Problem solving help!!

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Problem solving help!!

by ramonsa » Tue Jun 02, 2009 4:00 pm
12. A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55
Thanks
Source: — Data Sufficiency |

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by ssmiles08 » Tue Jun 02, 2009 4:27 pm
Set up the rate distance time chart:

Original situation: t1 = 450/s

Hypothetical situation: t2 = 450/(s+5)

original trip takes 1 hour longer than the hypothetical trip

original = hypothetical + 1

450/s = 450/(s+5) + 1


s = 45 or s = -50

s can not be a negative number so s = 45

(C)

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by VP_Jim » Tue Jun 02, 2009 4:45 pm
This is a good candidate for backsolving, as well. Just pick an answer and see if it "works".

If the speed were 45 mph, the trip would've taken 10 hours. If the speed were 5 miles per hour greater (50 mph), the trip would've taken 9 hours. That's our answer since we need the times to differ by an hour. Had these numbers not differed by an hour, we would try another answer choice - keeping in mind whether we wanted to go up or down from 45 to eliminate some others at the same time.

Another trick: I automatically assumed the speeds were 45 and 50 mph since both of those numbers divide evenly into 450 miles. But of course, doing the calculations to confirm your suspicions is always a good idea.
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by ssmiles08 » Tue Jun 02, 2009 5:12 pm
VP_Jim wrote:
Another trick: I automatically assumed the speeds were 45 and 50 mph since both of those numbers divide evenly into 450 miles. But of course, doing the calculations to confirm your suspicions is always a good idea.
Thanks Jim...that was pretty useful now that I notice it. The quadratic equation took up quite a bit of time on this one. Back solving seems like a more reasonable approach on this problem.