Welcome to Sid Classes. In this comprehensive master guide, we provide complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 2 End-of-Chapter Exercises. This review contains essential problem sets ranging from polynomial evaluation and word problems to coordinate graphing properties and algebraic matchstick progressions.
Our solutions use clean standard mathematical notation (\(\text{LaTeX}\)) and breakdown logic to ensure you score 100% marks in your school tests and board examinations.
🔑 Chapter Summary Checklist:
- Polynomial Degrees: The degree corresponds to the absolute highest exponent power present within the non-zero variable terms.
- Parallel Trajectories: Linear graphs run completely parallel to one another if and only if their computed slopes (\(a\)) are perfectly identical.
- Arithmetic Progression Rules: Visual geometric patterns addition can be represented algebraically by finding the starting core baseline combined with the repeating unit growth multiplier.
NCERT Class 9 Maths Ganita Manjari Chapter 2 End-of-Chapter Exercise Solutions
Question 1
Write a polynomial of degree 3 in the variable \(x\), in which the coefficient of the \(x^2\) term is \(-7\).
Solution:
A standard polynomial of degree 3 requires its highest exponent power to be \(x^3\). The coefficient of the \(x^2\) term must explicitly be \(-7\). Other terms can vary.
$$\text{Example Expression: } \mathbf{x^3 - 7x^2 + x + 5}$$
Question 2
Find the values of the following polynomials at the indicated values of the variables.
(i) \(5x^2 - 3x + 7\) if \(x = 1\)
(ii) \(4t^3 - t^2 + 6\) if \(t = a\)
Solution:
(i) Substitute \(x = 1\):
$$5(1)^2 - 3(1) + 7 = 5(1) - 3 + 7 = 5 - 3 + 7 = \mathbf{9}$$
(ii) Substitute \(t = a\):
$$4(a)^3 - (a)^2 + 6 = \mathbf{4a^3 - a^2 + 6}$$
Question 3
If we multiply a number by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, we get \(-\frac{7}{12}\). Find the number.
Solution:
Let the unknown target number be \(n\). According to the text constraints:
$$\frac{5}{2}n + \frac{2}{3} = -\frac{7}{12}$$
$$\frac{5}{2}n = -\frac{7}{12} - \frac{2}{3}$$
$$\text{Taking LCM of 12 and 3, which is 12:}$$
$$\frac{5}{2}n = \frac{-7 - 8}{12} \implies \frac{5}{2}n = -\frac{15}{12} = -\frac{5}{4}$$
$$n = -\frac{5}{4} \times \frac{2}{5} = -\frac{2}{4} = \mathbf{-\frac{1}{2}}$$
Final Answer: The required number is \(\mathbf{-\frac{1}{2}}\) (or \(-0.5\)).
Question 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 \(x\). Therefore, the larger number is \(5x\).
Adding 21 to both yields the new expressions: \((x + 21)\) and \((5x + 21)\).
According to the problem context, the larger revised value becomes twice the smaller revised value:
$$5x + 21 = 2(x + 21)$$
$$5x + 21 = 2x + 42$$
$$5x - 2x = 42 - 21$$
$$3x = 21 \implies x = \mathbf{7}$$
Smaller number \(= 7\); Larger number \(= 5(7) = 35\).
Final Answer: The two numbers are 7 and 35.
Question 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.
Solution:
Let \(m\) denote the number of months passed.
$$\text{Linear Formula expression: } \mathbf{\text{Amount} = 800 + 250m}$$
(i) After 6 months (\(m = 6\)):
$$\text{Amount} = 800 + 250(6) = 800 + 1500 = \mathbf{\text{₹}2300}$$
(ii) After 2 years (\(m = 24\text{ months}\)):
$$\text{Amount} = 800 + 250(24) = 800 + 6000 = \mathbf{\text{₹}6800}$$
Question 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 digit at the units place be \(x\) and the digit at the tens place be \(y\).
Original Number \(= 10y + x\); Interchanged Number \(= 10x + y\).
$$\text{Given: } (10y + x) + (10x + y) = 143 \implies 11x + 11y = 143 \implies x + y = 13$$
$$\text{Also given that the digits differ by 3: } y - x = 3 \text{ or } x - y = 3$$
Case 1: If \(y - x = 3\)
Adding \((x + y = 13)\) and \((y - x = 3)\) gives \(2y = 16 \implies y = 8\), so \(x = 5\). Number is 85.
Case 2: If \(x - y = 3\)
Adding \((x + y = 13)\) and \((x - y = 3)\) gives \(2x = 16 \implies x = 8\), so \(y = 5\). Number is 58.
