In the morning, Chris drives from Toronto to Oakville and in the evening he drives back from Oakville to Toronto on the same road. Was his average speed for the entire round trip less than 100 miles per hour?
(1) In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville.
(2) In the evening, Chris drove at an average speed which was no more than 50 miles per hour while travelling from Oakville to Toronto.
OA B
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In the morning, Chris drives from Toronto to Oakville and in
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Statement 1 is obviously not sufficient.
For Statement 2, it takes just as long to drive d miles at 50 miles per hour as it does to drive 2d miles at 100 miles per hour. So if he drove half the distance at exactly 50 mph, he'd need to travel the other half instantaneously, in 0 seconds, to have an average speed of exactly 100 mph for the round trip. Since it's presumably impossible to do that, it's impossible for his average speed to be 100 mph (or greater) for the whole trip, and Statement 2 is sufficient.
If someone preferred to do this algebraically, you could also notice that, if d is the distance one-way, and t the time for the first part of the trip,
s = d/t
d/t < 50
d/50 < t
So his time one way was at least d/50, and so his time for the round trip was strictly greater than d/50. The total distance of the round trip is 2d, and since time = distance/speed, if we let s be his average speed for the round trip, we have
total time > d/50
2d/s > d/50
100d/d > s
100 > s
which is what we wanted to prove.
For Statement 2, it takes just as long to drive d miles at 50 miles per hour as it does to drive 2d miles at 100 miles per hour. So if he drove half the distance at exactly 50 mph, he'd need to travel the other half instantaneously, in 0 seconds, to have an average speed of exactly 100 mph for the round trip. Since it's presumably impossible to do that, it's impossible for his average speed to be 100 mph (or greater) for the whole trip, and Statement 2 is sufficient.
If someone preferred to do this algebraically, you could also notice that, if d is the distance one-way, and t the time for the first part of the trip,
s = d/t
d/t < 50
d/50 < t
So his time one way was at least d/50, and so his time for the round trip was strictly greater than d/50. The total distance of the round trip is 2d, and since time = distance/speed, if we let s be his average speed for the round trip, we have
total time > d/50
2d/s > d/50
100d/d > s
100 > s
which is what we wanted to prove.
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