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A bus trip of 450 miles would have taken 1 hour less if the

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by BTGmoderatorDC » Tue Jul 24, 2018 5:31 pm
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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
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by [email protected] » Tue Jul 24, 2018 9:03 pm
Hi All,

We're told that 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. We're asked for the average speed S, in miles per hour, for the trip. This question can be solved in a couple of different ways, including by TESTing THE ANSWERS.

To start, it's worth noting that increasing the speed by 5 miles/hour would decrease the travel time by EXACTLY 1 HOUR. This implies that the original speed and the increased speed are both factors of 450 (which is why the decrease in time is a nice 'round' number: 1 hour). Thus, we should look to TEST the answers that are divisors of 450. Let's TEST Answer D first...

Answer D: 50 miles/hour
At 50 miles/hour, the trip would take 450/50 = 9 hours
At 55 miles/hour, the trip would take 450/55 = something between 8 hours and 9 hours. This would NOT lead to a difference of 1 hour (and notice how 55 is NOT a divisor of 450). However, 45 IS a divisor.... so let's TEST answer C next...

Answer C: 45 miles/hour
At 45 miles/hour, the trip would take 450/45 = 10 hours
At 50 miles/hour, the trip would take 450/50 = 9 hours
This is an exact different of 1 hour, so this must be the answer.

Final Answer: C

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by GMATGuruNY » Wed Jul 25, 2018 1:55 am
BTGmoderatorDC wrote: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
Since the time difference yielded by the two speeds is an INTEGER -- 1 hour less -- the actual speed and hypothetical greater speed must both divide evenly into the 450-mile distance.

We can PLUG IN THE ANSWERS, which represent the actual speed.
Since A, B and E do not evenly into 450, eliminate A, B and E.
The hypothetical greater speed is 5 miles per hour greater than the actual speed.
D implies that the hypothetical greater speed = 50+5 = 55.
Since 55 does not divide evenly into 450, eliminate D.

The correct answer is C.
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by Brent@GMATPrepNow » Wed Jul 25, 2018 4:59 am
BTGmoderatorDC wrote: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
Let's start with a word equation:
travel time at actual speed = travel time at faster speed + 1 hour
In other words: travel time at S mph = travel time at (S + 5) mph + 1 hour

travel time = distance/speed
So, we get: 450/S= 450/(S + 5) + 1
Multiply both sides by S to get: 450 = 450S/(S+5) + S
Multiply both sides by S+5 to get: 450(S + 5) = 450S + S(S+5)
Expand: 450S + 2250 = 450S + S² + 5S
Subtract 450S from both sides: 2250 = S² + 5S
Rewrite as: S² + 5S - 2250 = 0
Factor: (S + 50)(S - 45) = 0
So, EITHER S = 50, OR S = 45
Since the speed can't be negative, the correct answer must be S = 45

Answer: C

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by Jeff@TargetTestPrep » Thu Jul 26, 2018 3:25 pm
BTGmoderatorDC wrote: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
Let t = the time to complete the trip of 450 miles when the average speed is S. Thus, we have:

St = 450

and

(S + 5)(t - 1) = 450

Isolating t in the first equation, we have: t = 450/S. Substituting this in the second equation, we have:

(S + 5)(450/S - 1) = 450

450 - S + 2250/S - 5 = 450

-S + 2250/S - 5 = 0

S + 5 - 2250/S = 0

S^2 + 5S - 2250 = 0

(S + 50)(S - 45) = 0

S = -50 or S = 45

Since S can’t be negative, S = 45.

Answer: C

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