The Fort Collins city parking lot initially could park 120 large cars for a fee of $2 per day. The city council voted to establish a new
parking fee to encourage the use of smaller cars, which would help to conserve gasoline. The fee was $1 per day for small cars and $3 per day for
the larger ones.
If the smaller car only takes up 2/3 the space of the larger, how should the lot be laid out so as to maintain the lot's income at $240 per day?
In other words, how many spaces for small cars and how many spaces for large cars should be alotted?
You must show your algebra to get credit.
Solution to the Problem:
The answer is 40 large cars and 120 small cars.Let x = size of a large car.
Let y = size of a small car.
Let n = number of large car parking spaces in the new parking lot.
Let p = number of small car parking spaces in the new lot.
Then y = 2/3 x.
So the area of the original lot is 120 x.
This must be equal to the area of the new lot which is nx + py.
So, 120x = nx + py.
120x = nx + 2/3xp
Hence, 120 = n + 2/3p
We also know the costs must remain the same, so
$240 = $3n + p
Solve these two equations simultaneously to get:
p = 240 - 3n
120 = n + 2/3(240) - 3n
120 = n + 160 - 2n
So, n = 40 and p = 120.
Correctly solved by:
| 1. James Alarie | Flint, Michigan |
| 2. Ivy Joseph | Pune, Maharashtra, India |
| 3. Behrooz Ranjbar. | Iran |