A multi-storey car park has eight levels.
On the top seven levels there are eight rows of parking. Two of these rows hold 15 cars each whilst the others hold 10 cars each.
On the road level there are two rows holding 15 cars but only four rows holding 10 cars each.
The entry control system counts cars in and out. The system stops admitting cars once 90% of the total capacity is in use.
Four spaces on the road level are reserved for staff parking and these are not available to the public.
What is the maximum number of public cars which can be admitted?
A. 644
B. 500
C. 696
D. 630
E. 626
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ColeF1
2
Simple steps to solve word problems:
- Underline key information
- Determine what they are trying to ask, and what you will need to solve it
- Eliminate any non-essential information
- Draw a picture, graph, or equation
- In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.
- Solve.
What do we want: to find the number of public cars that can be admitted into this car park.
Givens:
On top 7 levels there are 8 rows of parking
2 rows x 15 cars
8 - 2 rows with 10 cars each
Bottom: only 4 rows with 10 cars but still 2 rows with 15
90% = cutoff
- 4 spots from employee parking
Solve:
2 rows x 15 cars = 30 cars
(8 - 2 rows)(10 cars/row) = 60 cars
Max cars per level (top 7) = 60 + 30 = 90
Bottom level = (15 spots)(2 rows) + (10 spots)(4 rows) = 70 spots
Total spaces = (90 spaces)(7 levels) + 70 bottom level spots = 700 spots total
(Note that (90)(7) is not the most comfortable so practice multiplication (ex. 9 times table).
Now we need to find what 90% capacity of 700 spots are:
(700 spots)(0.10) = 70 spots.
(We use 0.10 to find 10% of 700 and then subtract that from 700).
So we now know that the parking system will stop admitting cars after 700 - 70 = 630 cars. We cannot forget about the 4 employee spots. So:
630 - 4 = 626
\fcolorbox{red}{grey!30}{Therefore E is the correct answer.}
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