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# At 1 PM, Ship A leaves port heading due west at x miles per hour. Two

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At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

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06 Oct 2017, 02:36
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45% (medium)

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76% (06:30) correct 24% (02:01) wrong based on 38 sessions

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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. $$\sqrt{(4x)^2 + ( 100 - 2y)^2}$$

Source: NOVA GMAT

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At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

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06 Oct 2017, 08:01
Attachment:

ships.jpg [ 13.53 KiB | Viewed 763 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 = $$\sqrt{(4x)^2 + ( 100 - 2y)^2}$$

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At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

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06 Oct 2017, 08:34
fitzpratik wrote:
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. $$\sqrt{(4x)^2 + ( 100 - 2y)^2}$$

Source: NOVA GMAT

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.:

$$(4x)^2 + (100 - 2y)^2 = D^2$$

$$D = \sqrt{(4x)^2 + (100 - 2y)^2}$$

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Re: At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

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10 Oct 2017, 09:07
fitzpratik wrote:
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. $$\sqrt{(4x)^2 + ( 100 - 2y)^2}$$

Since one ship travels due west and the other travels due south, at 5 PM their distance apart will be equal to the hypotenuse of a right triangle. Let’s first determine the distance each ships travels.

Since Ship A has traveled 4 hours and its speed is x mph, Ship A’s distance traveled = 4x. Since Ship B has traveled 2 hours and its speed is y mph, Ship B’s distance traveled = 2y.

However, we have to measure the distance relative to the port because the port is the vertex of the right angle. Since Ship A leaves from the port and is heading due west, at 5 PM the distance of Ship A from the port is 4x. Since Ship B starts 100 miles due south from the port and heads due north, Ship B is getting closer to the port, and at 5 PM the distance of Ship B from the port is 100 - 2x. If we let the distance between the two ships at 5 PM be z, then by the Pythagorean theorem, we have:

(4x)^2 + (100 - 2y)^2 = z^2

√[(4x)^2 + (100 - 2y)^2] = z

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Re: At 1 PM, Ship A leaves port heading due west at x miles per hour. Two &nbs [#permalink] 10 Oct 2017, 09:07
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