a LOT easier than I first thought, but that's what I suppose the question was testing, comprehension, etc. A simple ratio would solve the problem.
Time is given as a ratio: \(2x:3x:5x\), sum \(10x\)
Work is given in hours: \(110\) hours
Thus, each person works \(11\) times their time ratio
Hardest working (most hours presumably) - person who has least amount of hours =
\(55 - 22 = 33\)
OR, differece of two ratios \(5-2 = 3\). Ratio \(3x * 11 = 33\). Since the integrity of the ratio is maintained, this method works too.
Thus, C.
Suppose back solving would be another approach, less efficient, mind you.
A.
55 hours, impossible since the range of the ratio limits the multiple to 3xB.
44 hours, impossible since the range of the ratio limits the multiple to 3xC. 33
hours, most correct answerD.
22 hours, trick answer. If range is sought, then the range must not be less than 3x.E.
11 hours, trick answer. If range is sought, then the range must not be less than 3x.Thanks,