SOLUTION: Two mechanics worked on a car. The first mechanic charged 105 per hour, and the second mechanic charged 85 per hour. The mechanics worked for a combined total of 20 hours, and toge

Question 1204237: Two mechanics worked on a car. The first mechanic charged 105 per hour, and the second mechanic charged 85 per hour. The mechanics worked for a combined total of 20 hours, and together they charged a total of 1800. How long did each mechanic work?

Found 3 solutions by ikleyn, MathLover1, greenestamps:
Answer by ikleyn(52891)   (Show Source): You can put this solution on YOUR website!
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Two mechanics worked on a car. The first mechanic charged 105 per hour,
and the second mechanic charged 85 per hour. The mechanics worked for a combined total of 20 hours,
and together they charged a total of 1800. How long did each mechanic work?
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Let x = hours of the first mechanics; y = hours of the second mechanics. Equations x + y = 20 hours (1) combined time 105x + 85y = 1800 dollars (2) combined dollars To solve, express y = 20-x from equation (1) and substitute it into equation (2). You will get then 105x + 85(20-x) = 1800 105x + 1700 - 85y = 1800 105x - 85y = 1800 - 1700 20x = 100 x = 100/20 = 5. ANSWER. First mechanics worked 5 hours; second mechanics worked 20-5 = 15 hours. CHECK. 105*5 + 85*15 = 1800 dollars total combined. ! correct !

Solved.

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

the first mechanic charged per hour, cost of his work is where is number of ours
and the second mechanic charged per hour, cost of his work is where is number of ours
if the mechanics worked for a combined total of hours, we have
....solve for
.....eq.1

and, if together they charged a total of , we have
......eq.2, substitute from eq.1
....solve for

go to eq.1

the mechanic who charged per hour, worked hours
the mechanic who charged per hour, worked hours

Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!

The two responses you have received show basically the same solution, using two equations with two unknowns solved using substitution.

The other standard algebraic way of solving a system of two equations in two unknowns is elimination.

Let x be the number of hours worked at $85 per hour and y be the number of hours worked at $105 per hour. Then

the total time was 20 hours:
the total charge was $1800:

Simplify the second equation by dividing by the greatest common factor, 5.

Multiply the first equation by 17:

Eliminate x by subtracting one equation from the other.

Use y=5 in the first equation to find x=15.

ANSWER: The mechanic who charges $85 per hour worked 15 hours on the car; the mechanic who charged $105 per hour worked 5 hours.

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Here is a quick and easy informal solution that uses exactly the same calculations as the formal elimination method above.

If all 20 hours were by the mechanic who charged $85 per hour, the total charge would be $1700. The actual charge was $100 more than that. Since the second mechanic charged $20 more per hour than the first, the number of hours he worked was $100/$20 = 5; so the first mechanic worked 15 hours.

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And here is another, very different, informal way of solving any 2-part "mixture" problem like this.

The average charge per hour was $1800/20 = $90. Use a number line if it helps to observe/calculate that $90 is one-fourth of the way from $85 to $105. That means the mechanic who charged $105 per hour worked one-fourth of the total of 20 hours -- i.e., 5 hours; meaning the other mechanic worked 15 hours.

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