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How to solve this distance problem algebraically?

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How to solve this distance problem algebraically?

by ksc1940 » Tue Aug 14, 2012 9:21 pm
This question is from jeff sackmann's 1000 problem set, which is superb. I know I can solve this by plugging in numbers, and that's actually the explanation that jeff gives. But I want to know how to do it algebraically. I tried figuring it out but couldn't get to the right answer.

Hayden began walking from F to G, a distance of 40 miles, at the
same time Ava began walking from G to F on the same road. If
HaydenÂ’'s walking speed was x miles per hour and AvaÂ’'s was y
miles per hour, how many miles away from F were they, in terms
of x and y, when they met?
(A) 40(x-y)/x+y
(B) 40x-y/x+y
(C) x-y/x+y
(D) 40y/x+y
(E) 40x/x+y
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by anuprajan5 » Tue Aug 14, 2012 10:56 pm
Speed * Time = Distance

Assume time to be t at which point x and y meet
Therefore, x*t + y*t = 40
Therefore t = 40/(x+y)

Distance from town F - Speed * Time = x*t = x*40/(x+y)

Hence Answer E.

My rationale is that distance from G would be - Speed*Time - y*t = y*40/(x+y).

Therefore if I add these 2 distances, the distance covered by both of them would be x*40/(x+y) + y*40/(x+y) which would equate to 40, which is the distance both of them covered to meet at time t.




Hope I am not way off base here.

Regards
Anup
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by niketdoshi123 » Tue Aug 14, 2012 11:06 pm
ksc1940 wrote:This question is from jeff sackmann's 1000 problem set, which is superb. I know I can solve this by plugging in numbers, and that's actually the explanation that jeff gives. But I want to know how to do it algebraically. I tried figuring it out but couldn't get to the right answer.

Hayden began walking from F to G, a distance of 40 miles, at the
same time Ava began walking from G to F on the same road. If
HaydenÂ’'s walking speed was x miles per hour and AvaÂ’'s was y
miles per hour, how many miles away from F were they, in terms
of x and y, when they met?
(A) 40(x-y)/x+y
(B) 40x-y/x+y
(C) x-y/x+y
(D) 40y/x+y
(E) 40x/x+y
Since Hayden and Ava are walking in opposite direction, towards each other.
Their relative walking speed will be = x+y

Total distance walked when they met = 40
# of hours they walked just before they met = 40/(x+y) = # of hours each of them walked.
Since Hayden started walking form F, the distance he walked in 40/(x+y) hrs = x*40/(x+y)

Hence, they were 40x/(x+y) miles away from F when they met
The correct answer is E
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by GMATGuruNY » Wed Aug 15, 2012 2:32 am
ksc1940 wrote:This question is from jeff sackmann's 1000 problem set, which is superb. I know I can solve this by plugging in numbers, and that's actually the explanation that jeff gives. But I want to know how to do it algebraically. I tried figuring it out but couldn't get to the right answer.

Hayden began walking from F to G, a distance of 40 miles, at the
same time Ava began walking from G to F on the same road. If
HaydenÂ’'s walking speed was x miles per hour and AvaÂ’'s was y
miles per hour, how many miles away from F were they, in terms
of x and y, when they met?
(A) 40(x-y)/x+y
(B) 40x-y/x+y
(C) x-y/x+y
(D) 40y/x+y
(E) 40x/x+y
Each hour, Hayden travels x miles and Ava travels y miles.
Thus, of the x+y miles traveled each hour, the fraction traveled by Hayden = x/(x+y).
Thus, of the 40 miles between Hayden and Ava, the portion traveled by Hayden = x/(x+y) * 40 = 40x/(x+y).

The correct answer is E.
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