Asked by Bridget Davis
two guys painting a barn. one of them paints twice as fast as the other. On the first day, they have worked for 6 h and completed 1/3 of the job, when the 2nd guy gets hurt. If the 1st guy has to complete by himself how many additional hours will it take?
Answers
Answered by
Henry2
1/(1/3) * 6h = 18h = Time for 2 to do the full job.
T*2T/(T+2T) = 18.
2T^2/3T = 18,
2T/3 = 18,
2T = 54 Hours.
T = 27 Hours = Time for the faster painter to do the full job alone.
2/3 * 27 = 18 h = Time for faster painter to do the final 2/3 of the job alone.
T*2T/(T+2T) = 18.
2T^2/3T = 18,
2T/3 = 18,
2T = 54 Hours.
T = 27 Hours = Time for the faster painter to do the full job alone.
2/3 * 27 = 18 h = Time for faster painter to do the final 2/3 of the job alone.
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