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19 Aug 2011, 01:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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0% (00:00) correct
100%
(03:28)
wrong
based on 3
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Alex swims from Point X to Point Y, and back in 10 hrs. He drops his goggles at point X the goggles take 15 hrs to float to point Y. What is the ratio of the speed of Alex in still water to that of the current?
A) 10:1
B) 6:1
C) 3:1
D) 3+\sqrt{13}:2
E) 4:1
Any Help.....
A) 10:1
B) 6:1
C) 3:1
D) 3+\sqrt{13}:2
E) 4:1
Any Help.....
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jamifahad
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19 Aug 2011, 03:33
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Let the speed in still water be W
Let the speed of current be C
Upstream rate = W-C
Downstream rate = W+C
Let the distance between X and Y be d.
Question: What is w/c?
Time taken to swim upstream + Time taken to swim downstream = 10
\(d/w-c + d/w+c =10\)
\(2wd/w^2-c^2 =10\) ----(1)
Also, goggle travels from point X to Y (distance d) with the speed of current.(floats) in 15hours
15=d/c
d=15c
Substituting d=15c in (1) and solving
\(2w*15c/w^2-c^2=10\)
\(30wc/w^2-c^2=10\)
\(3wc=w^2-c^2\)
Dividing by wc
\(3=w/c - c/w\)
\(w/c = 3+c/w\)
Now, we know speed in still water(w) must be greater than speed of current for alex to swim upstream.
i.e w > c
or 1>c/w
or c/w<1
or 0<c/w<1
i.e c/w lies between 0 and 1.
Hence w/c is a number between 3 and 4
let c/w=0.5
w/c=3+0.5=3.5
or w=3.5c
Now check answer choices
1)w/c=10:1
w=10c. Eliminate
2) w/c=6:1
w=6c. Eliminate
3)w/c=3:1
w=3c. Eliminate
4) Hold.
5) w/c =5:1
w=5c. Eliminate
4) \(w/c = 3+\sqrt{13} / 2\)
\(\sqrt{13}\) is between 3 and 4.
w/c = 3+(number between 3 and 4) /2
w/c= 1.5 + (number between 1.5 and 2)
w/c=number between 3 and 4.
OA D.
P.S : As much as I like solving your problems, somewhere i think it deviates from what is asked in GMAT! But great practice nonetheless.
Let the speed of current be C
Upstream rate = W-C
Downstream rate = W+C
Let the distance between X and Y be d.
Question: What is w/c?
Time taken to swim upstream + Time taken to swim downstream = 10
\(d/w-c + d/w+c =10\)
\(2wd/w^2-c^2 =10\) ----(1)
Also, goggle travels from point X to Y (distance d) with the speed of current.(floats) in 15hours
15=d/c
d=15c
Substituting d=15c in (1) and solving
\(2w*15c/w^2-c^2=10\)
\(30wc/w^2-c^2=10\)
\(3wc=w^2-c^2\)
Dividing by wc
\(3=w/c - c/w\)
\(w/c = 3+c/w\)
Now, we know speed in still water(w) must be greater than speed of current for alex to swim upstream.
i.e w > c
or 1>c/w
or c/w<1
or 0<c/w<1
i.e c/w lies between 0 and 1.
Hence w/c is a number between 3 and 4
let c/w=0.5
w/c=3+0.5=3.5
or w=3.5c
Now check answer choices
1)w/c=10:1
w=10c. Eliminate
2) w/c=6:1
w=6c. Eliminate
3)w/c=3:1
w=3c. Eliminate
4) Hold.
5) w/c =5:1
w=5c. Eliminate
4) \(w/c = 3+\sqrt{13} / 2\)
\(\sqrt{13}\) is between 3 and 4.
w/c = 3+(number between 3 and 4) /2
w/c= 1.5 + (number between 1.5 and 2)
w/c=number between 3 and 4.
OA D.
P.S : As much as I like solving your problems, somewhere i think it deviates from what is asked in GMAT! But great practice nonetheless.
21 Aug 2011, 13:57
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total time it took alex to travel upstream and down stream = 10
\(d/(v+c) + d/(v-c) = 10\)
where d is the distance between X and Y.
v is the speed of Alex
c is the speed of the current.
When goggles floated from X to Y
goggles speed = current speed => c = d/15
solving the above two equations we can find out v/c as (3+sqrt(13))/2
Answer is D.
\(d/(v+c) + d/(v-c) = 10\)
where d is the distance between X and Y.
v is the speed of Alex
c is the speed of the current.
When goggles floated from X to Y
goggles speed = current speed => c = d/15
solving the above two equations we can find out v/c as (3+sqrt(13))/2
Answer is D.