Final Answer: The numbers can be either 85 or 58.
Question 7
Draw the graph of the following equations, identify their slopes and \(y\)-intercepts, and find where they cut the \(y\)-axis.
(i) \(y = -3x + 4\)
(ii) \(2y = 4x + 7 \implies y = 2x + 3.5\)
(iii) \(5y = 6x - 10 \implies y = 1.2x - 2\)
(iv) \(3y = 6x - 11 \implies y = 2x - \frac{11}{3}\)
Are any of the lines parallel?
Solution:
By arranging each line into the general slope intercept form \(y = ax + b\), we isolate values:
- Line (i): Slope \((a) = -3\), \(y\)-intercept \((b) = 4\). Cuts \(y\)-axis at \(\mathbf{(0, 4)}\).
- Line (ii): Slope \((a) = 2\), \(y\)-intercept \((b) = 3.5\). Cuts \(y\)-axis at \(\mathbf{(0, 3.5)}\).
- Line (iii): Slope \((a) = 1.2\), \(y\)-intercept \((b) = -2\). Cuts \(y\)-axis at \(\mathbf{(0, -2)}\).
- Line (iv): Slope \((a) = 2\), \(y\)-intercept \((b) = -\frac{11}{3}\). Cuts \(y\)-axis at \(\mathbf{(0, -\frac{11}{3})}\).
Parallel Status: Lines (ii) and (iv) are parallel because they share the exact same slope value of \(a = 2\).
Question 8
The relation between Kelvin (\(x\)) and Fahrenheit (\(y\)) systems is given by: \(y = \frac{9}{5}(x - 273) + 32\).
(i) Find Fahrenheit if temperature is 313 K.
(ii) If temperature is 158 °F, find the value in Kelvin.
Solution:
(i) Substitute \(x = 313\):
$$y = \frac{9}{5}(313 - 273) + 32 \implies y = \frac{9}{5}(40) + 32$$
$$y = 9(8) + 32 = 72 + 32 = \mathbf{104\text{ °F}}$$
(ii) Substitute \(y = 158\):
$$158 = \frac{9}{5}(x - 273) + 32 \implies 126 = \frac{9}{5}(x - 273)$$
$$126 \times \frac{5}{9} = x - 273 \implies 14 \times 5 = x - 273$$
$$70 = x - 273 \implies x = 70 + 273 = \mathbf{343\text{ K}}$$
Question 9
Express Work (\(w\)) as the product of a constant force (3 units) and distance (\(d\)). Draw its graph and verify work done when distance travelled is 2 units.
Solution:
Linear Equation where force is 3:
$$\mathbf{w = 3d}$$
When distance \(d = 2\text{ units}\):
$$w = 3(2) = \mathbf{6\text{ units of work}}$$
Graph Verification Point: Plot distance on the \(x\)-axis and work on the \(y\)-axis. The coordinate point line maps cleanly through coordinates \((0,0)\), \((1,3)\), and \(\mathbf{(2,6)}\).
Question 10
The graph of a linear polynomial \(p(x) = ax + b\) passes through points \((1, 5)\) and \((3, 11)\).
(i) Find \(p(x)\).
(ii) Find coordinates where it cuts both axes.
Solution:
(i) Find \(p(x)\):
Using point \((1,5) \implies 5 = a(1) + b \implies a + b = 5\)
Using point \((3,11) \implies 11 = a(3) + b \implies 3a + b = 11\)
Subtracting the equations: \(2a = 6 \implies a = 3\); which means \(b = 2\).
Therefore, the polynomial is: \(\mathbf{p(x) = 3x + 2}\).
(ii) Intersections with axes:
* Cuts \(y\)-axis when \(x=0 \implies p(0) = 2 \rightarrow \mathbf{(0, 2)}\)
* Cuts \(x\)-axis when \(p(x)=0 \implies 3x + 2 = 0 \implies x = -\frac{2}{3} \rightarrow \mathbf{(-\frac{2}{3}, 0)}\)
Question 11
Let \(p(x) = ax + b\) and \(q(x) = cx + d\) satisfy: (i) \(p(0) = 5\), (ii) \(p(x) - q(x)\) cuts \(x\)-axis at \((3,0)\), (iii) \(p(x) + q(x) = 6x + 4\). Find \(p(x)\) and \(q(x)\).
Solution:
From condition (i): \(p(0) = b = 5\).
From condition (iii): \((ax+b) + (cx+d) = 6x + 4 \implies (a+c)x + (b+d) = 6x + 4\).
