Car parking is a major problem in urban areas in both developed and developing countries. In the Jaipur city, GT mall is shopping place charged the parking charges of a car in a parking lot is ₹ 30 for the first one hour and ₹ 10 per hour for subsequent hours, then answer the following questions. * a) 10x + y -10 =0 b) 10x-y +20 =0 c) 8x + y -10 =0 d) 8x -y +10 =6 2) In the linear equation formed in (i) , if y =30, then the value of x is ___________. * a) 1 b) 2 c)3 d) 4 3. In the linear equation formed in (i) ,if x=2, then value of y is ________ * a) 10 b) 20 c) 30 d) 40 4. In the given question, if charges of a car in a parking lot is ₹ 50 for the first one hour, then find the linear equation. * a) 10x -y +40 =0 b) 10x +y +30 =0 c) 10x +y -30 =0 d) 30x +y -10 =0 This is a required question 5. In the given question, if charges for subsequent hours is ₹ 8 per hour, then write linear equation. * a) 10x -y +8 =0 b) 10x -y +10 =0 c) 8x -y +22=0 d) 8x -y +8 =0
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Answers
Answer:
Given, parking charges for the first two hours = ₹30
and for subsequent hours = ₹ 10
Total parking time = xh and total charges = ₹y
Then, according to the given condition,
30 + 10(x-2) = y=> 30 + 10x -20 = y
=>10x + 10-y = 0
=>10x-y+10 = 0 (i)
which is the required linear equation in two variables.
It can also be written as
y = 10x + 10 ,…(i)
Now, for drawing the graph, we need atleast two solutions of the equation.
When x = 0, then y = 10(0) +10 = 10
When x = -1, then y = -10 + 10 = 0
When x = 1, then y = 10(1) + 10 = 20
So, we have the following table to draw the graph
Here, we have three points A(-1, 0), B (0,10) and C(1,20). By plotting these points on the graph paper and joining them, we get a straight line AC, which represents the required graph of linear equation.
From the graph, charges for the hours 1,2, 3, 4 and 5 are ₹ 20, ₹ 30,₹ 40, ₹ 50 and ₹ 60, respectively