Bunuel wrote:
A painter was paid $800 as labor cost for painting the exterior of the front of a house. This amount was the painter’s estimated labor cost based on a regular rate of R, in dollars per hour, for an estimated time T, in hours. However, the actual time it took for the painter to do the job was 4 hours longer than the estimated time, thereby reducing the hourly rate for the job by $10 per hour. What is R, the painter’s regular rate, in dollars per hour?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60
Answer choicesFor each answer's rate and decreased rate, find time taken.
The difference in hours should be 4.
Rate * Time = Pay, P
Thus, Time,
\(T = \frac{P}{R}\)Start with C) $40 per hour (estimated). $30 per hour (actual)
Time at $40:
\(\frac{$800}{$40per.hr}= 20\) hours
Time at $30:
\(\frac{$800}{$30per.hr}=\frac{80}{3}= 26.xx\) hours
800/30 is not an integer. I calculated to find out
whether $40 per hour = too high or too low
6.xx extra hours. Too great. Needs to decrease.
R and T are inversely proportional
(one increases, the other decreases)
T needs to come down. R must go up.
$40 per hour is too low
Try D) $50 per hour (estimated). $40 per hour (actual)
Time at $50:
\(\frac{800}{50} = 16\) hours
Time at $40, from above =
\(20\) hours
Difference? 4 hours. That's correct.
Answer D
Algebra(Hourly pay rate) * (# hours worked) = Total pay in dollars
Same basis as
\(RT = W\)\(RT = W\), so \(T = \frac{W}{R}\)W = 800
Estimated rate =
\(r\)Actual rate =
\(r-10\)Estimated time =
\(\frac{800}{r}\)Actual time =
\(\frac{800}{r-10}\)To find rate, \(r\), use time, which is defined in terms of \(r\)
(Actual time) - (estimated time) = \(4\) hours\(\frac{800}{r-10} - \frac{800}{r} = 4\)
\(\frac{800r - 800(r-10)}{r(r-10)} = 4\)
\(800r - 800r + 8000 = 4* r(r-10)\)
\(8000 = 4 * r(r-10)\)
\(2000 = r^2 - 10r\)
\(r^2 - 10r - 2000 = 0\)
\((r - 50)(r + 40) = 0\)
\(r = 50\)
Answer D
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