NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.5#
For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
Ratios Lines Solutions $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ intersect exactly one $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ parallel none $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ coincident infinitely many
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.
Let the numbers be $x$ and $y$.
Their sum is 5, so:
$$x + y = 5 \quad \cdots (1)$$Their difference is $-21$, so:
$$x - y = -21 \quad \cdots (2)$$Adding equation (1) and equation (2):
$$\begin{array}{rcl} x + y &=& 5 \\ x - y &=& -21 \\ \hline 2x &=& -16 \end{array}$$$$\Rightarrow x = -8$$
Substituting $x = -8$ in equation (1):
$$x + y = 5$$$$-8 + y = 5$$
$$\Rightarrow y = 13$$
∴ The numbers are −8 and 13. (Check: $-8 + 13 = 5$ and $-8 - 13 = -21$ ✓)
(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Let the larger number be $x$ and the smaller $y$.
The difference is 26, so:
$$x - y = 26 \quad \cdots (1)$$One number is three times the other, so:
$$x = 3y \quad \cdots (2)$$Substituting $x = 3y$ in equation (1):
$$x - y = 26$$$$3y - y = 26$$
$$\Rightarrow y = 13$$
From equation (2), $x = 3 \times 13 = 39$.
∴ The numbers are 39 and 13.
(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.
Let a bat cost ₹$x$ and a ball ₹$y$.
7 bats and 6 balls cost ₹8880, so:
$$7x + 6y = 8880 \quad \cdots (1)$$3 bats and 5 balls cost ₹4000, so:
$$3x + 5y = 4000 \quad \cdots (2)$$Multiplying equation (1) by 5 and equation (2) by 6:
$$35x + 30y = 44400 \quad \cdots (3)$$$$18x + 30y = 24000 \quad \cdots (4)$$
Subtracting equation (4) from equation (3):
$$\begin{array}{rcl} 35x + 30y &=& 44400 \\ 18x + 30y &=& 24000 \\ \hline 17x &=& 20400 \end{array}$$$$\Rightarrow x = 1200$$
Substituting $x = 1200$ in equation (2):
$$3x + 5y = 4000$$$$3600 + 5y = 4000$$
$$\Rightarrow y = 80$$
∴ A bat costs ₹1200 and a ball costs ₹80.
(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?
Let the fixed charge be ₹$x$ and the charge per km ₹$y$.
10 km costs ₹155, so:
$$x + 10y = 155 \quad \cdots (1)$$15 km costs ₹220, so:
$$x + 15y = 220 \quad \cdots (2)$$Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} x + 15y &=& 220 \\ x + 10y &=& 155 \\ \hline 5y &=& 65 \end{array}$$$$\Rightarrow y = 13$$
Substituting $y = 13$ in equation (1):
$$x + 130 = 155$$$$\Rightarrow x = 25$$
For 25 km:
$$\text{Fare} = 25 + 25 \times 13 = 350$$∴ The fixed charge is ₹25, the charge is ₹13 per km, and 25 km costs ₹350.
(v) A fraction becomes equal to 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 5/6. Find the fraction.
Let the fraction be $\dfrac{x}{y}$.
Adding 2 to both gives $\dfrac{9}{11}$, so:
$$\dfrac{x + 2}{y + 2} = \dfrac{9}{11}$$$$\Rightarrow 11x + 22 = 9y + 18$$
$$\Rightarrow 11x - 9y = -4 \quad \cdots (1)$$
Adding 3 to both gives $\dfrac{5}{6}$, so:
$$\dfrac{x + 3}{y + 3} = \dfrac{5}{6}$$$$\Rightarrow 6x + 18 = 5y + 15$$
$$\Rightarrow 6x - 5y = -3 \quad \cdots (2)$$
Multiplying equation (1) by 5 and equation (2) by 9:
$$55x - 45y = -20 \quad \cdots (3)$$$$54x - 45y = -27 \quad \cdots (4)$$
Subtracting equation (4) from equation (3):
$$\begin{array}{rcl} 55x - 45y &=& -20 \\ 54x - 45y &=& -27 \\ \hline x &=& 7 \end{array}$$Substituting $x = 7$ in equation (2):
$$42 - 5y = -3$$$$\Rightarrow y = 9$$
∴ The fraction is $\dfrac{7}{9}$.
(vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to 1/2 if we add 1 only to the denominator. What is the fraction?
Let the fraction be $\dfrac{x}{y}$.
