1. For each line, read two points it passes through from the grid on the cover.
2. Find each line’s slope $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ and y-intercept $c$, and write its equation $y = mx + c$.
3. Solve the two equations together (by substitution or elimination). The solution $(x, y)$ is the point of intersection.
4. Check that the point you found lies on both lines in the picture.

The x-intercept is where the line meets the x-axis, that is, where $y = 0$:

$$0 = mx + c$$

$$\Rightarrow x = -\dfrac{c}{m}$$

∴ The x-intercept is $-\dfrac{c}{m}$ (for $m \neq 0$). If $m = 0$ and $c \neq 0$, the line is horizontal and never meets the x-axis.

For $x$ minutes, the monthly costs are:

$$y_A = 50 + 0.2x$$

$$y_B = 30 + 0.3x$$

They are equal when:

$$50 + 0.2x = 30 + 0.3x$$

$$\Rightarrow 20 = 0.1x$$

$$\Rightarrow x = 200$$

Both plans then cost $50 + 0.2 \times 200 = ₹90$.

Plan B starts cheaper (₹30 fee) but rises faster (₹0.30 per minute), so:

• Plan B is cheaper for fewer than 200 minutes;
• Plan A is cheaper for more than 200 minutes;
• they cost the same (₹90) at 200 minutes.

(i) Infinitely many. Lines with the same slope are parallel; for slope $m$, every line $y = mx + c$ (for any $c$) works.

(ii) Exactly one. Through a given point there is only one line parallel to a given direction.

A unique solution needs $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$:

$$\dfrac{4}{2} \neq \dfrac{p}{2}$$

$$\Rightarrow p \neq 4$$

∴ There is a unique solution for every value of $p$ except $p = 4$.

For infinitely many solutions, all three ratios must be equal:

$$\dfrac{a + b}{3} = \dfrac{-2b}{-1} = \dfrac{5a + 2b + 1}{14}$$

From the first two ratios:

$$\dfrac{a + b}{3} = 2b$$

$$\Rightarrow a + b = 6b$$

$$\Rightarrow a = 5b \quad \cdots (1)$$

From the second and third ratios:

$$2b = \dfrac{5a + 2b + 1}{14}$$

$$\Rightarrow 28b = 5a + 2b + 1$$

$$\Rightarrow 5a - 26b = -1 \quad \cdots (2)$$

Substituting $a = 5b$ in equation (2):

$$25b - 26b = -1$$

$$\Rightarrow b = 1$$

From equation (1), $a = 5$.

∴ $a = 5$, $b = 1$. (Check: the first equation becomes $6x - 2y = 28$, which is 2 times $3x - y = 14$ ✓)

For coincident lines:

$$\dfrac{1}{k - 1} = \dfrac{2}{k + 1} = \dfrac{3}{k + 3}$$

From the first two ratios:

$$k + 1 = 2(k - 1)$$

$$\Rightarrow k = 3$$

Check the third ratio with $k = 3$:

$$\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{3}{6}$$

All three are equal to $\dfrac{1}{2}$ ✓.

∴ $k = 3$ (the second equation becomes $2x + 4y = 6$, twice the first).

Each robot traces a “staircase” that follows a straight line through its corner points.

• Robot 1 reaches the corners $(3k, 4k)$, which lie on $y = \dfrac{4}{3}x$ (slope $\dfrac{4}{3}$).
• Robot 2 reaches the corners $(10 + j, 2j)$, which lie on $y = 2(x - 10)$ (slope 2).

The slopes differ, so the lines meet. Solving:

$$\dfrac{4}{3}x = 2x - 20$$

$$\Rightarrow \dfrac{2}{3}x = 20$$

$$\Rightarrow x = 30, \quad y = 40$$

$(30, 40)$ is a corner of both paths: Robot 1 after 10 steps ($k = 10$) and Robot 2 after 20 steps ($j = 20$).

∴ Yes, the paths meet at (30, 40). (Following the staircases exactly, they first touch a little earlier, at the corner $(24, 28)$, and then touch at several points around $(30, 40)$, where their general directions cross.)

The general directions of the two paths, $y = x - 3$ (up) and $y = -\dfrac{3}{5}(x - 7)$ (down), would cross at $(4.5, 1.5)$ — but Robot 2 starts at $x = 7$ and only moves right, so it never gets back to $x = 4.5$.

Look at the actual paths. Both robots begin by moving right along the x-axis: Robot 1 from $(3, 0)$ to $(8, 0)$, and Robot 2 from $(7, 0)$ to $(12, 0)$. After that, Robot 1 only goes up (above the x-axis) and Robot 2 only goes down (below it), so they never meet again.

∴ Yes, the paths overlap — along the stretch of the x-axis from (7, 0) to (8, 0), right at the start.

$a$ minutes after 2:00, plus 15 minutes, is $b$ minutes after 3:00, so:

$$a + 15 - 60 = b$$

$$\Rightarrow a - b = 45 \quad \cdots (1)$$

$a$ is six times $b$, so:

$$a = 6b \quad \cdots (2)$$

Substituting $a = 6b$ in equation (1):

$$6b - b = 45$$

$$\Rightarrow b = 9$$

From equation (2), $a = 54$.

∴ The first time was 2:54, and the second time was 3:09.

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

The sum of the digits is 15, so:

$$x + y = 15 \quad \cdots (1)$$

The reversed number exceeds the number by 9, so:

$$(10y + x) - (10x + y) = 9$$

$$\Rightarrow 9y - 9x = 9$$

$$\Rightarrow y - x = 1 \quad \cdots (2)$$

Adding equation (1) and equation (2):

$$\begin{array}{rcl} x + y &=& 15 \\ -x + y &=& 1 \\ \hline 2y &=& 16 \end{array}$$

$$\Rightarrow y = 8$$

From equation (1), $x = 7$.

∴ The number is 78. (Check: $87 - 78 = 9$ ✓)

Opposite angles of a cyclic quadrilateral add up to 180°.

$\angle A + \angle C = 180°$, so:

$$(x + 7) + (3y + 23) = 180$$

$$\Rightarrow x + 3y = 150 \quad \cdots (1)$$

$\angle B + \angle D = 180°$, so:

$$(y + 8) + (4x + 12) = 180$$

$$\Rightarrow 4x + y = 160 \quad \cdots (2)$$

Multiplying equation (2) by 3:

$$12x + 3y = 480 \quad \cdots (3)$$

Subtracting equation (1) from equation (3):

$$\begin{array}{rcl} 12x + 3y &=& 480 \\ x + 3y &=& 150 \\ \hline 11x &=& 330 \end{array}$$

$$\Rightarrow x = 30$$

Substituting $x = 30$ in equation (2):

$$120 + y = 160$$

$$\Rightarrow y = 40$$

So:

$$\angle A = 30 + 7 = 37°, \quad \angle B = 40 + 8 = 48°$$

$$\angle C = 120 + 23 = 143°, \quad \angle D = 120 + 12 = 132°$$

∴ ∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°. (Check: $37 + 143 = 180$ ✓, $48 + 132 = 180$ ✓)