MathRevolution wrote:
[GMAT math practice question]
(Speed) The distance between Jane’s home and her school is \(24km\). It takes \(4\) hours and \(50\) minutes for Jane to walk from home to school and it takes \(5\) hours to come back. The road consists of an uphill section, a downhill section and a flat section. How long is the flat section?
1) The speed on the uphill section is \(4km/hr\), on the downhill section is \(6km/hr\) and on the flat section is \(5km/h.\)
2) The speed on the flat section is the arithmetic average of the speeds on the uphill and downhill sections.
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Assume \(f, u\) and \(d\) are distances on the flat sections, uphill sections and downhill sections when she travels from home to school.
Then we have\( f + u + d = 24.\)
Even though we have \(3\) variables and \(1\) equation, condition 1) has \(3\) equations. So we should check condition 1) first.
Condition 1)
The time she travels from home to school is \(\frac{f}{5} + \frac{u}{4} + \frac{d}{6} = 4(\frac{50}{60}) = \frac{29}{6}.\)
The time she travels from school to home is \(\frac{f}{5} + \frac{u}{6} + \frac{d}{4} = 5\) since uphill sections becomes downhill sections and downhill sections becomes uphill when she travels back.
Adding the equations together, we have \((\frac{2}{5})f + \frac{(5u + 5d)}{12} = (\frac{2}{5})f + (\frac{5}{12})(u+d) = \frac{29}{6} + 5 = \frac{59}{6}.\)
Substituting in \(u + d = 24 – f,\) we have \((\frac{2}{5})f +(\frac{5}{12})(24 - f) = \frac{59}{6}, (\frac{2}{5})f + 10 - (\frac{5}{12})f = \frac{59}{6},\) or \((\frac{24}{60})f - (\frac{25}{60})f = \frac{59}{6} - \frac{60}{6}.\) Then we have \((\frac{-1}{60})f = \frac{-1}{6}\), or \(f = 10.\)
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Since the speed on the flat section is the arithmetic average of the speeds on the uphill and downhill sections, we can assume that s is the speed on the flat sections, \(s – a\) is the speed on the uphill sections and \(s + a\) is the speed on the downhill sections.
Then we have the time she takes when she travels from home to school, \(\frac{f}{s} + \frac{u}{(s-a)} + \frac{d}{(s+a)} = \frac{29}{6}\) and we have the time she takes when she travels from home to school, \(\frac{f}{s} + \frac{u}{(s+a)} + \frac{d}{(s-a)} = 5.\)
When we add those equations, we have \((\frac{2f}{s}) + (u+d)(\frac{1}{(s-a)}+\frac{1}{(s-b)}) = \frac{59}{6}.\)
We can notice that there must be many possibilities for solutions.
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, A is the answer.
Answer: A
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C or E.
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