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15 Nov 2021, 03:45
Kudos
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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:
85%
(hard)
Question Stats:
49% (02:48) correct
51%
(02:31)
wrong
based on 103
sessions
History
Date
Time
Result
Not Attempted Yet
Bob, David, and Edward ran a 17.5 miles race. The combined running time of Bob and David exceeded twice Edward's running time by exactly an hour. None of the three ran at a speed exceeding 10 miles per hour. The winner of the race can be
(I) Bob
(II) David
(III) Edward
A. Only II
B. Only I
C. Only I or II
D. Only III
E. I or II or III
(I) Bob
(II) David
(III) Edward
A. Only II
B. Only I
C. Only I or II
D. Only III
E. I or II or III
16 Nov 2021, 20:12
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Bunuel
Since the speed is less than 10 miles/ hr. Time taken will be more than 1.75 hr for each person
Let us assume that Edward takes 3 hr to complete the race. Time taken for Bob and David combined will be 7 hr. Bob can take 2 hr, David can take 5 hr or vice versa. In that case , either of the 2 can be the winner. Bob and David both can take 3.5 hr each. In this case, Edward will be the winner. Anyone can be the winner
Option E
17 Nov 2021, 08:48
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Bunuel
We can calculate that everybody took longer than \(\frac{17.5 \text{ miles}}{10 \text{ mi/h}} = 1.75 \text{ hours}\) first.
The relation given is \(t_{Bob} + t_{David} = 2*t_{Edward} + 1\). By plugging in the minimum time, we also know \(t_{Bob} + t_{David} > 4.5\).
Edward could possible be the fastest by splitting the total time between Bob and David equally, hence \(t_{Bob} = t_{David} = \frac{1}{2}(2*t_{Edward} + 1) > t_{Edward} \).
We may also let either Bob or David have a faster time than Edward and let the remaining person take the rest of the time. For example, \(t_{Edward} = 2, t_{Bob} \text{ or } t_{David} = 1.9 \text{, and } t_{David} \text{ or } t_{Bob} = 3.1\).
Then anyone could have been the fastest.
Answer: E
