Lesson: Multiple Trips or Multiple Travelers

Comment on Multiple Trips or Multiple Travelers

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/al-and-ben-are-drivers-for-sd-trucking-...

You can assign the variable t to either person.

If we let t = Ben's travel time, then we must say that t-3 = Al's travel time (since the question tells us that Ben travelled for 3 extra hours)
So the equation should be: 40(t-3) + 20(t) = 240
When we solve the equation we get t = 6, which means Ben travelled for 6 hours, and Al travelled for 3 hours.

Either way we still get the same answer

Brent, how did you get to rewrite 4 as 4(b-5)/etc...?

"So, we get: 240/B = 240/(B + 5) + 4
Rewrite 4 as 4(B + 5)/(B + 5) to get: 240/B = 240/(B + 5) + 4(B + 5)/(B + 5)"
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/bill-and-ted-each-competed-in-a-240-mil...
Please be sure to include a link to the question you're asking about.

There are a few different ways to solve the equation 240/B = 240/(B + 5) + 4
In my solution, I decided to combine 240/(B + 5) and 4, by rewriting 4 as a fraction with denominator (B + 5)

Keep in mind that 4 = 4/1
So, we can take the fraction 4/1 and multiply the numerator and denominator by (B + 5) to get the equivalent fraction 4(B + 5)/(B + 5)
So we now have: 40/(B + 5) + 4(B + 5)/(B + 5)
Simplify to get: 40/(B + 5) + (4B + 30)/(B + 5)

Since both terms that share the same denominator, we can have the numerators to get: [40 + (4B + 30)]/(B + 5)
Does that help?

The distance between Newark and Boston by train is 245 miles. train A departs Newark, at 12:00 PM traveling to Boston at a constant rate of 50 miles per hour. Train B departs Boston at 12: 30 pm traveling at a constant rate of 60 mph. If the trains are travelling on parallel tracks, at what time will they pass each other?

So when I did this this was my process. The aspect I was struggling with was for what train to make as t+.5 or t-.5.

I set up a word equation DNB = DBN (NB) = Newark to Boston and Boston to Newark (BN)

tA = tB-.5 or I could use tB = tA+.5
so my question looked like this 50tA = 60tB, now couldnt I plug in either tA or tB and get the same answer? for tA or tB we would plug in either equation into my word equation and solve for either tA or tB and then plug that tA or tB into the 12 or 12:30 and get the answer as 2:30?
Unfortunately I did this but that did not seem correct as I got a negative time?
greenlight-admin's picture

I suggest you rewrite the given information so that both trains are leaving at the same time.

Train A leaves at 12:00pm, and it travels at a speed of 50 miles per hour.
So, in 30 minutes, train A travels 25 miles.
So, at 12:30pm, the two trains are 220 miles apart (245 - 25 = 220)

So the new information looks something like this:
At 12:30pm, trains A and B are 220 miles apart and heading towards each other. If train A travels at 50 mph and train B travels at 60 mph, at what time do the two trains meet?

I believe you'll find that this question is much easier to answer.

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As far as your solution goes, can you tell me what t represents?

tA and tB are respective times of train A and train B what I was trying to do here was draw a relationship between them based off the problem which said Train A departs at 12 and train B departs at 12:30 So I was trying to sat tA= tB-.5 once I had this and the fact the distance from N-B and B-N is 245, I would setup a word equation and say 50tA = 60tB. From this logic it makes sense, what is odd is the time equation. Apparently that is not right, because I played around with the time equation and then got the right answer which is 2:30.
greenlight-admin's picture

I suggest that you avoid notation like tA and tB. Since we know the relationship between the 2 travel times, we can stick with one variable.

Let t = train A's travel time (in hours)
So, t - 0.5 = train B's travel time (in hours)

As far as word equations go, I suggest the following:
(train A's distance traveled) + (train B's distance traveled) = 245

Since distance = (rate)(time), we can plug the given values into our word equation to get:
(50)(t) + (60)(t - 0.5) = 245

Solve to get t = 2.5 (hours)

So train A travels for 2.5 hours.
Since train A leaves at noon. The two trains meet at 2:30pm.

Ok makes sense. Is there a way to solve it where we can set the Distance from Newark to Boston = the distance from Boston to Newark?
That was the initial way I tried it but couldnt get the right answer.
greenlight-admin's picture

Unfortunately, that information won't help us because neither train travels the entire distance.

That is, we can't say that (train A' travel distance) = (train B' travel distance)

Jacob drove from home to school in the morning at an average rate of 40 mph. In the evening he drove back home from school at a rate of 60 mph. If he spent a total of 2 hours traveling and he took the same roads each way what is the total distance that Jack traveled.

Approach 1: Distance from H-S = Distance from S-H
40t = 60(2-t) solve for t and plug back so we get t = 1.2 and total distance as 96

Approach 2: T (Home to School) +T (School to Home) = 2
D/40 +D/60 = 2 solve for D and get 48, but the D is only represented by 1 leg , so we should add 48 to the other side since its the same distance stated in the problem therefore 96.

Please let me know if this is all correct?
greenlight-admin's picture

Both solutions are perfect. Nice work, Ravin!

So this is a major problem I have noticed in these problems that I am doing. I can setup the word equation but when I am setting up the time it gets a little bit odd for me. Take a look at this problem

From point A Tom begins running at a constant of 8 miles per hour. Shandra starts from the same location a half hour later and runs at a constant of 12 miles per hour. If Shandra runs on the same path as Tom, how long will it take for Shandra to catch up to Tom?

So I set the Distance of Tom = Distance of Shandra. and I let t = Tom's time
therefore Shandra's time = t+.5

So we would have Distance of Tome = Distance of Shandra
8t= 12(t+.5), doing this we get a negative answer, which is incorrect. From this logic I am under the impression that the word equation is correct, but the equation of time is wrong. Is there a set rule for who we can set t to? Where am I going wrong?
greenlight-admin's picture

The error occurred when you assigned a variable to Shandra's travel time.

If t = Tom's travel time, then t - 0.5 = Shandra's travel time (since she started traveling "a half hour LATER")

From point A Tom begins running at a constant of 8 miles per hour. Shandra starts from the same location a half hour later and runs at a constant of 12 miles per hour. If Shandra runs on the same path as Tom, how long will it take for Shandra to catch up to Tom?

So for this problem I set up the word equation as Tom Time = Shandra's TIme -1/2.. Knowing they run the same distance, D. said D/8 = D/12 -1/2. and solved for D. Why is that the wrong approach?
greenlight-admin's picture

That's close but we're told Shandra starts from the same location a half hour LATER.
This means Shandra's travel time is 0.5 hours LESS THAN Tom's travel time.
So, to make their travel times equal, we must add 0.5 hours to Shandra's travel time
In other words: Tom's travel time = Shandra's travel time + 0.5

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