Write an Admission Experience

Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5

adi1212

contributor

33 published articles

Read

Discuss

Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

Solution:

Here,
a₁ = 1, b₁ = -3, c₁ = -3
a₂ = 3, b₂ = -9, c₂ = -2
So,
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$
As, $\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}$
Hence, the given pairs of equations have no solution.

(ii) 2x + y = 5, 3x + 2y = 8

Solution:

Rearranging equations, we get
2x + y -5 = 0
3x + 2y -8 = 0
Here,
a₁ = 2, b₁ = 1, c₁ = -5
a₂ = 3, b₂ = 2, c₂ = -8
So,
$\frac{a_1}{a_2} = \frac{2}{3}$
$\frac{b_1}{b_2} = \frac{1}{2}$
As, $\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}$
Hence, the given pairs of equations have unique solution.
For cross multiplication,
$\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}$
$\frac{x}{(1)(-8)-(2)(-5)} = \frac{y}{(-5)(3)-(-8)(2)} = \frac{1}{(2)(2)-(3)(1)}$
$\frac{x}{(-8)+10} = \frac{y}{(-15)+16} = \frac{1}{4-3}$
$\frac{x}{2}$ = y = 1
Hence,
$\frac{x}{2}$ = 1
x = 2
and, y = 1
Hence, the required solution is x = 2 and y = 1.

(iii) 3x – 5y = 20, 6x – 10y = 40

Solution:

Rearranging equations, we get
3x – 5y – 20 = 0
6x – 10y – 40 = 0
Here,
a₁ = 3, b1 = -5, c1 = -20
a₂ = 6, b₂ = -10, c₂ = -40
So,
$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2}$
As,
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
Hence, the given pairs of equations have infinitely many solutions.

(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

Solution:

Here,
a₁ = 1, b1 = -3, c1 = -7
a₂ = 3, b₂ = -3, c₂ = -15
So,
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-3}{-3} = 1$
As,
$\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}$
Hence, the given pairs of equations have unique solution.
For cross multiplication,
$\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}$
$\frac{x}{(-3)(-15)-(-3)(-7)} = \frac{y}{(-7)(3)-(-15)(1)} = \frac{1}{(1)(-3)-(3)(-3)}$
$\frac{x}{45-21} = \frac{y}{(-21)+15} = \frac{1}{-3+9}$
$\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}$
Hence,
x = $\frac{24}{6}$
x = 4
and, y = $\frac{-6}{6}$
y = -1
Hence, the required solution is x = 4 and y = -1.

Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

Solution:

Here,
a₁ = 2, b1 = 3, c1 = -7
a₂ = a-b, b₂ = a+b, c₂ = -(3a+b-2)
For having infinite number of solutions, it has to satisfy below conditions:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
$\frac{2}{a-b} = \frac{3}{a+b} = \frac{7}{(3a+b-2)}$
Now, on comparing
$\frac{2}{a-b} = \frac{3}{a+b}$
2(a+b) = 3(a-b)
2a+2b = 3a – 3b
a – 5b = 0 …………………(1)
And, now on comparing
$\frac{3}{a+b} = \frac{7}{(3a+b-2)}$
3(3a+b-2) = 7(a+b)
9a+3b-6 = 7a+7b
2a-4b-6=0
Reducing form, we get
a-2b-3=0 …………………(2)
Now, new values for
a₁ = 1, b₁ = -5, c₁ = 0
a₂ = 1, b₂ = -2, c₂ = -3
Solving Eq(1) and Eq(2), by cross multiplication,
$\mathbf{\frac{a}{b_1c_2-b_2c_1} = \frac{b}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}$
$\frac{a}{(-5)(-3)-(-2)(0)} = \frac{b}{(0)(1)-(-3)(1)} = \frac{1}{(1)(-2)-(1)(-5)}$
$\frac{a}{15-0} = \frac{b}{0+3} = \frac{1}{-2+5}$
$\frac{a}{15} = \frac{b}{3} = \frac{1}{3}$
Hence,
a = $\frac{15}{3}$
a = 5
and, b = $\frac{3}{3}$
b = 1
Hence, For values a = 5 and b = 1 pair of linear equations have an infinite number of solutions

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Solution:

Here,
a₁ = 3, b₁ = 1, c₁ = -1
a₂ = (2k-1), b₂ = k-1, c₂ = -(2k+1)
For having no solution, it has to satisfy below conditions:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}$
$\frac{3}{2k-1} = \frac{1}{k-1} ≠ \frac{1}{2k+1}$
Now, on comparing
$\frac{3}{2k-1} = \frac{1}{k-1}$
3(k-1) = 2k-1
3k-3 = 2k-1
k = 2
And, now on comparing
$\frac{1}{k-1} ≠ \frac{1}{2k+1}$
2k+1 ≠ k-1
k ≠ -2
Hence, for k = 2 and k ≠ -2 the pair of linear equations have no solution.

