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fitzpratik
Posts: 224
Updated on: 10 Jul 2019, 04:34
Originally posted by fitzpratik on 06 Oct 2017, 02:36.
Last edited by Sajjad1994 on 10 Jul 2019, 04:34, edited 1 time in total.
Last edited by Sajjad1994 on 10 Jul 2019, 04:34, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (02:33) correct
21%
(02:24)
wrong
based on 91
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At 1 PM, Ship A leaves port heading due west at x miles per hour. Two hours later, Ship B is 100 miles due south of the same port and heading due north at y miles per hour. At 5 PM, how far apart are the ships?
A. \(\sqrt{(4x)^2 + ( 100 + 2y)^2}\)
B. x + y
C. \(\sqrt{x^2 + y^2}\)
D. \(\sqrt{(4x)^2 + (2y)^2}\)
E. \(\sqrt{(4x)^2 + ( 100 - 2y)^2}\)
A. \(\sqrt{(4x)^2 + ( 100 + 2y)^2}\)
B. x + y
C. \(\sqrt{x^2 + y^2}\)
D. \(\sqrt{(4x)^2 + (2y)^2}\)
E. \(\sqrt{(4x)^2 + ( 100 - 2y)^2}\)
Source: Nova GMAT
Difficulty Level: 650
Difficulty Level: 650
06 Oct 2017, 08:01
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Attachment:
ships.jpg [ 13.53 KiB | Viewed 5700 times ]
By 5 P.M, ship A would have traveled for \(4\) hours, hence distance \(= 4x\)
and as ship B started two hours late, hence distance \(= 2y\) and reached point C
The initial distance between ship B and port P was 100, so PC \(= 100-2y\)
From the figure it is clear that we need to calculate AC
So AC = \(\sqrt{(4x)^2 + ( 100 - 2y)^2}\)
Option E
06 Oct 2017, 08:34
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fitzpratik
To find the distance between the two ships at 5 p.m., construct two legs of what will be a right triangle. Ship A travels away from port. Ship B travels toward port.
Compute each ship's distance covered by 5 p.m. without reference to the other.
Ship A starts at 1 p.m. and travels for t = 4 hours at rate \(x\) miles per hour.
Ship A's distance = r*t = 4x (miles). Ship A heads due west from port. If port is D = 0, Ship A's distance is 0 + 4x = 4x
Ship B starts at 3 p.m. and travels for 2 hours at rate \(y\) miles per hour.
Ship B is 100 miles due south from the port. It travels due north to the port. It covers part of that 100 miles. For 2 hours at \(y\) mph, rt= D, it covers 2y miles.
That partial distance must be subtracted from the given distance of 100 miles. Port is D = 0. Ship B started at 100 miles away from 0. It has covered 2y miles of the 100 miles.
Ship B's 5 p.m. location, in miles, therefore will be (100 - 2y)
Ship A's 5 p.m. location, in miles: 4x
The distance each has traveled is the leg of a right triangle. The distance between them is the hypotenuse.
Using Pythagorean theorem first, \(D\) is distance between them at 5 p.m.:
\((4x)^2 + (100 - 2y)^2 = D^2\)
\(D = \sqrt{(4x)^2 + (100 - 2y)^2}\)
Answer E
