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09 Feb 2022, 10:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:50) correct
30%
(01:39)
wrong
based on 148
sessions
History
Date
Time
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Not Attempted Yet
Suzy estimated both the distance of her trip to her hometown, in miles, and the average speed, in miles per hour. Was the estimated time within 30 minutes of the actual time of the trip?
(1) Suzy’s estimate for the distance was within 10 miles of the actual distance.
(2) Suzy’s estimate for her average speed was within 5 miles per hour of her actual average speed.
(1) Suzy’s estimate for the distance was within 10 miles of the actual distance.
(2) Suzy’s estimate for her average speed was within 5 miles per hour of her actual average speed.
17 Feb 2022, 11:34
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Bunuel
Breaking Down the Info:
We are mentioning distance and time, so we should prepare the formula \(d = v*t\) first. We are asking for time, so write this as \(t = \frac{d}{v}\). We need to know something about d and v first then.
Statement 1 Alone:
Nothing on \(v\), insufficient.
Statement 2 Alone:
Nothing on \(d\), insufficient.
Both Statements Combined:
Combined, we can write the new t = \(\frac{d \pm 10 }{ v \pm 5}\). If \(d\) and \(v\) are both big enough compared to 10, then the \(\pm\) parts won't make much of a difference, and the new t will not be much different from the old t. Similarly, we could have \(v\) being very small, so that the \(\pm 5\) will change the value of \(\frac{d}{v}\) by a lot. Therefore both cases are present, and combined it would be insufficient.
Answer: E
kungfury42
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17 Feb 2022, 12:48
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Bunuel
Agree with Jerry's solution. In fact, just like it happened with me, it might not be very clear upfront to a lot of people where does \(\frac{d±10}{s±5}\) lie with regard to \(\frac{d}{s}\) and if it can be determined at all.
I believe taking a practical case right off the bat can help arrive at the answer much quickly. The key is to test for time taken by the opposite extremes of the D and S values.
For ex: D = 100km, S = 40kmph and T = 2.5hrs
D extremes are 90 and 110
S extremes are 35 and 45
Test for 90 and 45, T=90/45=2
Test for 110 and 35, T=110/35=3.14
As we can see 2 lies within 30 mins of 2.5hrs but 3.14 lies outside of 30 mins of 2.5hrs, hence, even using both the statements we cannot arrive at the answer, and option E it is.
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