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by louvre » Tue Jul 15, 2008 6:32 am
Answer "A"?

Total time with variable speeds = (x+5)/60
Total time with constant Speed = (x/)/60

difference = 1/12 ~ 8.5%

as you see, just getting started ;)
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by gxliu » Tue Jul 15, 2008 6:35 am
Answer is actually E.
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by parallel_chase » Tue Jul 15, 2008 7:13 am
gxliu wrote:Answer is actually E.
The answer is indeed E.

Variable time = (x+5)/60
Constant time = x/60

We have to find out the percentage difference, the question already tells us that variable time is greater.

[(x+5)/60 - x/60] / x/60 => (5/x) * 100 = 500%/x

Kindly correct me if I have missed anything.

Thanks,
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by sudhir3127 » Tue Jul 15, 2008 8:21 am
answer E ...i did exactly as its done above.
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by AleksandrM » Wed Jul 16, 2008 9:58 am
parallel_chase wrote:
gxliu wrote:Answer is actually E.
The answer is indeed E.

Variable time = (x+5)/60
Constant time = x/60

We have to find out the percentage difference, the question already tells us that variable time is greater.

[(x+5)/60 - x/60] / x/60 => (5/x) * 100 = 500%/x

Kindly correct me if I have missed anything.

Thanks,
How come you add 5, whereas you should subtract the five miles already traveled from the entire distance x, to give you the remainding distance that is traveled at 60 m/h

I set it up as

First 5 miles:

rate = 30 and distance is 5 miles

The remaining trip:

rate = 60 and distance is x - 5

The second situation:

rate = 60 for the entire trip x.
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by parallel_chase » Wed Jul 16, 2008 10:15 am
AleksandrM wrote:
parallel_chase wrote:
gxliu wrote:Answer is actually E.
The answer is indeed E.

Variable time = (x+5)/60
Constant time = x/60

We have to find out the percentage difference, the question already tells us that variable time is greater.

[(x+5)/60 - x/60] / x/60 => (5/x) * 100 = 500%/x

Kindly correct me if I have missed anything.

Thanks,



How come you add 5, whereas you should subtract the five miles already traveled from the entire distance x, to give you the remainding distance that is traveled at 60 m/h

I set it up as

First 5 miles:

rate = 30 and distance is 5 miles

The remaining trip:

rate = 60 and distance is x - 5

The second situation:

rate = 60 for the entire trip x.
Well here it is:

In order to find the percentage difference you have to compare the complete distances.

for 5 miles the rate is 30 = 5/30

for x-5 miles the rate is 60 = x-5/60

add both the rates to find the rate for the complete distance i.e. x

5/30 + (x-5)/60 = (10+x-5)/60 = (x+5)/60

second situation

x/60

now find the percentage difference.

Let me know if you still have any questions.
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by notgoodinmath » Wed Jul 16, 2008 10:26 am
5/30 + (x-5)/60 = (10+x-5)/60 = (x+5)/60

quick question-how did you get (10+x-5)/60 to (x+5)/60

thanks
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by parallel_chase » Wed Jul 16, 2008 10:39 am
notgoodinmath wrote:5/30 + (x-5)/60 = (10+x-5)/60 = (x+5)/60

quick question-how did you get (10+x-5)/60 to (x+5)/60

thanks
10-5 = 5

we already have x as a variable

therefore,

(10-5+x)/60 => (x+5)/60

Hope its clear now.
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by dalwow » Mon Aug 25, 2008 6:38 pm
An easier way to solve this is just plug in a number. If we say x is 10 miles,

It would have been 10 min. + 5 min. = total of 15 min.

If it had been 60 mph the entire time it would be = total of 10 min.

15 is 50% greater than 10.

Now all we have to do is put 10 in for x in each of the options and when we get 50 that's the answer.
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