Chapter 6.3, Problem 48STP

To determine

To calculate: how long the parking lane will be if 1 fire truck and 5 ambulances placed

Answer to Problem 48STP

1 fire truck and 5 ambulances will acquire 105 feet distance

Explanation of Solution

Given information:

3 fire trucks and 4 ambulances can fit into a parking lane 152 ft long

2 fire trucks and 5 ambulances can fit into a parking lane 136 ft long

The space between each vehicle is 1 ft

  

Calculation:

Let, x represents the length occupied by the one fire truck and y represents the length occupied by the one ambulance

According to the question:

  $\begin{array}{l}3x+4y+6=152\\ 2x+5y+6=136\end{array}$

Here, 6 in both equations represent the 1 ft distance 6 times between the vehicles

On simplifying the equations:

  $\begin{array}{l}3x+4y=146\mathrm{...}\left(i\right)\\ 2x+5y=130\mathrm{...}\left(ii\right)\end{array}$

Multiply equation (i) by 2 and equation (ii) by 3

  $\begin{array}{l}6x+8y=292\mathrm{...}\left(iii\right)\\ 6x+15y=390\mathrm{...}\left(iv\right)\end{array}$

Subtract equation (iii) by (iv)

  $\begin{array}{l}\left(6x+8y\right)-\left(6x+15y\right)=292-390\\ -7y=-98\\ y=14\end{array}$

Put the value of y in equation (i)

  $\begin{array}{l}3x+4\left(14\right)=146\\ 3x+56=146\\ 3x=146-56\\ 3x=90\\ x=30\end{array}$

1 fire truck and 5 ambulance equation will be:

  $\begin{array}{l}x+5y+5=30+5\left(14\right)+5\\ x+5y+5=30+70+5\\ x+5y+5=105ft\end{array}$