GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jul 2018, 10:56

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

Hide Tags

Manager
Joined: 17 Oct 2016
Posts: 93
Location: India
Concentration: General Management, Healthcare
GMAT 1: 640 Q40 V38
GMAT 2: 680 Q48 V35
GPA: 3.05
WE: Pharmaceuticals (Health Care)
At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

Show Tags

06 Oct 2017, 02:36
00:00

Difficulty:

45% (medium)

Question Stats:

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

HideShow timer Statistics

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

_________________

Sometimes you have to burn yourself to the ground before you can rise like a phoenix from the ashes.

Dr. Pratik

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1183
Location: India
GPA: 3.82
At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

Show Tags

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

Option E
SC Moderator
Joined: 22 May 2016
Posts: 1831
At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

Show Tags

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.:

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

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

_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2969
Location: United States (CA)
Re: At 1 PM, Ship A leaves port heading due west at x miles per hour. Two  [#permalink]

Show Tags

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

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

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

Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.