Equating the constants: \(b + d = 4 \implies 5 + d = 4 \implies d = -1\).
Equating coefficients of \(x\): \(a + c = 6\).
From condition (ii): The line \((a-c)x + (b-d)\) has a zero at \(x = 3\):
$$(a-c)(3) + (5 - (-1)) = 0 \implies 3(a-c) + 6 = 0 \implies 3(a-c) = -6 \implies a - c = -2$$
Solving system \(a+c = 6\) and \(a-c = -2\): Adding them gives \(2a = 4 \implies a = 2\), so \(c = 4\).
Final Answer: \(\mathbf{p(x) = 2x + 5}\) and \(\mathbf{q(x) = 4x - 1}\).
Question 12 (Matchstick Progression)
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.
Stage 1 (6 sticks) ⬢⬢
Stage 2 (11 sticks) ⬢⬢⬢
Stage 3 (16 sticks)
Solution:
(i) Draw next stages and stick counts:
Stage 4 adds one more connected hexagon. Count \(= 16 + 5 = \mathbf{21\text{ sticks}}\).
Stage 5 adds another hexagon. Count \(= 21 + 5 = \mathbf{26\text{ sticks}}\).
(ii) Completed Data Table:
| Stage Number | 1 | 2 | 3 | 4 | 5 | ... | \(n\) |
|---|---|---|---|---|---|---|---|
| Number of Matchsticks | 6 | 11 | 16 | 21 | 26 | ... | \(5n + 1\) |
(iii) Find the general rule for \(n^{\text{th}}\) stage:
The pattern starts with 6 sticks and increments by 5 each time. Formula: \(6 + 5(n - 1) = \mathbf{5n + 1}\).
(iv) Matchsticks for 15th stage:
$$\text{Sticks} = 5(15) + 1 = 75 + 1 = \mathbf{76\text{ sticks}}$$
(v) Can 200 matchsticks form a valid stage?
$$5n + 1 = 200 \implies 5n = 199 \implies n = 39.8$$
Since the stage number \(n\) must be a whole counting integer, 200 matchsticks cannot form a stage in this pattern.
Question 13
Graph of \(p(x)\) passes through \((2, 3)\) and \((6, 11)\). Graph of \(q(x)\) passes through \((4, -1)\) and is parallel to \(p(x)\). Find both polynomials and their \(x\)-axis intersection points.
Solution:
First, let's find the slope of \(p(x)\):
$$\text{Slope } a = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$$
Using \((2,3)\) for \(p(x) = 2x + b \implies 3 = 2(2) + b \implies b = -1\). So, \(\mathbf{p(x) = 2x - 1}\).
Since \(q(x)\) is parallel, it shares the same slope (\(a=2\)). Passing through point \((4,-1)\):
$$-1 = 2(4) + d \implies -1 = 8 + d \implies d = -9\text{. So, }\mathbf{q(x) = 2x - 9}\text{.}$$
Intersection points on the \(x\)-axis (where \(y = 0\)):
* For \(p(x)\): \(2x - 1 = 0 \implies x = 0.5 \rightarrow \mathbf{(0.5, 0)}\)
* For \(q(x)\): \(2x - 9 = 0 \implies x = 4.5 \rightarrow \mathbf{(4.5, 0)}\)
Question 14
What do all linear functions of the form \(f(x) = ax + a\), \(a > 0\), have in common?
Solution:
Let's determine the \(x\)-intercept by setting \(f(x) = 0\):
$$ax + a = 0 \implies a(x + 1) = 0$$
Since we are given that \(a > 0\), we can safely divide both sides by \(a\):
$$x + 1 = 0 \implies x = -1$$
Final Answer: All functions of this form share the exact same \(x\)-intercept at the point \((-1, 0)\). No matter how you scale the positive slope value \(a\), the resulting lines will all pivot through that same point on the horizontal axis.
Frequently Asked Questions (FAQs)
Q1: Why does adding a shared hexagon side only add 5 matchsticks instead of 6?
A: The first hexagon uses 6 full sticks. Every subsequent hexagon shares exactly 1 side boundary with the previous one, meaning you only need to add 5 new outer sticks to complete the next loop.
Q2: How do you verify if an equation is parallel without graphing it?
A: Reduce the equations to their slope-intercept forms (\(y = ax+b\)). If the computed coefficients for \(x\) match perfectly, the paths are parallel and will never collide.
Q3: What does an x-intercept point represent?
A: The \(x\)-intercept marks the exact location where a line crosses the horizontal axis, which always occurs when the value of \(y\) (or \(p(x)\)) equals zero.
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