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At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship

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At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship [#permalink]

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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
[Reveal] Spoiler: OA

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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship [#permalink]

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New post 28 May 2017, 03:37
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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


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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship [#permalink]

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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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 29 May 2017, 11:09
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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 :-D

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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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

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Re: At 1 PM, Ship A leaves port traveling 15 mph. Three hours later, Ship   [#permalink] 13 Aug 2017, 13:56
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