SOLUTION: a baseball team held a fundraising car wash. They charged $4 for each car and $6 for each truck. They washed 50 cars and trucks and collected $210. How many of each type of vehicle

Question 1208864: a baseball team held a fundraising car wash. They charged $4 for each car and $6 for each truck. They washed 50 cars and trucks and collected $210. How many of each type of vehicle did they wash?
Found 3 solutions by josgarithmetic, timofer, greenestamps:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
c number of care
t number of trucks

The description becomes this system:

You could simplify the revenue equation.

You can finish that...
Answer by timofer(104)   (Show Source): You can put this solution on YOUR website!
One single variable, x trucks
50-x cars

revenue collected, $210.

Five trucks and 45 cars
Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

Here are a couple of non-algebraic methods for solving the problem.

(1) by logical reasoning and simple arithmetic....

If all 50 vehicles had been cars, the amount they would have collected is 50($4) = $200
but the actual amount was $10 more than that
they collect $2 more for each truck than for each car
the number of trucks they had to wash to make the additional $10 is $10/$2 = 5.

ANSWER: 5 trucks (and 50-5 = 45 cars)

(2) as an unusual type of "mixture" problem....

You are "mixing" cars at $4 each and trucks at $6 each and getting an average of $210/50 = $4.20

Use a number line if it helps to see that $4.20 is 1/10 of the way from $4 to $6.

That means 1/10 of the total 50 vehicles were trucks.

ANSWER (again, of course): 5 trucks and 45 cars

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