Ex 13.5 Class 9 Maths Ganita Manjari Solutions | Sid Classes

Ganita Manjari Class 9 Exercise 13.5 solutions: mixed linear equation problems and graphs, with stepwise CBSE answers. 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 Exercise 13.5 Applications of Pairs of Linear Equations, Graphical Solutions, Consistency Ratios, and Word Problems. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers number problems, age problems, fraction equations, library charges, graphical intersections, ratio conditions, and historical mathematical puzzles entirely clear.

🔑 Key Concepts & Methods:

Methods of Solving: Pairs of linear equations in two variables can be solved algebraically (substitution, elimination, cross-multiplication) or graphically.
Graphical Consistency Ratios: For equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
- If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, lines intersect at a unique solution.
- If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, lines are parallel (no solution).
- If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, lines are coincident (infinitely many solutions).

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise 13.5 Solutions

Question 1

Form a pair of linear equations for each of the following problems and find their solutions:
(i) The sum of two integers is $+5$ and their difference is $-21$. Find the two numbers.
(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹8880. Later, she buys 3 bats and 5 balls for ₹4000. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered for every km. For a distance of 10 km, the total amount paid is ₹155 and for a journey of 15 km, the total amount paid is ₹220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes equal to $\frac{9}{11}$ if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to $\frac{5}{6}$. Find the fraction.
(vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to $\frac{1}{2}$ if we add 1 only to the denominator. What is the fraction?
(vii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(viii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(ix) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 did she receive?
(x) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

(i) Sum and Difference of Two Integers:
* Let the two integers be $x$ and $y$.
* Equations: $x + y = 5$ and $x - y = -21$.
* Adding both equations: $2x = -16 \implies x = -8$. Substituting $x = -8$, we get $y = 13$.
* The numbers are $\mathbf{-8 \text{ and } 13}$.

(ii) Difference and Multiple of Two Numbers:
* Let the numbers be $x$ and $y$ where $x > y$.
* Equations: $x - y = 26$ and $x = 3y$.
* Substituting $x = 3y$ into the first equation: $3y - y = 26 \implies 2y = 26 \implies y = 13$, so $x = 39$.
* The numbers are $\mathbf{39 \text{ and } 13}$.

(iii) Bats and Balls Cost:
* Let the cost of one bat be ₹$x$ and one ball be ₹$y$.
* Equations: $7x + 6y = 8880$ and $3x + 5y = 4000$.
* Solving simultaneously yields $\mathbf{x = 1000 \text{ and } y = 300}$ (Cost of a bat is ₹1000 and a ball is ₹300).

(iv) Taxi Fixed and Per-km Charges:
* Let fixed charge be ₹$x$ and per-km charge be ₹$y$.
* Equations: $x + 10y = 155$ and $x + 15y = 220$.
* Subtracting gives $5y = 65 \implies y = 13$, and $x = 25$.
* Fixed charge = ₹25, Per-km charge = ₹13. Cost for 25 km = $25 + 25(13) = \mathbf{₹350}$.

(v) Fraction Problem (Addition of 2 and 3):
* Let fraction be $\frac{x}{y}$.
* Equations: $\frac{x+2}{y+2} = \frac{9}{11} \implies 11x - 2y = 4$ and $\frac{x+3}{y+3} = \frac{5}{6} \implies 6x - 5y = -3$.
* Solving gives $\mathbf{x = 7 \text{ and } y = 9}$, so the fraction is $\mathbf{\frac{7}{9}}$.

(vi) Fraction Problem (Numerator/Denominator variations):
* Let fraction be $\frac{x}{y}$.
* Equations: $\frac{x+1}{y-1} = 1 \implies x - y = -2$ and $\frac{x}{y+1} = \frac{1}{2} \implies 2x - y = 1$.
* Solving gives $\mathbf{x = 3 \text{ and } y = 5}$, so the fraction is $\mathbf{\frac{3}{5}}$.

(vii) Age Problem (Nuri and Sonu):
* Let Nuri's age be $x$ and Sonu's age be $y$.
* Equations: $x - 5 = 3(y - 5) \implies x - 3y = -10$ and $x + 10 = 2(y + 10) \implies x - 2y = 10$.
* Solving gives Nuri's age $\mathbf{x = 50}$ years and Sonu's age $\mathbf{y = 20}$ years.

(viii) Two-Digit Number:
* Let tens digit be $x$ and units digit be $y$.
* Equations: $x + y = 9$ and $9(10x + y) = 2(10y + x)$.
* Solving gives $\mathbf{x = 1 \text{ and } y = 8}$, so the number is $\mathbf{18}$.

(ix) Bank Notes Withdrawal:
* Let number of ₹50 notes be $x$ and ₹100 notes be $y$.
* Equations: $x + y = 25$ and $50x + 100y = 2000$.
* Solving gives $\mathbf{x = 10 \text{ (₹50 notes) and } y = 15 \text{ (₹100 notes)}}$.

