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01 Jun 2013, 02:38
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If Danielle ran a race at a constant speed, at what time did she finish?
(1) Danielle started the race at 8:00 a.m.
(2) At 9:30 a.m. Danielle was halfway through the race, and at 10:00 a.m., she was 2/3 of the way through the race.
Please provide detailed explanation!
(1) Danielle started the race at 8:00 a.m.
(2) At 9:30 a.m. Danielle was halfway through the race, and at 10:00 a.m., she was 2/3 of the way through the race.
Please provide detailed explanation!
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01 Jun 2013, 02:43
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If Danielle ran a race at a constant speed, at what time did she finish?
(1) Danielle started the race at 8:00 a.m.
No info about the speed, or time.
Not sufficient
(2) At 9:30 a.m. Danielle was halfway through the race, and at 10:00 a.m., she was 2/3 of the way through the race.
This means that in 30 minutes (10:00-9:30) she runs \(\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\) of the race.
At 10:00 she has \(\frac{1}{3}\) left to run or \(\frac{2}{6}\) and, according to the upper calculus, she will run it in \(1\) hour (1/6 every 30 minutes).
She will finish at 11:00 am.
Sufficient
B
(1) Danielle started the race at 8:00 a.m.
No info about the speed, or time.
Not sufficient
(2) At 9:30 a.m. Danielle was halfway through the race, and at 10:00 a.m., she was 2/3 of the way through the race.
This means that in 30 minutes (10:00-9:30) she runs \(\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\) of the race.
At 10:00 she has \(\frac{1}{3}\) left to run or \(\frac{2}{6}\) and, according to the upper calculus, she will run it in \(1\) hour (1/6 every 30 minutes).
She will finish at 11:00 am.
Sufficient
B
22 Jun 2013, 10:49
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solved in 32 secs. exactly the approach by zarrolou. between 9.30 and 10 she covers 2/3 -1/2 part of the journey. Therefore we can calculate the time required to travel 1/2 of the distance and adding this time to 9.30 we get the time of finish.
