SOLUTION: The debate team washed cars to raise money for a trip. They charged $8 for a large car and $5 for a small car. All together they raised $550 and washed 80 cars. How many of each ty

Question 1202009: The debate team washed cars to raise money for a trip. They charged $8 for a large car and $5 for a small car. All together they raised $550 and washed 80 cars. How many of each type of car did they wash?

Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
The debate team washed cars to raise money for a trip.
They charged $8 for a large car and $5 for a small car.
All together they raised $550 and washed 80 cars.
How many of each type of car did they wash?
~~~~~~~~~~~~~~~~~

Let x be the number of large cars; then (80-x) is the number of small cars. Write the total money equation (for the revenue) 8x + 5*(80-x) = 550 dollars. Simplify and find x 8x + 400 - 5x = 550 8x - 5x = 550 - 400 3x = 150 x = 150/3 = 50. ANSWER. 50 large cars and 80-50 = 30 small cars.

Solved.

Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!

CHARGE RATE QUANTITY REVENUE LARGE 8 v 8v SMALL 5 80-v 5(80-v) TOTALS 80 550

Money charged as H dollars for large and L dollars for small
total items handled, M
total revenue, R

CHARGE RATE QUANTITY REVENUE LARGE H v Hv SMALL L M-v L(M-v) TOTALS M R

Substitute your given values which are
H=8
L=5
M=80
R=550

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

If you just really, REALLY love magic formulas and don't care about actually understanding how to solve the problem, use the method shown by tutor @josgarithmetic.

If you want a typical formal algebraic solution, use something like the method shown by tutor @ikleyn.

But if a formal algebraic method is not required, and if the speed of obtaining the solution is important, then here is a quick and easy informal method for solving any problem like this mentally.

(1) If all 80 cars had been small cars, the cost would have been 80($5) = $400; but the actual cost was $550, which is $150 more.
(2) The cost for each large car was $8-$5 = $3 more than for each small car.
(3) The number of large cars, to make the additional $150, was $150/$3 = 50.

ANSWER: 50 large cars; 30 small cars

CHECK: 50($8)+30($5) = $400+$150 = $550

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