Question

A state park charges an entrance fee based on the number of people in a vehicle. A car containing 2 people is charged 14$, a car containing 4 people is charged 20$, and a van containing 8 people is charged 32$. How much would a bus containing 30 people be charged?

Answer 1

look at the step by step

Explanation:

Answer: 98$

Step By Step explaination:

People Parking Charge ( 14$,20$,32)

2

4

8

Let x be charge per person and y be base changes. So for m people parking charges = x ( x + y )

So for m people parking charges = mx+y

When m = 2, parking charges = 14$

2x + y = 14

When m = 4, parking charges = 20$

4x + y = 20

Solve eq1 and eq2

eq(2) - eq(1)

4x + y - 2x - y = 20

2x = 6

x = 3

from eq(1)

2x3+ y = 14

y = 14 - 6

y = 8

Parking Charges = 3x + 8

where is the x number of people

When number of people = 30 => x 30

Parking Charges = 3 x 30 + 8

Parking Charges = 90 + 8

Answer: 98$

Answer 2

`\$ 98`

Explanation:

`People\ \ \ \ \ \ \ \ \ \ \ \ \ Parking\ charge\\\\ 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ 14\\\\4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ 20\\\\8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ 32`

`Let\ x\ be\ charge\ per\ person\ and\ y\ be\ base\ charges\\\\So\ for\ m\ people\ parking\ charges=Number\ of\ people* x+y\\\\So\ for\ m\ people\ parking\ charges=mx+y\\\\When\ m=2,\ parking\ charges=\$ 14\\\\2x+y=14................................................eq(1)\\\\When\ m=4,\ parking\ charges=\$ 20\\\\4x+y=20................................................eq(2)\\\\Solve\ eq1\ and\ eq2\\\\eq(2)-eq(1)\\\\4x+y-2x-y=20-14\\\\2x=6\\\\x=3\\\\from\ eq(1)\\2* 3+y=14\\\\y=14-6`

`y=8`

`Hence\ Parking\ charges=3x+8\\\\where\ x\ is\ number\ of\ people\\\\When\ number\ of\ people=30\Rightarrow x=30\\\\Parking\ charges=3* 30+8\\\\Parking\ charges=90+8\\\\Parking\ charges=\$ 98`