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

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A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 18 Aug 2015, 01:19
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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

Kudos for a correct solution.

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Re: A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 18 Aug 2015, 01:53
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Bunuel 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

Kudos for a correct [i]solution.[/i]



(S+5)(T-1) = 450
S*T = 450

Solving both the equations, we get: S = 45 or -50
Since Speed should be positive, S = 45

Hence, option C
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Re: A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 18 Aug 2015, 04:44
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Bunuel 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

Kudos for a correct solution.

Per the question,

\(\frac{450}{s}\) - \(\frac{450}{s+5}\) = 1

Instead of solving, use the options.

Choose s= 45 and see if the statement is satisfied and it does. So C is the correct answer.
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Re: A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 18 Aug 2015, 21:29
Bunuel 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

Kudos for a correct solution.


IMO : C

Avg Speed (s) = \(\frac{Total Distance Traveled(d)}{Total Time taken(t)}\)

Given s = \(\frac{450}{t}\)
s*t = 450 --(i)

s-5 = \(\frac{450}{t-1}\)

(s-5)*(t-1) = 450 --(ii)

Solving equations (i) & (ii)
we get s = 45 and s = -50
Since speed cannot be negative.
s = 45
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A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 12 Oct 2016, 19:25
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

if S*T=450 miles, then either S or T is a multiple of 9
45 is the only multiple of 9 given as choice for S
450/45mph=10 hours=T
checking: (45+5)(10-1)=50*9=450 miles
45 fits
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Re: A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 13 Oct 2016, 02:04
Bunuel 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

Kudos for a correct solution.


Quote:
450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour.

So, S must completely divide 450

Hence among the given options only options (A) , (C) & (D) is possible...

Further ( S + 5 ) must completely divide 450 , hence among (A) , (C) & (D) only options (A) & (C) is possible...

Now, check option (A) or option (C) , I am taking option (A) -

450/10 = 45
450/15 = 30

This doesn't hold good, so answer will be (C)....

Check option (C) to be double sure -

450/45 = 10
450/50 = 9

See difference is 1 hour, this is definitely our answer...


Hence Answer will be (C)
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Re: A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 05 Jan 2018, 17:00
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Hi All,

This particular prompt has a built-in shortcut that you can take advantage of. Since the total Distance is 450 miles - and the DIFFERENCE in TIMES is exactly 1 hour, we will almost certainly need two different speeds that BOTH divide evenly into 450... and that DIFFER by 5 miles/hour and would lead to a 1 hour difference in time-traveled. Looking at the answer choices, the only possibility that stands out would be if the speeds were 45 and 50 (since both of those values divide evenly into 450 and differ by 5). If we 'TEST' those speeds against that distance, we find....

450 mi = (45 mi/hour)(10 hours)
450 mi = (50 mi/hour)(9 hours)

This is an exact match for what we were told. We're asked for the slower speed...

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Re: A bus trip of 450 miles would have taken 1 hour less if the average  [#permalink]

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New post 25 Jul 2018, 06:04
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Bunuel 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

Kudos for a correct solution.


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

Cheers,
Brent
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Re: A bus trip of 450 miles would have taken 1 hour less if the average &nbs [#permalink] 25 Jul 2018, 06:04
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