NCERT Solutions for Class 9 Maths Chapter 2 — End-of-Chapter Exercises#
These problems combine everything from the chapter: writing polynomials of a chosen degree, evaluating polynomials, translating word problems into linear equations, and graphing $y = a x + b$. Key formula used often: for a line through two points $(x_1, y_1)$ and $(x_2, y_2)$, slope $a = \dfrac{y_2 - y_1}{x_2 - x_1}$.
1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7. #
Solution (answers vary)
Any polynomial whose highest power of $x$ is 3 and whose $x^2$ term is $-7x^2$ works. For example:
$$p(x) = x^{3} - 7x^{2} + 2x + 5$$2. Find the values of the following polynomials at the indicated values of the variables. #
(i) 5x² − 3x + 7 if x = 1
We find the value of $5x^{2} - 3x + 7$ at $x = 1$:
$$= 5(1)^{2} - 3(1) + 7$$$$= 5 - 3 + 7$$
$$= 9$$
∴ The value is 9.
(ii) 4t³ − t² + 6 if t = a
Putting $t = a$ in $4t^{3} - t^{2} + 6$, the value is:
$$4a^{3} - a^{2} + 6$$∴ The value is $4a^{3} - a^{2} + 6$.
3. If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number. #
Solution
Let the number be $x$.
Multiplying it by $\dfrac{5}{2}$ and adding $\dfrac{2}{3}$ gives $-\dfrac{7}{12}$, so:
$$\dfrac{5}{2}x + \dfrac{2}{3} = -\dfrac{7}{12}$$$$\Rightarrow \dfrac{5}{2}x = -\dfrac{7}{12} - \dfrac{2}{3}$$
$$\Rightarrow \dfrac{5}{2}x = -\dfrac{7}{12} - \dfrac{8}{12}$$
$$\Rightarrow \dfrac{5}{2}x = -\dfrac{15}{12}$$
$$\Rightarrow \dfrac{5}{2}x = -\dfrac{5}{4}$$
$$\Rightarrow x = -\dfrac{5}{4} \times \dfrac{2}{5}$$
$$\Rightarrow x = -\dfrac{1}{2}$$
∴ The number is $-\dfrac{1}{2}$.
4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? #
Solution
Let the smaller number be $x$, so the larger number is $5x$.
After adding 21 to both, the larger new number is twice the smaller new number, so:
$$5x + 21 = 2(x + 21)$$$$\Rightarrow 5x + 21 = 2x + 42$$
$$\Rightarrow 3x = 21$$
$$\Rightarrow x = 7$$
Larger number $= 5 \times 7 = 35$
Check: $7 + 21 = 28$ and $35 + 21 = 56$, which is $2 \times 28$ ✓
(The other way round, $x + 21 = 2(5x + 21)$, gives $x = -\dfrac{7}{3}$, but the numbers are positive, so that case is ruled out.)
∴ The numbers are 7 and 35.
5. If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern. #
You start with ₹800 and save ₹250 every month, so after $m$ months you have:
$$A(m) = 800 + 250m$$This is the linear pattern.
(i) After 6 months
We find $A(6)$:
$$= 800 + 250(6)$$$$= 800 + 1500$$
$$= 2300$$
∴ After 6 months you have ₹2300.
(ii) After 2 years
2 years is 24 months. We find $A(24)$:
$$= 800 + 250(24)$$$$= 800 + 6000$$
$$= 6800$$
∴ After 2 years you have ₹6800.
*6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers. #
Solution
Let the tens digit be $a$ and the units digit be $b$. The number is $10a + b$, and with the digits interchanged it is $10b + a$.
The two numbers add up to 143, so:
$$(10a + b) + (10b + a) = 143$$$$\Rightarrow 11a + 11b = 143$$
$$\Rightarrow a + b = 13 \quad \cdots (1)$$
The digits differ by 3, and we take the tens digit to be the larger one, so:
$$a - b = 3 \quad \cdots (2)$$Adding equation (1) and equation (2):
$$\begin{array}{rcl} a + b &=& 13 \\ a - b &=& 3 \\ \hline 2a &=& 16 \end{array}$$$$\Rightarrow a = 8$$
Substituting $a = 8$ in equation (1):
$$a + b = 13$$$$8 + b = 13$$
$$\Rightarrow b = 5$$
So the number is 85. If the units digit is the larger one instead, the digits swap places and the number is 58.
Check: $85 + 58 = 143$ ✓
∴ The two numbers are 85 and 58.
*7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis. (i) y = −3x + 4 (ii) 2y = 4x + 7 (iii) 5y = 6x − 10 (iv) 3y = 6x − 11. Are any of the lines parallel? #
Solution
We write each equation in the form $y = ax + b$, where $a$ is the slope and $b$ the y-intercept.
