Question

Gibson's Bodywork does automotive collision work. An insurance agency has determined that the standard time to replace a fender is 0.16 hours​ (i.e., "standard ​output" equals 6.25 fenders per​hour) and is willing to pay Gibson ​$95per hour for labor​ (parts and supplies are billed​ separately). Gibson pays its workers ​$75 per hour. Suppose​ Gibson's workers take 7 hours to replace a fender.

a. ​Gibson's labor hour efficiency is _____​%. ​(Enter your response rounded to two decimal​ places.)

Given​ Gibson's labor​ costs, the company ______will /or will not make money on the job.

b. ​Gibson's labor hour efficiency has to be _____​% for Gibson to break even on the job. ​(Enter your response rounded to two decimal​ places.)

Answer 1

Answer A.
Gibson labor hour efficiency = (actual output/ standard output) *100
Where actual output of labor = 1/(Time to replace a single fender) = 1/7 = 0.143

And, standard output of labor = 1/ (Standard time to replace a fender) = 1/0.16 =6.25

Gibson labor hour efficiency = (actual output of labor) / (standard output of labor) = 0.143 / 6.25 = 2.29%

Answer b.

​Gibson's labor hour efficiency has to be _____​% for Gibson to break even on the job.

The insurance agency pays $95 per hour for a labor, therefore for 0.16 hours of work they pay 0.16 *$95 = $15.2 per labor-hour

Gibson pays to his workers $75 per hour for each labor for 7 hours work, therefore he pay total 7 *$75 = $525 per labor-hour

So Gibson is losing $525 - $15.2 = $509.8 per hour

At breakeven point, the labor cost for Gibson should be $15.2 and needs ($15.2/$75=) 0.2027 hours to replace a fender or actual output of labor = 1/0.2027 = 4.93 per hour

Gibson's labor hour efficiency has to be ((4.93/ 6.25) *100) =78.88% for Gibson to break-even on the job