NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.5)

NCERT solutions for class 10 Maths


Chapter 3


Pair of Linear Equations in Two Variables


Exercise 3.5


Ex 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 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8

(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Solutions:

(i) x – 3y – 3 =0

3x – 9y -2 =0

a1/a2=1/3 ,

b1/b2= -3/-9 =1/3,

c1/c2=-3/-2 = 3/2

(a1/a2) = (b1/b2) ≠ (c1/c2)

∴The given pair of equations has no solution.

(ii) Given, 2x + y = 5

3x +2y = 8

a1/a2 = 2/3

b1/b2 = 1/2

c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

∴ The given pair of equations has unique solution.

By cross multiplication method:

x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 1/(a1b2-a2b1)

x/(-8-(-10)) = y/(15+16) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) 3x – 5y = 20

6x – 10y = 40

(a1/a2) = 3/6 = 1/2

(b1/b2) = -5/-10 = 1/2

(c1/c2) = 20/40 = 1/2

a1/a2 = b1/b2 = c1/c2

The given pair of equations has infinite solutions.

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

(a1/a2) = 1/3

(b1/b2) = -3/-3 = 1

(c1/c2) = -7/-15

a1/a2 ≠ b1/b2

∴ The given pair of equations has unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1.


Ex 3.5 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

Solution:

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a1/a2 = 2/(a-b)

b1/b2 = 3/(a+b)

c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4  ……….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 …….….(ii)

Subtracting equation (i) from equation (ii),

4b = 4

b =1

Substituting the value of b equation in (ii),

a -5 x 1= 0

a = 5

Thus  a = 5 and b = 1 will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x  +  (k-1)y – 2k -1 = 0

a1/a2 = 3/(2k -1)

b1/b2 = 1/(k-1)

c1/c2 = -1/(-2k -1) = 1/( 2k +1)

For no solutions

a1/a2 = b1/b2 ≠ c1/c2

3/(2k-1) = 1/(k -1)   ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

k =2

∴ The value of k=2 has no solutions


Ex 3.5 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 ……..(i)

3x + 2y = 4 …….….(ii)

From equation (ii)

x = (4 – 2y )/ 3  ………. (iii)

Using this value of x in equation (i),

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5

Substituting this value of y in equation (ii),

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

By Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.


Ex 3.5 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 Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 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.

Solutions:

(i) Let x be the fixed charge and y be the charge of food per day.

x + 20y = 1000…….. (i)

x + 26y = 1180……..(ii)

Subtracting equation (i) from equation (ii)

6y = 180

y = Rs.30

Substituting the value of y in equation (i),

x = 1180 -26 x 30

x= Rs.400.

∴ The fixed charges are Rs.400, and the charge per day is Rs.30.

(ii) Let the fraction be x/y.

(x -1)/y = 1/3 => 3x – y = 3………(i)

x/(y + 8) = 1/4  => 4x –y =8 ……..(ii)

Subtracting equation (i) from (ii) , we get

x = 5

Substituting the value of x in equation (i),

(3×5)– y = 3

y= 12

∴The fraction is 5/12.

(iii) Let the number of right answers and wrong answers be x and y respectively.

3x−y=40……..(i)

4x−2y=50

⇒2x−y=25…….(ii)

Subtracting equation (ii) from equation (i)

x = 15

Substituting the value of x in equation (ii)

30 – y = 25

Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

(iv) Let the speed of first car and second car be u km/h and v km/h respectively.

Speed of both cars while they are travelling in the same direction = (u – v) km/h

Speed of both cars while they are travelling in opposite directions i.e. when they are travelling towards each other = (u + v) km/h

Distance travelled = Speed x Time

5(u – v) = 100

⇒ u – v =20 …….(i)

1(u + v) = 100

⇒ (u + v) = 100………(ii)

Adding equation (i) and (ii)

2u = 120

u = 60

Substituting the value of u in equation (ii),

v = 40

Hence, speed of the first car is 60 km/h and speed of the second car is 40 km/h

(v) Let,the length of rectangle = x unit and breadth of the rectangle = y unit

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0…………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0………..(2)

By cross multiplication method,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

∴ x = 17 and y = 9.

Thus, the length of the rectangle = 17 units and the breadth of the rectangle = 9 units.


 

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