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kevincan
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kevincan,

You mean I got the intial problem construction incorrect or the ans incorrect
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kevincan
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kevincan,

You mean I got the intial problem construction incorrect or the ans incorrect


I see them at opposite ends of the hypotenuse of a right triangle!
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hummmm.... makes sense; cause in the way i have assumed the road turns, and u mention that they donnot change directions!

kevincan
Lionel and Ursula are both cycling in the direction of Fenton on straight roads (I took this to mean 2 separate roads, may may not be parallel) at constant speeds. Lionel, who is heading north, is currently twice as far away from Fenton as is Ursula, who is cycling to the west. (90 degree difference in orientation) If neither stops nor changes direction in the course of the day, what is the ratio of the speed of Lionel's bike to that of Ursula's?

(1) In four hours, Lionel and Ursula will both be 10 miles from Fenton.
(2) In terms of straight line distance, Ursula's car is currently 25*sqrt(2) miles from Lionel's.
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If the answer is not C then what is it please ?
From (1), I get the ratio of speeds to be (2x-10)/(x-10).
From (2), I get to solve for x.
Thus, only by combining (1) and (2) I will arrive at a ratio of Ursula's and Leon's speeds.
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kevincan
prashrash
If the answer is not C then what is it please ?
From (1), I get the ratio of speeds to be (2x-10)/(x-10).
From (2), I get to solve for x.
Thus, only by combining (1) and (2) I will arrive at a ratio of Ursula's and Leon's speeds.

Must this be true?

Stmt 1 says that after 4 hrs, both Ursula and Leon are 10 miles from the destination.
Also the q says that Leon was twice as far away as Ursula at the start.
If we consider Ursula's distance from destination to be x, the Leon was 2x away from it.
After 4 hours, Ursula has travelled x-10 miles.
Leon has done 2x-10 miles. (from stmt 1)
Speed of Ursula = (x-10)/4
Speed of Leon = (2x-10)/4
Hence ratio = (2x-10)/(x-10).
All of this is, of course, assuming that U and L start from the 2 ends of the hypotenuse of a rt triangle with the right angle being formed at destination. That was the only feasible figure, representing their relative direction of travel, that I could come up with (also confirmed by you in your post above :) ).
Is there another possible eqn for the ratios ?
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How do we know whether, when they are at equal distances from Fenton, either has already gone through Fenton?
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tricky one))
I dunno but i think it is E
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do we need the slope of the hypotynuss of the right angled triangle????

Kevin??

Thanks
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I believe its (A)

From 1) Let the distance of L from kenvin be 2d and speed be u and that of U be d and speed be v
Now d = speed * time
hence for L after four hours
(2d - 10)/u = 4
hence d = (4u +10)/2 ------------- (1)
for U
(d-10)/v = 4

d = 4v +10 -----(2)

combining 1 and 2 we get the ratio of u and v

From 2)

the distance is a hypotenuse of a rt angled triagnle ..25sqrt(2) = sqrt ( 2d*d) = dsqrt(2)

hence d = 25 , 2d = 50

But this doesnt help much

Hence A
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Fel into the same trap.

On a second thought its (C) both are needed .
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OA E: From A, we cannot conclude anything about the ratio of speeds, because we don't know on what side on Fenton the two cities are. Knowing the current distances (from 2) of each will not help.



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