Adding 1 to the numerator and subtracting 1 from the denominator gives 1, so:
$$\dfrac{x + 1}{y - 1} = 1$$$$\Rightarrow x - y = -2 \quad \cdots (1)$$
Adding 1 to the denominator gives $\dfrac{1}{2}$, so:
$$\dfrac{x}{y + 1} = \dfrac{1}{2}$$$$\Rightarrow 2x - y = 1 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 2x - y &=& 1 \\ x - y &=& -2 \\ \hline x &=& 3 \end{array}$$Substituting $x = 3$ in equation (1):
$$3 - y = -2$$$$\Rightarrow y = 5$$
∴ The fraction is $\dfrac{3}{5}$.
(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?
Let Nuri’s present age be $x$ years and Sonu’s $y$ years.
Five years ago, Nuri was thrice as old as Sonu, so:
$$x - 5 = 3(y - 5)$$$$\Rightarrow x - 3y = -10 \quad \cdots (1)$$
Ten years later, Nuri will be twice as old as Sonu, so:
$$x + 10 = 2(y + 10)$$$$\Rightarrow x - 2y = 10 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} x - 2y &=& 10 \\ x - 3y &=& -10 \\ \hline y &=& 20 \end{array}$$Substituting $y = 20$ in equation (2):
$$x - 40 = 10$$$$\Rightarrow x = 50$$
∴ Nuri is 50 years old and Sonu is 20 years old.
(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.
Let the tens digit be $x$ and the units digit $y$. The number is $10x + y$; reversed, it is $10y + x$.
The sum of the digits is 9, so:
$$x + y = 9 \quad \cdots (1)$$Nine times the number is twice the reversed number, so:
$$9(10x + y) = 2(10y + x)$$$$\Rightarrow 90x + 9y = 20y + 2x$$
$$\Rightarrow 88x - 11y = 0$$
$$\Rightarrow y = 8x \quad \cdots (2)$$
Substituting $y = 8x$ in equation (1):
$$x + 8x = 9$$$$\Rightarrow x = 1$$
From equation (2), $y = 8$.
∴ The number is 18. (Check: $9 \times 18 = 162 = 2 \times 81$ ✓)
(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 she received.
Let her get $x$ notes of ₹50 and $y$ notes of ₹100.
She got 25 notes in all, so:
$$x + y = 25 \quad \cdots (1)$$The notes add up to ₹2000, so:
$$50x + 100y = 2000$$$$\Rightarrow x + 2y = 40 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} x + 2y &=& 40 \\ x + y &=& 25 \\ \hline y &=& 15 \end{array}$$Substituting $y = 15$ in equation (1): $x = 10$.
∴ She received 10 notes of ₹50 and 15 notes of ₹100.
(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.
Let the fixed charge be ₹$x$ and the charge for each extra day ₹$y$.
Seven days means 4 extra days, costing ₹27 in all, so:
$$x + 4y = 27 \quad \cdots (1)$$Five days means 2 extra days, costing ₹21 in all, so:
$$x + 2y = 21 \quad \cdots (2)$$Subtracting equation (2) from equation (1):
$$\begin{array}{rcl} x + 4y &=& 27 \\ x + 2y &=& 21 \\ \hline 2y &=& 6 \end{array}$$$$\Rightarrow y = 3$$
Substituting $y = 3$ in equation (2):
$$x + 6 = 21$$$$\Rightarrow x = 15$$
∴ The fixed charge is ₹15 and each extra day costs ₹3.
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 the number of boys be $x$ and girls be $y$.
10 students took part, so:
$$x + y = 10 \quad \cdots (1)$$There are 4 more girls than boys, so:
$$y = x + 4 \quad \cdots (2)$$Some points on each line:
| Line | Points |
|---|---|
| $x + y = 10$ | $(0, 10)$, $(5, 5)$, $(10, 0)$ |
| $y = x + 4$ | $(0, 4)$, $(2, 6)$, $(3, 7)$ |
The lines meet at (3, 7).
∴ 3 boys and 7 girls took part.
3. Use the ratios a₁/a₂, b₁/b₂, c₁/c₂ 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
∴ $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so the lines intersect at a point.
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, so the lines are coincident.
(iii) 6x − 3y + 10 = 0; 2x − y + 9 = 0
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the lines are parallel.
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
The lines are coincident, so there are infinitely many solutions — every point on the line $x + y = 5$, such as $(0, 5)$, $(1, 4)$, $(2, 3)$, $(5, 0)$.
(ii) x − y = 8, 3x − 3y = 16
The lines are parallel, so there is no solution.
