11. Answer the followingThe parking charges of a car in railway statio

1.	11. Answer the followingThe parking charges of a car in railway station for first two hours is 40Rs and 8Rs for each subsequenthour. Write down a equation and draw the graph. Find the following charges from the graph (1) For fourhours ( 2)For seven hours.
Answer» 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, $\begin{gathered}\boxed{\begin{array}{c\|c} \bf time(hrs) & \bf fare \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 56 \\ \\ \sf 7 & \sf 80 \end{array}} \\ \end{gathered}$

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

Answer»

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,

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