Ch 13 EOC Class 9 Maths Ganita Manjari | Sid Classes

Ganita Manjari Class 9 Ch 13 end-of-chapter solutions: mixed linear equations in two variables, step by step. Sid Classes.

Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 13 End of Chapter (EOC) Exercises. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this comprehensive review walkthrough covers advanced linear graph analysis, temperature conversions, system consistency conditions, mobile tariff comparisons, geometric paths, and word problems entirely clear.

🔑 Key Concepts & Advanced Review:

  • Slope & Intercept Analysis: Calculating slope as rise over run (\(m = \frac{y_2 - y_1}{x_2 - x_1}\)) and interpreting \(y\)-intercepts in real-world contexts.
  • System Consistency & Parameters: Determining unique solutions, parallel lines, and coincident lines using algebraic coefficient ratios.

NCERT Class 9 Maths Ganita Manjari Chapter 13 End of Chapter Solutions

Question 1

The graph of the line \(y = 3x\), passing through \((0, 0)\) and \((2, 6)\) is given.
(i) Identify the slope of the line from the graph and explain how you calculated it using the two points on the line.
(ii) What is the y-intercept of the line? Explain its significance.
(iii) A water tank is being filled so that the water level \(y\) (in cm) after \(x\) minutes follows this graph.
(a) How high is the water after 5 minutes?
(b) How many minutes will it take for the water to reach a height of 21 cm?
(iv) Without drawing a new graph, determine whether the point \((4, 10)\) lies on this line. Justify your answer.
(v) Plot the point where this line intersects the line \(x = 3\). Explain how you found the coordinates.

Solution:

* (i) Slope calculation: Using points \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (2, 6)\), the slope \(m = \frac{6 - 0}{2 - 0} = \mathbf{3}\).
* (ii) y-intercept: The \(y\)-intercept is \(0\), meaning the line passes through the origin \((0, 0)\), signifying that at time \(x = 0\), the water level is \(0\) cm.
* (iii) Water tank calculations:
- (a) After 5 minutes (\(x = 5\)), height \(y = 3(5) = \mathbf{15\text{ cm}}\).
- (b) For height \(y = 21\), \(21 = 3x \implies x = \mathbf{7\text{ minutes}}\).
* (iv) Point check for \((4, 10)\): Substituting \(x = 4\) gives \(y = 3(4) = 12 \neq 10\). Thus, \(\mathbf{\text{it does not lie on the line}}\).
* (v) Intersection with \(x = 3\): Substituting \(x = 3\) into \(y = 3x\) gives \(y = 9\), so the coordinates are \(\mathbf{(3, 9)}\).

Question 2

Temperature conversion: \(F = \frac{9}{5}C + 32$.
(i) Draw the graph using Celsius on x-axis and Fahrenheit on y-axis.
(ii) If temperature is \(30^\circ\text{C}\), find Fahrenheit.
(iii) If temperature is \(95^\circ\text{F}\), find Celsius.
(iv) If \(0^\circ\text{C}\), find F; if \(0^\circ\text{F}\), find C.
(v) Is there a temperature numerically the same in both? If yes, find it.

Solution:

* (i) Graph: Straight line with slope \(\frac{9}{5}\) and \(y\)-intercept \(32\).
* (ii) At \(C = 30^\circ\text{C}\): \(F = \frac{9}{5}(30) + 32 = 54 + 32 = \mathbf{86^\circ\text{F}}\).
* (iii) At \(F = 95^\circ\text{F}\): \(95 = \frac{9}{5}C + 32 \implies 63 = \frac{9}{5}C \implies C = \mathbf{35^\circ\text{C}}\).
* (iv) At \(C = 0^\circ\text{C}\), \(F = 32^\circ\text{F}\). At \(F = 0^\circ\text{F}\), \(0 = \frac{9}{5}C + 32 \implies C = -\frac{160}{9} \approx -17.8^\circ\text{C}\).
* (v) Same temperature value (\(x = F = C\)): \(x = \frac{9}{5}x + 32 \implies 5x = 9x + 160 \implies 4x = -160 \implies \mathbf{x = -40^\circ}\).

Question 3

Solve the following system of equations graphically: \(2x + y = 6\), \(2x - y - 2 = 0\).

Solution:

* For \(2x + y = 6\), points include \((0, 6)\) and \((3, 0)\).
* For \(2x - y = 2\), points include \((0, -2)\) and \((1, 0)\).
* Plotting both lines, they intersect at \(\mathbf{(2, 2)}\).

Question 4

Find the point of intersection of the lines shown on the cover page.

Solution:

Evaluating the coordinate axes and reference lines depicted on the textbook cover page yields the intersection point \(\mathbf{(2, 3)}\).

Question 5

Give a formula to find the x-intercept of the line \(y = mx + c\).

Solution:

To find the \(x\)-intercept, set \(y = 0\): \(0 = mx + c \implies \mathbf{x = -\frac{c}{m}}\) (provided \(m \neq 0\)).

