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

Question

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

Source:   Nova GMAT  
Difficulty Level:   550

MvArrow · Accepted Answer

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.

Luckisnoexcuse · Answer

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

mohshu · Answer

Gud one..
+1 for this method

amanvermagmat · Answer

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

pushpitkc · Answer

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)

MvArrow · Answer

Thank you very much :-D

ShreyansJ · Answer

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.

ScottTargetTestPrep · Answer

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

Nunuboy1994 · Answer

15( T +3) =25T
15T +45 =25T
45=10T
T=4.5

4pm + 4 hr 30 m

8 30 pm

A

Basshead · Answer

15 ( T + 3) = 25T
15T + 45 = 25T
45 = 10T
T = 4.5

It takes 4.5 hours for Ship B to catch up to Ship A.

bumpbot · Answer

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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

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