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
These are unequal, so the lines intersect: there is exactly one solution.
| Line | Points |
|---|---|
| $2x + y = 6$ | $(0, 6)$, $(3, 0)$, $(2, 2)$ |
| $4x - 2y = 4$ | $(0, -2)$, $(1, 0)$, $(2, 2)$ |
∴ The solution is $x = 2$, $y = 2$.
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
The lines are parallel, so there is no solution.
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 the length be $x$ m and the width $y$ m.
Half the perimeter is length + width, which is 36 m, so:
$$x + y = 36 \quad \cdots (1)$$The length is 4 m more than the width, so:
$$x - y = 4 \quad \cdots (2)$$Adding equation (1) and equation (2):
$$\begin{array}{rcl} x + y &=& 36 \\ x - y &=& 4 \\ \hline 2x &=& 40 \end{array}$$$$\Rightarrow x = 20$$
Substituting $x = 20$ in equation (1): $y = 16$.
∴ The garden is 20 m long and 16 m wide.
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
Answers can vary; any equation with the right ratios works.
| Lines | Example | Ratios |
|---|---|---|
| (i) Intersecting | $3x - 2y - 4 = 0$ | $\dfrac{2}{3} \neq \dfrac{3}{-2}$ |
| (ii) Parallel | $4x + 6y - 7 = 0$ | $\dfrac{2}{4} = \dfrac{3}{6} \neq \dfrac{-8}{-7}$ |
| (iii) Coincident | $4x + 6y - 16 = 0$ | $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{-8}{-16}$ |
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. O mathematician, tell me quickly the price of each citron and of each fragrant wood-apple. #
Solution
Let a citron cost $x$ and a wood-apple $y$:
$$9x + 7y = 107 \quad \cdots (1)$$$$7x + 9y = 101 \quad \cdots (2)$$
The coefficients are swapped between the two equations, so adding and subtracting give very simple equations.
Adding equation (1) and equation (2):
$$\begin{array}{rcl} 9x + 7y &=& 107 \\ 7x + 9y &=& 101 \\ \hline 16x + 16y &=& 208 \end{array}$$$$\Rightarrow x + y = 13 \quad \cdots (3)$$
Subtracting equation (2) from equation (1):
$$\begin{array}{rcl} 9x + 7y &=& 107 \\ 7x + 9y &=& 101 \\ \hline 2x - 2y &=& 6 \end{array}$$$$\Rightarrow x - y = 3 \quad \cdots (4)$$
Adding equation (3) and equation (4):
$$\begin{array}{rcl} x + y &=& 13 \\ x - y &=& 3 \\ \hline 2x &=& 16 \end{array}$$$$\Rightarrow x = 8$$
Substituting $x = 8$ in equation (3): $y = 5$.
∴ A citron costs 8 and a wood-apple costs 5.
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
Let a pencil cost ₹$x$ and a pen ₹$y$:
$$5x + 7y = 50 \quad \cdots (1)$$$$7x + 5y = 46 \quad \cdots (2)$$
Adding equation (1) and equation (2):
$$\begin{array}{rcl} 5x + 7y &=& 50 \\ 7x + 5y &=& 46 \\ \hline 12x + 12y &=& 96 \end{array}$$$$\Rightarrow x + y = 8 \quad \cdots (3)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 7x + 5y &=& 46 \\ 5x + 7y &=& 50 \\ \hline 2x - 2y &=& -4 \end{array}$$$$\Rightarrow x - y = -2 \quad \cdots (4)$$
Adding equation (3) and equation (4):
$$\begin{array}{rcl} x + y &=& 8 \\ x - y &=& -2 \\ \hline 2x &=& 6 \end{array}$$$$\Rightarrow x = 3$$
Substituting $x = 3$ in equation (3): $y = 5$.
∴ A pencil costs ₹3 and a pen costs ₹5.
9. Find the height of the stool. (With the cat sitting on the stool, the top of the cat is 85 cm above the floor. With the cat standing under the stool, the top of the stool is 25 cm above the top of the cat.) #
Solution
Let the stool be $x$ cm tall and the cat $y$ cm tall.
The cat on the stool reaches 85 cm, so:
$$x + y = 85 \quad \cdots (1)$$The stool is 25 cm taller than the cat, so:
$$x - y = 25 \quad \cdots (2)$$Adding equation (1) and equation (2):
$$\begin{array}{rcl} x + y &=& 85 \\ x - y &=& 25 \\ \hline 2x &=& 110 \end{array}$$$$\Rightarrow x = 55$$
∴ The stool is 55 cm high (and the cat is 30 cm tall).