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

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]
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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Expert Reply
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...

Final Answer:

GMAT assassins aren't born, they're made,
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Re: A bus trip of 450 miles would have taken 1 hour less if the average [#permalink]
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.


I solved it by testing the ans choices.

We are told D = 450 , Original speed = S and Time = T

We are als told the 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. So it means

450 = Speed * Time

Lets start testing and choice from C

Speed is 45 so if speed was 45 and D= 450 then time would be 10 as D=s*t

So now that we know the Time

We can also check if 450 = 45+5 * 10 -1 = 50*9 = 450 True . Ans = C
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Re: A bus trip of 450 miles would have taken 1 hour less if the average [#permalink]
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.



Okay I wouldn't have posted the solution but correct time for answer this one (so far) is above 2 minutes so here is the approach.

USE OPTIONS!

A. 450/10 = 45. 450/15=30 - No
B. 450/40 - Well, Let's just leave it here.
C. 450/45=10. 450/50=9. That's it.

You can test D and E to be more confident.

30 seconds approach.
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A bus trip of 450 miles would have taken 1 hour less if the average [#permalink]
BrentGMATPrepNow wrote:
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


Hello, why do we have to add 1 in the first equation? Shouldn't we subtract? Isn't the same distance reached while traveling 1 hour less, but 5 mph faster?
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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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Beko wrote:
BrentGMATPrepNow wrote:
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


Hello, why do we have to add 1 in the first equation? Shouldn't we subtract? Isn't the same distance reached while traveling 1 hour less, but 5 mph faster?


When you increase your speed, your travel time decreases.
So, (travel time at actual speed) > (travel time at faster speed)
So, to create an equation we must add 1 hour to the lesser value (i.e., travel time at faster speed)

Does that help?
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