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At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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28 May 2017, 03:32
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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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28 May 2017, 06:04
We can also use the GAP approach.
Ship A in 3hrs travels 45km Ship B has to close that GAP.
Rate(B)  Rate(A) = 25  15 = 10mph
Then we can set up the following equation
45 = 10T T= 45/10 > 9/2 hence 4,5hrs
So the time required is 1pm + 3 + 4,5 hrs = 8.30 therefore A.




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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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28 May 2017, 03:37
Ship A travels 112.5 Km( 15 *7.5) in 7:30 hrs (from 1 PM to 8:30 PM) Ship B travels 112.5(25*4.5) Km in 4:30 hrs hence option A Sent from my iPhone using GMAT Club Forum
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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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29 May 2017, 03:48
MvArrow wrote: We can also use the GAP approach.
Ship A in 3hrs travels 45km Ship B has to close that GAP.
Rate(B)  Rate(A) = 25  15 = 10mph
Then we can set up the following equation
45 = 10T T= 45/10 > 9/2 hence 4,5hrs
So the time required is 1pm + 3 + 4,5 hrs = 8.30 therefore A. Gud one.. +1 for this method



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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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29 May 2017, 04:13
Let them meet 'n' hours after 4 pm. From 1 to 4 pm, ship A has already travelled = 15*3 = 45 miles So in 'n' hours, A has total travelled = (45 + 15n) miles and B has travelled = 25n miles
Since they meet at that time, distances should be equal OR 45+15n = 25n Solving we get n=4.5
So the ships meet 4.5 hours after 4 pm, OR at 8.30 pm Option A is the answer



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At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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29 May 2017, 04:26
Since, the ship A leaves three hours early and travels 15 mph, it would have traveled 45 miles. Now the speed relative to Ship B, which travels at 25 mph is 10 miles/hour Since the distance it need to cover up is 45 miles, time take can be given by formula Time taken = Distance to cover/Relative Speed = \(\frac{45}{10}\) = 4.5 hours. Since the ship B leaves port at 4 PM, it would meet Ship A exactly 4.5 hours later(at 8:30 PM) (Option A)
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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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29 May 2017, 11:09
mohshu wrote: MvArrow wrote: We can also use the GAP approach.
Ship A in 3hrs travels 45km Ship B has to close that GAP.
Rate(B)  Rate(A) = 25  15 = 10mph
Then we can set up the following equation
45 = 10T T= 45/10 > 9/2 hence 4,5hrs
So the time required is 1pm + 3 + 4,5 hrs = 8.30 therefore A. Gud one.. +1 for this method Thank you very much



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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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30 May 2017, 02:01
Option A: 8.30 PM
A @ 1PM, 15 MPH B @4PM (3 hrs later), 25 MPH. by the time B starts, A would have travelled 45 miles (15MPH * 3 H) So, as A is moving away from B, the combined rate would be = 25  15 = 10 MPH and the distance to be covered is 45 miles. Time = 45/10 hrs = 4.5 so, 4 PM + 4.5 = 8.30 PM.



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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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01 Jun 2017, 10:21
Bunuel wrote: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship B leaves the same port in the same direction traveling 25 mph. At what time does Ship B pass Ship A?
(A) 8:30 PM (B) 8:35 PM (C) 9 PM (D) 9:15 PM (E) 9:30 PM We can let the time traveled by Ship A = t + 3 and time traveled Ship B = t. Since Ship A is traveling at a rate of 15 mph, the distance of Ship A = 15(t + 3) = 15t + 45, and since Ship B is traveling at a rate of 25 mph, the distance of Ship B = 25t. Let’s now determine t: Ship A’s distance = Ship B’s distance 15t + 45 = 25t 45 = 10t 45/10 = 9/2 = 4.5 hours = t Recall that Ship B left port at 1 p.m + 3 hours = 4 p.m. Thus, Ship B passes Ship A at 4 p.m + 4.5 hours = 8:30 p.m. Answer: A
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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship
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13 Aug 2017, 13:56
Bunuel wrote: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship B leaves the same port in the same direction traveling 25 mph. At what time does Ship B pass Ship A?
(A) 8:30 PM (B) 8:35 PM (C) 9 PM (D) 9:15 PM (E) 9:30 PM 15( T +3) =25T 15T +45 =25T 45=10T T=4.5 4pm + 4 hr 30 m 8 30 pm A




Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship &nbs
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13 Aug 2017, 13:56






