Question

Gibson's Bodywork does automotive collision work. An insurance agency has determined that the standard time to replace a fender is 2.5 hours​ (i.e., "standard ​output"equals0.4 fenders per​ hour) and is willing to pay Gibson ​$75 per hour for labor​ (parts and supplies are billed​ separately). Gibson pays its workers ​$75 per hour. Suppose​ Gibson's workers take 9 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

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

A. Gibson labor hour efficiency = (actual output/ standard output) *100

Where actual output of labor = 1/4 = 0.25 per hour

And standard output of labor = 1/2.5 = 0.4 per hour

Therefore

Gibson labor hour efficiency = (0.25/ 0.4) *100 = 62.5%

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

The insurance agency pays $50 per hour for a labor, therefore for 2.5 hours of work they pay 2.5*$50 = $125 per labor

Gibson pays to his workers $35 per hour for each labor for 4 hours work, therefore he pay total 4 *$35 = $140 per labor

But Gibson losing $140 -$125 = $15 per labor

Now for Gibson to break even on the job have to pay his labor only $125 and need $125/$35 = 3.57 hours to replace a fender or actual output of labor = 1/3.57 = 0.28 per hour

Gibson's labor hour efficiency has to be = (0.28/ 0.4) *100 = 70% for Gibson to break even on the job