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Speed and Distance problem

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Speed and Distance problem

by MI3 » Tue May 31, 2011 2:18 am
Q. Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour and the remainder at 50 miles per hour. If Martha's average speed for the entire trip is represented as a fraction in its reduced form, in terms of x, which of the following is the numerator?
A. 110
B. 300
C. 1,100
D. 3,000
E. 30,000

I got the answer as D, am I correct?

Thanks,
M.
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by Frankenstein » Tue May 31, 2011 2:41 am
Hi,
Let the total distance be 100d.
Time taken for his journey@60mph = xd/60 hours
Time taken for his journey@50mph = (100-x)d/50 hours

Average speed = total distance/total time = 100d/[(xd/60)+(100-x)d/50] = 30,000/(600-x).
So, I think E

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by sivaelectric » Tue May 31, 2011 2:42 am
Can you post the procedure you followed for solving. :)
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by cans » Tue May 31, 2011 6:12 am
let total distance = 100 miles
x miles at 60 miles per hour and thus time = x/60 hr
100-x miles at 50 miles per hour and thus time = 100-x/50 hr
total time = x/60 + (100-x)/50 = (600-x)/300
thus avg speed = distance/time = 100/[(600-x)/300] = 30000/600-x
Numerator = 30,000
IMO E
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by MI3 » Tue May 31, 2011 6:17 am
Thank you Cans and Frankenstein for the reply, but can you please explain to me, on what basis do we take the total distance as 100 or 100d?

Am clear with the rest of the procedures explained, but not clear why we are only using 100 as a possible distance (as it could be any other value as well?).

Please advise.
M
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by sivaelectric » Tue May 31, 2011 6:18 am
The distance covered is given in % and thats why for easy calculation we are considering 100 as the value. :) hope it helped
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by bblast » Tue May 31, 2011 7:41 am
yup, if i take distance as 1 and not 100. The numerator comes 300 !!

{x+(1-x)}/{(x/60)+(1-x)/50}
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by cans » Tue May 31, 2011 8:07 am
bblast wrote:yup, if i take distance as 1 and not 100. The numerator comes 300 !!

{x+(1-x)}/{(x/60)+(1-x)/50}
There is some error in your calculation. If you take total distance = 1,
x/100 (note x% of the distance is x/100) miles traveled at 60 miles/hr, t1 = x/6000
(1-x)/100 distance traveled at 50 miles/hr, t2=(1-x)/5000
t = t1+t2 = x/6000 + (1-x)/5000 = (6-x)/30000
distance = 1
speed = d/t = 30000/(6-x) => numerator is still 30,000
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by Frankenstein » Tue May 31, 2011 8:09 am
bblast wrote:yup, if i take distance as 1 and not 100. The numerator comes 300 !!

{x+(1-x)}/{(x/60)+(1-x)/50}
Hi,
even if u take distance as '1' instead of '100', you will still get 30,000 in the numerator. Please check your calculations. Your denominator should be ((x/100)/60)+(1-x/100)/50} because the ques states x %.

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by SoCan » Tue May 31, 2011 8:48 am
bblast wrote:yup, if i take distance as 1 and not 100. The numerator comes 300 !!

{x+(1-x)}/{(x/60)+(1-x)/50}
It may help you to think of distance as a variable d, instead of plugging in an actual value for it. This looks a lot better when it's written down, hopefully I can keep my parentheses straight here:

Average time = d/((xd/100)/60+(d(100-x)/100)/50)
=d/(xd/6000+(100d-xd)/5000)
=d/(5xd/30000+(600d-6xd)/30000)
=d/(600d-xd)/30000
=d/d(600-x)/30000 ---> the d cancels out
=30000/(600-x)

Since the d cancels out, it doesn't matter what you choose for its value, the expression remains the same.
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by bblast » Tue May 31, 2011 8:58 am
thanks cans, frankenstein and SoCan. you guys r doing great on the math forums these days.

Basic lesson-> x percent means we need to take x/100 and not just x.
Cheers !!

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