In looking back on the answer the solution makes perfect sense, and I can see both ways of answering the questions: 1 algebraically 2 backsolving.


My question is I couldn't see how to approach the question and made the question MUCH MUCH harder than what it really is! How can I deal with this?

At 3:20, train A has traveled 10 miles (30mph for 20 minutes or 1/3 hour), so at that point in time, the distance between the trains is 80 miles.

Train A is moving at 30mph and Train B is moving at 10 mph. Every hour, therefore, together the trains are covering 40 miles per hour. Since there are 80 miles between them, they will cross each other exactly 2 hours after 3:20, or 5:20.

Problems that entail bodies moving toward each other can usually be solved quickly with this method.

Rey

Rey offers some good suggestions.

In general, we're faced with 2 different situations in these mulitple body rate problems.

(1) 2 objects moving in opposite directions (either directly toward or directly away from each other). To find the total rate, we ADD the rates.

(2) 2 objects moving in the same direction. To find the relative rate, we SUBTRACT the rates.

The question above falls into category (1). To find the total distance covered in one hour by the two trains, we need to add their rates together.

The question has an additional twist - the trains start moving at different times. Therefore, the first thing we need to do is equalize the times. We do so by figuring out where the earlier train will be when the latter train leaves the station. Since train #2 leaves at 3:20, we move train #1 ahead 20 minutes in time. We basically rewrite the question as:

If two trains, 80 miles apart and traveling directly toward each other at a combined rate of 40mph, leave their stations at 3:20, at what time do they meet?

Now we just apply the distance formula:

d = r * t
80 = 40t
t = 2

So, 3:20 plus 2 hours = 5:20 meeting time.

these questions are usually fun for me but this one had me going through very tough computations and I made it much harder than needed. I do know that combining the rates is an awesome approach.

In regards to the two object in the same direction, you reccomend that I SUBTRACT the rates. This baffles me! My approach is to apply the distance to the speed of body A and subtract that from the distance to the speed of body B. In other words, I would subtract the action (time-distance-rate) from eachother but only after treating them separately. Is there an example that you can provide to help explain this idea?

Enginpasa1 wrote: In regards to the two object in the same direction, you reccomend that I SUBTRACT the rates. This baffles me! My approach is to apply the distance to the speed of body A and subtract that from the distance to the speed of body B. In other words, I would subtract the action (time-distance-rate) from eachother but only after treating them separately. Is there an example that you can provide to help explain this idea?

Sure - let's even look at one that doesn't involve traditional distance/rate/time.

Bob starts reading War and Peace, a 1000 page book, at 9:00 a.m., at a rate of 50 pages per hour. Fred starts reading an identical copy of War and Peace at 11:00 a.m., at a rate of 70 pages per hour. At what time will Bob and Fred be starting to read the same page of the book?

First, we need to figure out where Bob is when Fred starts reading. Bob has a 2 hour head start and is reading 50 pages per hour, so after 2 hours Bob will be 100 pages into the book. Therefore, Fred is 100 pages behind.

Fred reads at 70 pages per hour and Bob reads at 50 pages per hour. To find the relative page/hour rate, we SUBTRACT the individual rates. 70-50=20. Therefore, each hour Fred reads an additional 20 pages.

The distance that Fred needs to catch up is 100 pages and each hour Fred reads an extra 20 pages.

Time = distance/rate = 100/20 = 5 hours.

So, Bob and Fred will be on the same page 5 hours after Fred starts reading. 11:00 a.m. + 5 hours = 4:00 p.m.

Can I try it this way?

lcm of 50 and 70 is 350. It will take fred 5 hours starting at 11am and ending at 4pm. This coincides with your answer. Also bob starts at 9am and it takes him 7 hours and he will end up at 4pm also.

Is there a place to recommend me to get some practice on simultaneous traveling bodies. I can do the work problems but I am stuck on the two objects traveling in certain distances. Both types of questions are the same idea but I am missing one key for the distance questions!

I just tried one of the hardest challenge manhattan gmat work problems and I nailed it. My clear weakness is the simulatneous distance problems. I bet, I need to do some ground work in order to fix this!

Train A:

Rate = 30mph

Time = t + 1/3 [we have to convirt minutes - 20 - into hours]

Distance = 90 - d

Therefore: 90 - d = 30(t + 1/3)

Train B:

Rate = 10mph

Time = t

Distance = d

Therefore: d = 10t

Put it together:

90 - 10t = 30t + 10
80 = 40t
2 = t

3:20 + 2 [hours] = 5:20

However, I certainly like Stuart's approach. I'll try it out next time I hit a similar problem.