A certain car dealership sells only Full-size and Mid-size c

He gets $15,000 in salary, so to earn $25,000, he needs $10,000 in sales.

Focus on the $800 cars. In order to minimize the total number of cars, we should maximize the number of more expensive full size cars, F -- we need fewer of them to get a high total dollar amount.

Another reason to focus in the $800 cars: divisibility. $800 and $500 are awkward together. We need the $800 car number to make a round total that is divisible by $500 for medium cars, M.

He needs exactly $10,000 from commissions.

To be a "round" number divisible by $500, $800 must be multiplied by 5, 10, or 15

$800 * 5F = $4,000
$6,000 left
$6,000/$500 per M = 12M
Total cars: 17

$800 * 10F = $8,000
$2,000 left
$2,000/$500 per M = 4M
Total cars: 14

The next number of $800F that is round is 15, which is a greater number of cars than the 14 we just calculated. Reject.

Least number of cars: 14
10 are F
4 are M

Answer D

Since the sales rep makes $15,000 per year, he needs to make exactly $10,000 in commision.

Let f = the number of full-size cars and m = the number of mid-size cars he sells. We have:

800f + 500m = 10,000

8f + 5m = 100

m = (100 - 8f)/5

Since we want the least number of cars he must sell to make the exact $10,000 commision, he must sell as many full-size cars as possible. We see that the maximum value of f can be 10 where (100 - 8f) is still a multiple of 5. Therefore, he sells 10 full-size cars and (100 - 8(10))/5 = 20/5 = 4 mid-size cars, for a total of 14 cars.


Answer: D

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