GRE 6- At a local appliance manufacturing facility,...

At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?
A) 83%
B) 80%
C) 20%
D) 17%
E) 12%

1. Assume that the worker in question, call him John, earned $50 by working 10 hours for $5 per hour.
2. After the pay-raise, the total amount earned will be the same at $50, but the hourly rate will increase by 20% to $5(1.2) = $6 per hour.
3. Using the intuition that salary=(hours)(wage), you can find that $50=(hours)($6) and therefore the number of hours worked after the change in pay will equal (50/6) = 8.33 hours
4. Since John worked 10 hours before the change in wage and he now works 8.33 hours, John reduced the number of hours he worked by (10-8.33)/10 = .1666666% = 17%
5. Note: It does not matter what numbers you use, as long as the original pay equals the post wage-change pay and the wage is increased by 20%.

or

1. Assign variables to pieces of the problem:
   Let p = the percentage of old hours worked after the raise
   if p = 95%, the worker labors 95% of the old hours after the raise (i.e., 5% fewer hours after the raise)
   Let r = hourly rate that employees were paid before the raise
   Let R = hourly rate that employees were paid after the raise
   Let n = number of hours worked before pay-raise
   Let N = number of hours worked after pay-raise
2. We know that the total pay before and after the pay-raise equal each other. The pay before was based on the rate r with hours n while the pay afterwards was based on rate R and hours N:
   rn = RN
3. The rate after the raise was 20% more than the rate before the raise:
   R = 1.20r
   Plug this value into the equation for R:
   rn = RN
   rn = 1.20rN
4. We are looking for the percentage of hours worked after the raise, so substitute N in terms of n and percent p.
   N = pn; the new number of hours equals the old number of hours multiplied by the percent of old hours worked after the raise
   rn = (1.2r)(pn).
5. Divide both sides by r and n:
   1 = 1.2p
6. Divide by 1.2 to solve for p:
   p = 1/1.2 = 10/12 = 5/6 = .8333 = 83.33%.
7. Be careful: the question asks what percent the worker would reduce his hours, so subtract from 100% to yield 16.66%, or 17%. The correct answer is D.