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

8x + 5y – 9 = 0 …………………(1)
3x + 2y – 4 = 0 …………………(2)
Substitution Method
From Eq(2), we get
x = 4-2y/3
Now, substituting it in Eq(1), we get
8(4-2y/3) + 5y – 9 = 0
32-16y/3 + 5y – 9 = 0
32 – 16y + 15y – 27 = 0
y = 5
Now, substituting y = 5 in Eq(2), we get
3x + 2(5) – 4 = 0
3x = -6
x = -2
Cross Multiplication Method
Here,
a₁ = 8, b₁ = 5, c₁ = -9
a₂ = 3, b₂ = 2, c₂ = -4
So,
$\frac{a_1}{a_2} = \frac{8}{3}$
$\frac{b_1}{b_2} = \frac{5}{2}$
As,
$\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}$
Hence, the given pairs of equations have unique solution.
For cross multiplication,
$\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}$
$\frac{x}{(5)(-4)-(2)(-9)} = \frac{y}{(-9)(3)-(-4)(8)} = \frac{1}{(8)(2)-(3)(5)}$
$\frac{x}{(-20)+18} = \frac{y}{(-27)+32} = \frac{1}{16-15}$
$\frac{x}{-2} = \frac{y}{5}$ = 1
Hence,
$\frac{x}{-2}$ = 1
x = -2
and, $\frac{y}{-5}$ = 1
y = 5
Hence, the required solution is x = -2 and y = 5.

Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:

Let’s take,
Fixed charge = x
Charge of food per day = y
According to the given question,
x + 20y = 1000 ……………….. (1)
x + 26y = 1180 ………………..(2)
Subtracting Eq(1) from Eq(2) we get
6y = 180
y = 30
Now, substituting y = 30 in Eq(2), we get
x + 20(30) = 1000
x = 1000 – 600
x= 400.
Hence, fixed charges is ₹ 400 and charge per day is ₹ 30.

(ii) A fraction become $\frac{1}{3}$ when 1 is subtracted from the numerator, and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

Solution:

Let the fraction be $\frac{x}{y}$ .
So, as per the question given,
$\frac{x-1}{y} = \frac{1}{3}$
3x – y = 3 …………………(1)
$\frac{x}{y+8} = \frac{1}{4}$
4x –y =8 ………………..(2)
Subtracting Eq(1) from Eq(2) , we get
x = 5
Now, substituting x = 5 in Eq(2), we get
4(5)– y = 8
y = 12
Hence, the fraction is $\mathbf{\frac{5}{12}}$ .

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Solution:

Let’s take
Number of right answers = x
Number of wrong answers = y
According to the given question;
3x−y=40 ……….……..(1)
4x−2y=50
2x−y=25 ……………….(2)
Subtracting Eq(2) from Eq(1), we get
x = 15
Now, substituting x = 15 in Eq(2), we get
2(15) – y = 25
y = 30-25
y = 5
Hence, number of right answers = 15 and number of wrong answers = 5
Hence, total number of questions = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Solution:

Let’s take
Speed of car from point A = x km/he
Speed of car from point B = y km/h
When car travels in the same direction,
5x – 5y = 100
x – y = 20 ………………(1)
When car travels in the opposite direction,
x + y = 100 ………………..(2)
Subtracting Eq(1) from Eq(2), we get
2y = 80
y = 40
Now, substituting y = 40 in Eq(1), we get
x – 40 = 20
x = 60
Hence, the speed of car from point A = 60 km/h
Speed of car from point B = 40 km/h.

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

Let’s take
Length of rectangle = x unit
Breadth of the rectangle = y unit
Area of rectangle will be = xy sq. units
According to the given conditions,
(x – 5) (y + 3) = xy -9
3x – 5y – 6 = 0 ……………………(1)
(x + 3) (y + 2) = xy + 67
2x + 3y – 61 = 0 ……………………..(2)
Using cross multiplication method, we get,
$\frac{x}{(-5)(-61)-(3)(-6)} = \frac{y}{(-6)(2)-(-61)(3)} = \frac{1}{(3)(3)-(2)(-5)}$
$\frac{x}{305 +18} = \frac{y}{-12+183} = \frac{1}{9+10}$
$\frac{x}{323} = \frac{y}{171} = \frac{1}{19}$
Hence,
$\frac{x}{323} = \frac{1}{19}$
x = 17
and, $\frac{y}{171} = \frac{1}{19}$
y = 9
Hence, the required solution is x = 17 and y = 9.
Length of rectangle = 17 units
Breadth of the rectangle = 9 units

My Personal Notes

Last Updated : 12 Mar, 2021

Related Articles

Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5

Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

(ii) 2x + y = 5, 3x + 2y = 8

(iii) 3x – 5y = 20, 6x – 10y = 40

(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

(ii) A fraction become $\frac{1}{3}$ when 1 is subtracted from the numerator, and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5

Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

(ii) 2x + y = 5, 3x + 2y = 8

(iii) 3x – 5y = 20, 6x – 10y = 40

(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

(ii) A fraction become when 1 is subtracted from the numerator, and it becomes when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Please Login to comment...

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

(ii) A fraction become $\frac{1}{3}$ when 1 is subtracted from the numerator, and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.