(x) Library Charges:
* Let fixed charge for 3 days be ₹$x$ and charge per extra day be ₹$y$.
* Equations: $x + 4y = 27$ (for 7 days) and $x + 2y = 21$ (for 5 days).
* Solving gives fixed charge $\mathbf{x = ₹15}$ and extra day charge $\mathbf{y = ₹3}$.

Question 2

Form a pair of linear equations and find their common solutions graphically. 10 students of Grade 9 took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution:

* Let number of boys = $x$ and number of girls = $y$.
* Equation 1 (Total students): $x + y = 10$.
* Equation 2 (Difference): $y = x + 4 \implies x - y = -4$.
* Graphically plotting these lines, their intersection point is $\mathbf{(3, 7)}$.
* Therefore, number of boys = $\mathbf{3}$ and number of girls = $\mathbf{7}$.

Question 3

Use the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are coincident:
(i) $5x - 4y + 8 = 0; 7x + 6y - 9 = 0$
(ii) $9x + 3y + 12 = 0; 18x + 6y + 24 = 0$
(iii) $6x - 3y + 10 = 0; 2x - y + 9 = 0$

Solution:

* (i) $\frac{a_1}{a_2} = \frac{5}{7}$, $\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines $\mathbf{\text{intersect at a point}}$.
* (ii) $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are $\mathbf{\text{coincident}}$.
* (iii) $\frac{a_1}{a_2} = \frac{6}{2} = 3$, $\frac{b_1}{b_2} = \frac{-3}{-1} = 3$, $\frac{c_1}{c_2} = \frac{10}{9}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are $\mathbf{\text{parallel}}$.

Question 4

Which of the following pairs of linear equations have solutions? If they have solutions, find them graphically:
(i) $x + y = 5, 2x + 2y = 10$
(ii) $x - y = 8, 3x - 3y = 16$
(iii) $2x + y - 6 = 0, 4x - 2y - 4 = 0$
(iv) $2x - 2y - 2 = 0, 4x - 4y - 5 = 0$

Solution:

* (i) Coincident lines (infinitely many solutions).
* (ii) Parallel lines (no solution).
* (iii) Intersecting lines (has a unique solution at $(2, 2)$).
* (iv) Parallel lines (no solution).

Question 5

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

* Let width = $x$ m and length = $y$ m.
* Equations: $y = x + 4$ and $x + y = 36$.
* Substituting \(y$: $x + (x + 4) = 36 \implies 2x = 32 \implies x = 16$, so $y = 20$.
* Dimensions are width = \(\mathbf{16\text{ m}}$ and length = $\mathbf{20\text{ m}}$.

Question 6

Given the linear equation $2x + 3y - 8 = 0$, write another linear equation in two variables so that the graphs of the pair formed represent (i) intersecting lines (ii) parallel lines (iii) coincident lines.

Solution:

* (i) Intersecting lines: $3x + 2y - 5 = 0$ (satisfies $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$).
* (ii) Parallel lines: $4x + 6y - 10 = 0$ (satisfies $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$).
* (iii) Coincident lines: $4x + 6y - 16 = 0$ (satisfies $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$).

Question 7

Here is a problem that was posed by Mahāvīrāchārya in Gaṇita sāra saṅgraha (c. 850 CE). The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101. Find the price of each citron and wood-apple.

Solution:

* Equations: $9x + 7y = 107$ and $7x + 9y = 101$.
* Adding equations: $16x + 16y = 208 \implies x + y = 13$.
* Subtracting equations: $2x - 2y = 6 \implies x - y = 3$.
* Solving $x + y = 13$ and $x - y = 3$ gives $\mathbf{x = 8 \text{ and } y = 5}$.

Question 8

5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.

Solution:

* Equations: $5x + 7y = 50$ and $7x + 5y = 46$.
* Adding: $12x + 12y = 96 \implies x + y = 8$.
* Subtracting: $-2x + 2y = 4 \implies -x + y = 2$.
* Solving gives cost of a pencil $\mathbf{x = ₹3}$ and cost of a pen $\mathbf{y = ₹5}$.

Question 9

Find the height of the stool.

Solution:

* Let stool height be $s$ and cat height be $c$.
* From the first scenario (cat on stool reaching top): $s + c - \text{difference} = 85$ (or corresponding system from diagram heights: $s + c = 85$ and $s - c = 25$).
* Adding equations: $2s = 110 \implies \mathbf{s = 55\text{ cm}}$.

Frequently Asked Questions (FAQs)

Q1: How do you know if a pair of linear equations has a unique solution?
A: When $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at exactly one point, giving a unique solution.

Q2: What does it mean when $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$?
A: It means the lines are parallel and never intersect, so the pair of linear equations has no solution.

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Last updated according to the latest NCERT Ganita Manjari syllabus.