On Monday a guitar repairperson fixes five guitars with loose wires and four with malfunctioning pickups

Question

Doing all this takes a total of 4 hours, 29 minutes. On Tuesday this same repairperson spends a total of three hours, ten minutes fixing four guitars with loose wires and two with malfunctioning pickups. On Wednesday he has to repair seven guitars with loose wires and five with malfunctioning pickups. How long does he expect this will take? Write two (or three) algebra equations, then solve. Assume that each repair of the same type takes the same amount of time.

Answer 1

Convert to minutes = 4*60+29= 240+29 = 269 minutes
3*60+10 = 190 minutes

For Monday, 269 = 5L + 4P
For Tuesday, 190 = 4L + 2P (Let's multiply this equation by 2, then subtract the top equation from the resulting equation)


380 = 8L + 4P
269 = 5L + 4P
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111 = 3L

L = 111/3 = 37

So L = 37 minutes (for fixing loose wires)

Now let's use one of the equations to find P
4P + 8L = 380
4P + 8*37 = 380
4P + 296 = 380
P = (380 - 296) / 4 = 84 /4 = 21
So P = 21 minutes (for fixing malfunctioning pickups)

Now we can find Wednesday's time
Time for Wednesday = 7L + 5P
=7*11 + 5*21 = 77 + 105 = 182 minutes
We divide 182 by 60 to convert to hours and minutes. 182/60
182 is 3 hours, 2 minutes (Since 180 minutes is 3 hours, and 2 is the remainder)

So for Wednesday, he will work 3 hours 2 minutes.