# Linear Equations

## Exercise 3.5 Part 2

Question 5: Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(a) 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.

Solution: Let us assume that fixed charges is x and per day charges is y

Amount paid by student A = x + 20y = 1000

Amount paid by student B = x + 26y = 1180

x+26y=1180

(a_1)/(a_2)=1 and (b_1)/(b_2)=(20)/(26)=(10)/(13)

It is clear that;

(a_1)/(a_2)≠(b_1)/(b_2)

Hence, there will be unique solution for the given pair of linear equations.

Subtracting first equation from second equation, we get;

x + 26y – x – 20y = 1180 – 1000

Or, 6y = 180

Or, y = 30

Substituting the value of y in first equation, we get;

x + 20y = 1000

Or, x + 20 x 30 = 1000

Or, x + 600 = 1000

Or, x = 1000 – 600 = 400

Hence, fixed charges = Rs. 400 and per day charges = Rs. 30

(b) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ when 8 is added to its denominator. Find the fraction.

Solution: Let us assume numerator is x and denominator is y.

First condition:

(x-1)/(y)=1/3

Or, 3x-3=y
Or, 3x-y-3=0 --------(1)

Second condition:

(x)/(y+8)=1/4

Or, 4x = y + 8
Or, 4x – y – 8 = 0 ……… (2)

(a_1)/(a_2)=3/4

(b_1)/(b_2)=1

It is clear that;

(a_1)/(a_2)≠(b_1)/(b_2)

Hence, there will be unique solution for the given pair of linear equations.

Subtracting first equation from second equation, we get;

4x – y – 8 – 3x + y + 3 = 0
Or, x – 5 = 0
Or, x = 5

Substituting the value of x in second equation, we get;

4x – y – 8 = 0
Or, 4 xx 5 – y – 8 = 0
Or, 20 – y – 8 = 0
Or, 12 – y = 0

Or, y = 12

Hence, x = 5 and y = 12

(c) 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 us assume that the number of correct answers = x and number of incorrect answers = y

First condition: 3x - y = 40

Second condition: 4x - 2y = 50

(a_1)/(a_2)=3/4 and (b_1)/(b_2)=1/2

It is clear that;

(a_1)/(a_2)≠(b_1)/(b_2)

Hence, there will be unique solution for the given pair of linear equations.

Multiply the first equation by 2 and subtract second equation from the resultant:

6x - 2y – 4x + 2y = 80 – 50
Or, 2x = 30
Or, x = 15

Substituting the value of x in second equation, we get;

4x - 2y = 50
Or, 4 xx 15 - 2y = 50
Or, 60 – 2y = 50
Or, 2y = 60 – 50 = 10
Or, y = 5

Hence, number of correct answers = 15 and number of incorrect answers = 5

(d) 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 us assume that the speed of car A is x and that of car B is y.

First condition: When both the cars are going in the same direction, their relative speed is; x – y

Hence,
(100)/(x-y)=5

Or, 5x-5y=100
Or, x-y=20

Second condition: When the cars are moving towards each other, their relative speed is; x + y

Hence;

(100)/(x+y)=1

Or, x+y=100

Adding the two equations, we get;

x – y + x + y = 20 + 100
Or, 2x = 120
Or, x = 60

Substituting the value of x in first equation, we get;

x – y = 20
Or, 60 – y = 20
Or, y = 60 – 20 = 40

Hence, speed of car A = 60 km/h and speed of car B = 40 km/h

(e) 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 us assume length is x and breadth is y.

Area of rectangle = xy

First condition: Length = x - 5 and breadth = y + 3

Area of rectangle:

(x – 5)(y + 3) = xy – 9
Or, xy – 5y + 3x – 15 = xy – 9
Or, 3x – 5y – 15 = - 9
Or, 3x – 5y – 15 + 9 = 0
Or, 3x – 5y – 6 = 0

Second condition: Length = x +3 and breadth = y + 2

Area of rectangle:

(x + 3)(y + 2) = xy + 67
Or, xy + 3y + 2x + 6 = xy + 67
Or, 2x + 3y + 6 = 67
Or, 2x + 3y – 61 = 0

Multiplying the first equation by 2, we get;

6x – 10y – 12 = 0

Multiplying the second equation by 3, we get;

6x + 9y – 183 = 0

Subtracting these two equations, we get;

6x + 9y – 183 – 6x + 10y + 12 = 0
Or, 19y – 171 = 0
Or, 19y = 171
Or, y = 9

Substituting the value of y in any of the above equations, we get;

6x + 9y – 183 = 0
Or, 6x + 9 x 9 – 183 = 0
Or, 6x + 81 – 183 = 0
Or, 6x – 102 = 0
Or, 6x = 102
Or, x = 17

Hence, length = 17 unit and breadth = 9 unit