Quesiton 4. Form the Pair of Linear Equations in the Following Pr Exercise 3.5 Chapter 3 Pair of Linear Equations in

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 Atakes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, paysRs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution: Let fixed charge = Rs x
Variable charge  = Rs y per day
When a student Atakes food for 20 days she has to pay Rs 1000 as hostel charges
So we get
X  + 20 y = 1000       …(1)
Subtract 20 y both side we get
X = 1000 – 20 y        …(2)
who takes food for 26 days, paysRs 1180 as hostel charges
we get
x + 26 y = 1180
plug the value of x we ge t

1000 – 20 y + 26y = 1180
6y = 180
Y = 180/6 = 30

Plug the value of y in equation second we get
X = 1000 – 20 * 30
X = 1000 – 600
X = 400
Hence fixed charge x  = Rs 400
And variable charge y = Rs 30

(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.
Let fraction is x/y
A fraction becomes 1/3 when 1 is subtracted from the numerator
So that
 
Cross multiply we get
3x -3  = y                    …(1)
and it becomes 1/ 4 when 8 is added to its denominator
so that

4x = y + 8
Plug the value of y from equation first we get
4x = 3x – 3 + 8
X = 5
Plug this value back in equation first we get
3*5 – 3  = y
15 – 3 = y
12 = y
Hence fraction is x/y = 5/12

(iii) Yash scored 40marks in a test, getting 3marks for each right answer and losing1 mark for each wrong answer. Had 4marks 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?
Let correct question = x
And incorrect question  = y
Yash scored 40 marks in a test, getting 3marks for each right answer and losing1 mark for each wrong answer.
3x -  y = 40
Subtract 3x both sides we get
-Y = 40 – 3x
Change the sign we get
Y = 3x – 40 …(1)
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
We get
4x – 2y = 50
Plug the value of y we get
4x – 2(3x - 40) =50
4x – 6x + 80 = 50
-2x = - 30
X = 15
Plug this value back in equation first we get
Y = 3x – 40
Y = 3* 15- 40
Y = 45 – 40
Y = 5
So total question are x + y =  15 + 5 = 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?
Let speed of faster car = x Km/ h
Speed of slower car = y km /h
Relative speed when both car travels in same direction = x - y
If the cars travel in the same direction at different speeds, they meet in 5 hours
Speed = distance/ time
Total distance  = 100 km
Speed x – y = 100/5
X – y = 20
Add y both side we get
X = 20 + y                              …(1)
Relative speed when both car travels in opposite direction = x + y
If they travel towards each other, they meet in 1 hour.
Speed = distance/ time
Total distance  = 100 km
Speed x  + y = 100/1
Plug the value of x form equation first we get
20 + y + y = 100
2y = 80
Y = 40
Plug the value of y in equation first we get
X  = 20 + 40
X = 60
Hence speeds are 40 Km /h and 60 Km /h

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units andbreadth is increased by 3 units. If we increase the length by 3units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Let length of rectangle = x units
And width of rectangle = y units
Area of rectangle = length * width = x*y
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.
So
Decrease the length by 5 unit so new length = x - 5
Increase the width by 3 unit so new width = y + 3
New area is reduced by 9 units
So new area  = xy – 9
Plug the value in formula length * width = area we get
(x  - 5)(y + 3) = xy  - 9
Xy  + 3x – 5y – 15  = xy – 9
Subtract xy both side we get
3x - 5y  = 6                 …(1)
If we increase the length by 3units and the breadth by 2 units, the area increases by 67 square units.
Increase the length by 3 unit so new length = x +3
Increase the width by 2 unit so new width = y + 2
New area is increased  by 67 units
So new area  = xy + 67
Plug the value in formula length * width = area we get
(x  +3)(y + 2) = xy  +  67
Xy  + 2x  +  3y  +  6  = xy + 67
Subtract xy both side we get
2x  +  3y = 61                         …(2)*3
3x - 5y  = 6                 …(1)*2
Cross multiply the coefficient of x we get
6x + 9 y = 183
6x -10y  =12
Subtract now we get
19 y = 171
Y = 171/19 = 9
Plug this value of y in equation first we get
2x + 3* 9 = 61
2x  = 61 – 27
2x = 34
X = 34/2 = 17
So length is 17 units and width is 9 units