Question

11. Answer the following
The parking charges of a car in railway station for first two hours is 40Rs and 8Rs for each subsequent
hour. Write down a equation and draw the graph. Find the following charges from the graph (1) For four
hours ( 2)For seven hours.​

Answer 1

40 is divided by 8 so equal is 5

1hour-60min

2hour-60Multipled by 2 so exual is 120

2hour-120min

so, A.T.Q

8rs-120divided by 5 so exual is 24

so,24min is answer

Answer 2

Basic Concept Used :-

Writing System of Linear Equation from Word Problems

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.

$\large\underline{\sf{Solution-}}$

Let suppose that

• Number of hours for which car is parked = 'x' hours

and

• Total fare of parking = Rs 'y'.

So,

According to statement,

• The charges for first two hours = Rs 40

and

• The charges for subsequent hours = Rs 8

This implies,

• Charges for first two hours = = Rs 40

Now,

• As total number of hours = x

• Remaining hours is now = x - 2 hours

So,

• Charges for (x - 2) hours = 8 × (x - 2) = Rs 8x -16

So, Total fare for parking car is

$\rm :\longmapsto\:y = 40 + 8x - 16$

$\bf\implies \:y \: = \: 8x + 24 \: where \: x \geqslant 2$

1. Substituting 'x = 2' in the given equation, we get

$\rm :\longmapsto\:y = 8 \times 2 + 24$

$\rm :\longmapsto\:y = 16 + 24$

$\bf\implies \:y = 40$

2. Substituting 'x = 3' in the given equation, we get

$\rm :\longmapsto\:y = 8 \times 3 + 24$

$\rm :\longmapsto\:y = 24 + 24$

$\bf\implies \:y = 48$

3. Substituting 'x = 4' in the given equation, we get

$\rm :\longmapsto\:y = 8 \times 4 + 24$

$\rm :\longmapsto\:y = 32 + 24$

$\bf\implies \:y = 56$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 40 \\ \\ \sf 3 & \sf 48 \\ \\ \sf 4 & \sf 56 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (2 , 40), (3 , 48) & (4 , 56)

➢ See the attachment graph.

From graph we concluded that,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf time(hrs) & \bf fare \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 56 \\ \\ \sf 7 & \sf 80 \end{array}} \\ \end{gathered}$