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by okigbo » Wed Oct 21, 2009 11:32 pm
5) If it took Carlos 1/2hr from home to library, is distance greater than 6 miles? (1mile = 5280 feet)

a. Average speed was greater than 16 feet/second
b. Average speed was less than 18 feet/second
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by briantime » Wed Oct 21, 2009 11:45 pm
R*T = D

T = 30min = 30 * 60 seconds = 1800 seconds

The question askes if D > 6 miles (= 6*5280 feet)

Calculate the rate for 6miles in 30minutes.
R = D/T
R = 6*5280/1800 = 5280 / 300 = 17.6

If his rate was > 17.6, he would have walked a longer way and therefore the distance would be > 6 miles.
If his rate was < 17.6, he would have walked a shorter way and therefore the distance would be < 6 miles

(1) R > 16:
This could for example be 17, which would mean that D was < 6. But it could also mean that his rate was 18, which would mean that D was > 6.

(2) R < 18:
The same concept. This could be greater or smaller than 17.6.

(1)+(2):
Doesn't help us. For example, R could be 16.5 or 17.8.

Answer: E
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