(i) $y = -3x + 4$ is already in this form.
(ii) Dividing $2y = 4x + 7$ by 2:
$$y = 2x + \dfrac{7}{2}$$(iii) Dividing $5y = 6x - 10$ by 5:
$$y = \dfrac{6}{5}x - 2$$(iv) Dividing $3y = 6x - 11$ by 3:
$$y = 2x - \dfrac{11}{3}$$| Equation | Slope $a$ | y-intercept $b$ | Cuts the y-axis at |
|---|---|---|---|
| (i) $y = -3x + 4$ | $-3$ | $4$ | $(0, 4)$ |
| (ii) $y = 2x + \dfrac{7}{2}$ | $2$ | $\dfrac{7}{2}$ | $\left(0, \dfrac{7}{2}\right)$ |
| (iii) $y = \dfrac{6}{5}x - 2$ | $\dfrac{6}{5}$ | $-2$ | $(0, -2)$ |
| (iv) $y = 2x - \dfrac{11}{3}$ | $2$ | $-\dfrac{11}{3}$ | $\left(0, -\dfrac{11}{3}\right)$ |
Comparing the slopes of lines (ii) and (iv):
$$\Rightarrow a_{\text{(ii)}} = a_{\text{(iv)}} = 2$$∴ $a_{\text{(ii)}} = a_{\text{(iv)}}$ and their y-intercepts are different, so lines (ii) and (iv) are parallel. No other pair of lines has equal slopes.
*8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = (9/5)(x − 273) + 32. #
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
We find $y$ when $x = 313$:
$$= \dfrac{9}{5}(313 - 273) + 32$$$$= \dfrac{9}{5}(40) + 32$$
$$= 72 + 32$$
$$= 104$$
∴ The temperature is 104 °F.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
The temperature is 158 °F, so:
$$158 = \dfrac{9}{5}(x - 273) + 32$$$$\Rightarrow 126 = \dfrac{9}{5}(x - 273)$$
$$\Rightarrow x - 273 = 126 \times \dfrac{5}{9}$$
$$\Rightarrow x - 273 = 70$$
$$\Rightarrow x = 343$$
∴ The temperature is 343 K.
*9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph. #
Solution
Work $=$ force $\times$ distance, and the force is 3 units, so:
$$w = 3d$$| $d$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $w$ | 0 | 3 | 6 | 9 |
When the distance is 2 units:
$$w = 3 \times 2 = 6$$The point $(2, 6)$ lies on the line, which verifies this on the graph.
∴ The work done is 6 units.
*10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). #
(i) Find the polynomial p(x).
Let $p(x) = ax + b$.
The graph passes through $(1, 5)$, so:
$$a + b = 5 \quad \cdots (1)$$The graph passes through $(3, 11)$, so:
$$3a + b = 11 \quad \cdots (2)$$Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 3a + b &=& 11 \\ a + b &=& 5 \\ \hline 2a &=& 6 \end{array}$$$$\Rightarrow a = 3$$
Substituting $a = 3$ in equation (1):
$$a + b = 5$$$$3 + b = 5$$
$$\Rightarrow b = 2$$
∴ $p(x) = 3x + 2$.
(ii) Find the coordinates where the graph of p(x) cuts the axes.
The graph cuts the y-axis where $x = 0$:
$$p(0) = 3(0) + 2 = 2$$The graph cuts the x-axis where $p(x) = 0$, so:
$$3x + 2 = 0$$$$\Rightarrow 3x = -2$$
$$\Rightarrow x = -\dfrac{2}{3}$$
∴ The graph cuts the y-axis at $(0, 2)$ and the x-axis at $\left(-\dfrac{2}{3}, 0\right)$.
(iii) Draw the graph of p(x) and verify your answers.
The line passes through both given points, $(1, 5)$ and $(3, 11)$, and cuts the axes at $(0, 2)$ and $\left(-\dfrac{2}{3}, 0\right)$, which verifies the answers.
