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# During a trip, Don drove a total of x miles. His average

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During a trip, Don drove a total of x miles. His average [#permalink]

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23 Jan 2008, 13:18
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During a trip, Don drove a total of x miles. His average speed on a certain 5 mile section was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A)8,5%
B)50%
C)x/12%
D)60/x%
E)500/x%
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23 Jan 2008, 19:43
You can right out the formula (ala Walker ) or you can pick simple numbers and do it that way. I prefer the number approach:

Let X = 60

60 miles at 60mph = 1 hour for trip at constant speed

Other trip would be 55 miles at 60mph + 5 miles at 30mph.
55 miles at 60 = 55 minutes
5 miles at 30 = 10 minutes
Total Trip: 65 minutes

65/60 = 13/12

The travel time is 1/12 greater which is equal to 8.33%

Of the answer choices only E is equal to 8.33% because 500/60 = 8 (to get to 480) and then 20/60 = 1/3.

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23 Jan 2008, 20:05
eschn3am wrote:
You can right out the formula (ala Walker ) or you can pick simple numbers and do it that way. I prefer the number approach:

Let X = 60

60 miles at 60mph = 1 hour for trip at constant speed

Other trip would be 55 miles at 60mph + 5 miles at 30mph.
55 miles at 60 = 55 minutes
5 miles at 30 = 10 minutes
Total Trip: 65 minutes

65/60 = 13/12

The travel time is 1/12 greater which is equal to 8.33%

Of the answer choices only E is equal to 8.33% because 500/60 = 8 (to get to 480) and then 20/60 = 1/3.

I did this same approach, but I said x=120.

For E 500/120 doesnt work. I still get 2.16667-2.083333 = .08333....

hrmmmmmm ?_?
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23 Jan 2008, 21:28
GMATBLACKBELT wrote:
eschn3am wrote:
You can right out the formula (ala Walker ) or you can pick simple numbers and do it that way. I prefer the number approach:

Let X = 60

60 miles at 60mph = 1 hour for trip at constant speed

Other trip would be 55 miles at 60mph + 5 miles at 30mph.
55 miles at 60 = 55 minutes
5 miles at 30 = 10 minutes
Total Trip: 65 minutes

65/60 = 13/12

The travel time is 1/12 greater which is equal to 8.33%

Of the answer choices only E is equal to 8.33% because 500/60 = 8 (to get to 480) and then 20/60 = 1/3.

I did this same approach, but I said x=120.

For E 500/120 doesnt work. I still get 2.16667-2.083333 = .08333....

hrmmmmmm ?_?

It will work for any number X. Let's work through it using 120 because that's easy to work with.

Let X = 120

120 miles at 60mph = 2 hours for trip at constant speed (or 120 minutes)

Other trip would be 115 miles at 60mph + 5 miles at 30mph
115 miles at 60 = 115 minutes
5 miles at 30 = 10 minutes
Total Trip: 125 minutes

125/120=25/24

The travel time is 1/24 greater, which is equal to 4.1667% or 4 + 1/6

500/120 = 4 + 20/120 = 4+1/6 = 4.1667%

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23 Jan 2008, 21:47
netcaesar wrote:
During a trip, Don drove a total of x miles. His average speed on a certain 5 mile section was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A)8,5%
B)50%
C)x/12%
D)60/x%
E)500/x%

Let say t5 =time travelling 5 mile at 30mph
t(x-5) = time travelling the rest of the trip at 60mph
tx = time travelling all the trip at 60 mph
t5 + t(x-5) = 5/30 + (x-5)/60 = (x+5)/60
tx = x/60

% greater = {(x+5)/60 - x/60}/[x/60] = 500/x % = E
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24 Jan 2008, 08:23
eschn3am wrote:
GMATBLACKBELT wrote:
eschn3am wrote:
You can right out the formula (ala Walker ) or you can pick simple numbers and do it that way. I prefer the number approach:

Let X = 60

60 miles at 60mph = 1 hour for trip at constant speed

Other trip would be 55 miles at 60mph + 5 miles at 30mph.
55 miles at 60 = 55 minutes
5 miles at 30 = 10 minutes
Total Trip: 65 minutes

65/60 = 13/12

The travel time is 1/12 greater which is equal to 8.33%

Of the answer choices only E is equal to 8.33% because 500/60 = 8 (to get to 480) and then 20/60 = 1/3.

I did this same approach, but I said x=120.

For E 500/120 doesnt work. I still get 2.16667-2.083333 = .08333....

hrmmmmmm ?_?

It will work for any number X. Let's work through it using 120 because that's easy to work with.

Let X = 120

120 miles at 60mph = 2 hours for trip at constant speed (or 120 minutes)

Other trip would be 115 miles at 60mph + 5 miles at 30mph
115 miles at 60 = 115 minutes
5 miles at 30 = 10 minutes
Total Trip: 125 minutes

125/120=25/24

The travel time is 1/24 greater, which is equal to 4.1667% or 4 + 1/6

500/120 = 4 + 20/120 = 4+1/6 = 4.1667%

DOH!!!! arghhh mistake from my notes: I meant 2 and 1/12 hrs - 2hrs --> .083333 hrs/2hrs = .0416667

I was takin the .08333 only... forgot to divide by 2.

BAH! Thx.
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24 Jan 2008, 08:34
x - distance

if drive like described time is 5/30 + (x-5)/60 = (x+5)/60 = x/60 + 1/12

if constant 60, x/60

Extra travel time = 1/12, percentage of x /60 = (1/12)/(x/60) = 5/x -> E
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24 Jan 2008, 10:14
You are right.

OA is E
Re: PS: Distances   [#permalink] 24 Jan 2008, 10:14
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