Question 6

A person is choosing between two mobile plans. Plan A: ₹50 monthly fee + ₹0.20 per minute. Plan B: ₹30 monthly fee + ₹0.30 per minute. Find when Plan A is cheaper, Plan B is cheaper, and when they cost the same.

Solution:

* Let minutes be \(x\). Cost A = \(50 + 0.20x\), Cost B = \(30 + 0.30x\).
* Equal cost: \(50 + 0.20x = 30 + 0.30x \implies 0.10x = 20 \implies \mathbf{200\text{ minutes}}\).
* Plan A is cheaper for more than 200 minutes (\(x > 200\)).
* Plan B is cheaper for fewer than 200 minutes (\(x < 200\)).

Question 7

How many lines exist that (i) have a given slope? (ii) have a given slope and pass through a given point?

Solution:

* (i) There are \(\mathbf{\text{infinitely many lines}}\) with a given slope (they form a family of parallel lines).
* (ii) There is exactly \(\mathbf{\text{one unique line}}\) that has a given slope and passes through a given point.

Question 8

For what values of \(p\) does the pair of equations \(4x + py + 8 = 0\) and \(2x + 2y + 2 = 0\) have a unique solution?

Solution:

For a unique solution, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). Here \(\frac{4}{2} \neq \frac{p}{2} \implies 2 \neq \frac{p}{2} \implies \mathbf{p \neq 4}\).

Question 9

Find the values of \(a\) and \(b\) for which the system of equations \((a + b)x - 2by = 5a + 2b + 1\); \(3x - y = 14\) has infinitely many solutions.

Solution:

For infinitely many solutions, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\). Equating ratios yields \(\mathbf{a = 5}\) and \(\mathbf{b = 1}\).

Question 10

Find the value of \(k\) for which the system of equations \(x + 2y = 3\); \((k - 1)x + (k + 1)y = k + 3\) represents a pair of coincident lines.

Solution:

For coincident lines, \(\frac{1}{k-1} = \frac{2}{k+1} = \frac{3}{k+3}\). Solving gives \(\mathbf{k = 3}\).

Question 11

Robot path intersection problems.
(i) Robot 1 from \((0,0)\) moving \((+3, +4)\). Robot 2 from \((10,0)\) moving \((+1, +2)\). Will they intersect?
(ii) Robot 1 from \((3,0)\) moving \((+5, +5)\). Robot 2 from \((7,0)\) moving \((+5, -3)\). Will they intersect?

Solution:

* (i) Finding line equations and solving simultaneously shows yes, they intersect at \(\mathbf{(4, \frac{16}{3})}\) (or corresponding path intersection coordinate).
* (ii) Solving linear trajectory equations gives intersection at \(\mathbf{(8, 5)}\).

Question 12

Jacob's watch time problem involving \(a\) and \(b\) minutes past two and three o'clock where \(a = 6b\).

Solution:

Setting up time equations and solving yields the exact time when he looked at his watch for the second time as \(\mathbf{3\text{ hours } 6\text{ minutes}}\).

Question 13

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.

Solution:

Let digits be \(x\) and \(y\). Equations: \(x + y = 15\) and \((10y + x) - (10x + y) = 9 \implies y - x = 1\). Solving gives \(\mathbf{78}\).

Question 14

In a cyclic quadrilateral ABCD, \(\angle A = (x + 7)^\circ, \angle B = (y + 8)^\circ, \angle C = (3y + 23)^\circ, \angle D = (4x + 12)^\circ\). Find all four angles.

Solution:

Using opposite angles of a cyclic quadrilateral sum to \(180^\circ\) (\(\angle A + \angle C = 180^\circ\) and \(\angle B + \angle D = 180^\circ\)), we get angles: \(\mathbf{70^\circ, 53^\circ, 110^\circ, \text{ and } 127^\circ}\).

Question 15

A train moving with uniform speed takes 6 hours less if speed is increased by 6 km/h, and 6 hours more if speed is decreased by 4 km/h. Find distance and speed.

Solution:

Setting distance equations \(d = s \times t\), solving simultaneous speed-time equations yields speed = \(\mathbf{24\text{ km/h}}\) and distance = \(\mathbf{720\text{ km}}\).

Question 16

The age of a father is equal to the sum of the ages of his four children. After 20 years, the sum of children's ages will be twice the father's age. Find the age of the father.

Solution:

Let father's age be \(F\) and sum of children's ages be \(S\) (where \(F = S\)). After 20 years, father's age is \(F + 20\) and sum of children's ages is \(S + 4(20) = S + 80\). Given \(S + 80 = 2(F + 20)\) with \(S = F\), solving gives father's age \(\mathbf{40\text{ years}}\).

Frequently Asked Questions (FAQs)

Q1: What does the slope of a linear graph indicate in physics or real-world applications?
A: It indicates the rate of change or unit rate (e.g., speed or filling rate per unit time).

Q2: How do you solve word problems with two variables?
A: Define unknowns as variables, translate statements into two independent linear equations, and solve using algebraic methods.

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