*11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: (i) p(0) = 5. (ii) The polynomial p(x) − q(x) cuts the x-axis at (3, 0). (iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x. Find the polynomials p(x) and q(x). #
Solution
From condition (i), $p(0) = 5$, so:
$$a(0) + b = 5$$$$\Rightarrow b = 5$$
From condition (iii), $p(x) + q(x) = (a + c)x + (b + d)$ equals $6x + 4$ for all $x$. The coefficients must match, so:
$$a + c = 6 \quad \cdots (1)$$$$b + d = 4 \quad \cdots (2)$$
Substituting $b = 5$ in equation (2):
$$b + d = 4$$$$5 + d = 4$$
$$\Rightarrow d = -1$$
From condition (ii), $p(x) - q(x) = (a - c)x + (b - d)$ is 0 when $x = 3$, so:
$$3(a - c) + (b - d) = 0$$$$\Rightarrow 3(a - c) + (5 - (-1)) = 0$$
$$\Rightarrow 3(a - c) + 6 = 0$$
$$\Rightarrow 3(a - c) = -6$$
$$\Rightarrow a - c = -2 \quad \cdots (3)$$
Adding equation (1) and equation (3):
$$\begin{array}{rcl} a + c &=& 6 \\ a - c &=& -2 \\ \hline 2a &=& 4 \end{array}$$$$\Rightarrow a = 2$$
Substituting $a = 2$ in equation (1):
$$a + c = 6$$$$2 + c = 6$$
$$\Rightarrow c = 4$$
Check: $p(3) = 11$ and $q(3) = 11$, so $p(3) - q(3) = 0$ ✓
∴ $p(x) = 2x + 5$ and $q(x) = 4x - 1$.
*12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage. #
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
Stage 1 is one hexagon of 6 matchsticks. Each new hexagon shares one side with the previous one, so it needs only 5 more matchsticks.
| Stage | Matchsticks |
|---|---|
| 3 | 16 |
| 4 | 21 |
| 5 | 26 |
∴ Stage 4 needs 21 matchsticks and Stage 5 needs 26 matchsticks.
(ii) Complete the following table.
| Stage number ($n$) | 1 | 2 | 3 | 4 | 5 | … | $n$ |
|---|---|---|---|---|---|---|---|
| Number of matchsticks | 6 | 11 | 16 | 21 | 26 | … | $5n + 1$ |
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
Stage 1 uses 6 matchsticks and every later stage adds 5, so stage $n$ uses $6 + 5(n - 1)$ matchsticks. We simplify $6 + 5(n - 1)$:
$$= 6 + 5n - 5$$$$= 5n + 1$$
∴ Stage $n$ needs $5n + 1$ matchsticks.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
We find $5n + 1$ when $n = 15$:
$$= 5(15) + 1$$$$= 75 + 1$$
$$= 76$$
∴ The 15th stage needs 76 matchsticks.
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
If 200 matchsticks formed stage $n$, then:
$$5n + 1 = 200$$$$\Rightarrow 5n = 199$$
$$\Rightarrow n = 39.8$$
A stage number must be a whole number, and 39.8 is not.
∴ No, 200 matchsticks cannot form a stage. Stage 39 uses 196 matchsticks and stage 40 uses 201.
*13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: (i) The graph of p(x) passes through the points (2, 3) and (6, 11). (ii) The graph of q(x) passes through the point (4, −1). (iii) The graph of q(x) is parallel to the graph of p(x). Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis. #
Solution
From condition (i), the graph of $p(x)$ passes through $(2, 3)$ and $(6, 11)$, so:
$$2a + b = 3 \quad \cdots (1)$$$$6a + b = 11 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 6a + b &=& 11 \\ 2a + b &=& 3 \\ \hline 4a &=& 8 \end{array}$$$$\Rightarrow a = 2$$
Substituting $a = 2$ in equation (1):
$$2a + b = 3$$$$2(2) + b = 3$$
$$\Rightarrow b = -1$$
So $p(x) = 2x - 1$.
From condition (iii), $q(x)$ is parallel to $p(x)$, so it has the same slope:
$$c = 2$$From condition (ii), the graph of $q(x)$ passes through $(4, -1)$, so:
$$4c + d = -1$$$$4(2) + d = -1$$
$$\Rightarrow d = -9$$
So $q(x) = 2x - 9$.
The line $y = p(x)$ meets the x-axis where $p(x) = 0$, so:
$$2x - 1 = 0$$$$\Rightarrow x = \dfrac{1}{2}$$
The line $y = q(x)$ meets the x-axis where $q(x) = 0$, so:
$$2x - 9 = 0$$$$\Rightarrow x = \dfrac{9}{2}$$
∴ $p(x) = 2x - 1$ and $q(x) = 2x - 9$. The lines are parallel and meet the x-axis at $\left(\dfrac{1}{2}, 0\right)$ and $\left(\dfrac{9}{2}, 0\right)$.
*14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common? #
Solution
We factorise $ax + a$:
$$= a(x + 1)$$So at $x = -1$, every such function is 0:
$$f(-1) = a(-1 + 1) = 0$$∴ Every line $y = ax + a$ passes through the point $(-1, 0)$, whatever the value of $a$. They also all rise from left to right (since $a > 0$), and each one cuts the y-axis at a height equal to its slope, $(0, a)$.