# Lesson: Multiple Trips or Multiple Travelers

## Comment on Multiple Trips or Multiple Travelers

### Two trains, X and Y, started

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5 Don't know what should be done after taking out per hour speed of x and y ### There are some very nice

Here's my solution on GMAT Club: https://gmatclub.com/forum/two-trains-x-and-y-started-simultaneously-fro...
Please let me know if you need any clarification.

### Speed of Train X 20kmph and

Speed of Train X 20kmph and Train Y 100/3 kmph.Suppose they meet after T hrs time.So,20T+100/3 T=100.We ger T=15/8 .
So,Train X would travel 20*15/8=37.5 km before meeting Train Y. I don't understand is how it came out to be 15/8 . I understand that we take out speed . After that I am confused
Thanks for help ### Let's take it from 20T + (100

Let's take it from 20T + (100/3)(T) = 100 (where T = the time each train travels until they meet)

Eliminate the fraction by multiplying both sides by 3.
We get: 60T + 100T = 300
Combine: 160T = 300
Solve: T = 300/160 = 30/16 = 15/8

So, each train travels for 15/8 hours

Since train X travels at 20 km per hours, its travel distance = (speed)(time) = (20)(15/8) = 300/8 = 150/4 = 37.5 km

### Jerry and Jim run a race of

Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a start of 200m and beats him by 30 seconds. Next, Jerry gives Jim a start of 3mins and is beaten by 1000m. Find the time in minutes in which Jerry and Jim can run the race seperately?

A. 8,10
B. 4,5
C. 5,9
D. 6,9
E. 7,10 ### Hey,

Hey,
1st off, the Jerry and Jim question still gives me nightmares at night. I straight up hate that question now.

https://gmatclub.com/forum/ann-and-bea-leave-townville-at-the-same-time-and-travel-237882.html
This question however was much more to my liking.
Furthermore, out of a silly mistake I ended up finding a much easier solution.

Upon reading the question I assumed, wrongly, that the 2k distance meant 2000km. Then I assumed a value for Bee's speed, 100kmph, then naturally Anne's speed was 400kmph.
This allowed some easy arithmetic which left me with 800km as an answer; Foolishly now I checked the answer choices and found that K was in fact a variable. So I just multiplied and divided 800 by 1000. (which is 2k without the 2) replaced the numerator 1000 with K, and got 4k/5 .
Though later I realised that a much easier option would've been to see which of the answers gave me an 800 if I replaced the answer K's with 1000. ### The Jim & Jerry question

The Jim & Jerry question (here https://gmatclub.com/forum/jerry-and-jim-run-a-race-of-2000-m-first-jerr...) is a killer. Hopefully, the nightmares will subside soon :-)

As for the Ann & Bea question (here https://gmatclub.com/forum/ann-and-bea-leave-townville-at-the-same-time-...), it never dawned on me (when I created the question) that 2K might be misinterpreted as 2000, but I can totally see it now. That said, the test-makers would never assume that test-takers are aware that we can use "2K" to represent 2000

Once you realized your error, I like how you were able to use your existing calculations to determine the correct answer. Good stuff!

### My quick intuition-based

My quick intuition-based response was that if Ata traveled at 100km/h and arrived 1hour ahead of Carl who did 75km/h, Ata must have gained 25km over Carl in each hour and at the end of the third hour when Ata must have traveled 300km, the cumulative distance gained over Carl should be 75km. Great logic!

### Hi Brent,

Hi Brent,

The current in a river is 4 mph. A boat can travel 20 mph in still water. How far up the river can the boat travel if the round trip is to take 10 hours?

A. 69 miles
B. 88 miles
C. 96 miles
D. 100 miles
E. 112 miles

My question concerns assigning the “time” variable: When I set up the equation this way,

24(T1) = 16(10−T1)

I am confused as to why (it seems like it should not matter) if you set it up this way

24(10-T) = 16(T1)

As the time is the same - 10 hours! But it does matter and I get a different answer for the Time.

Is there something I am missing? How do you know which time (10 – T or T) to assign to which rate? ### To avoid confusion, it helps

To avoid confusion, it helps to formally assign your variables.

In the equation 24(T1) = 16(10−T1), T1 represents the time spent traveling DOWNSTREAM at 24 miles per hour.

In the equation 24(10-T) = 16(T1), T1 represents the time spent traveling UPSTREAM at 16 miles per hour.

Does that help?

Cheers,
Brent

### Yes! I was under the

Yes! I was under the misguided impression that as long as the hours added to the "desired time" the variable assignment did not matter. Ok, got it, Thanks Brent! I had seen it done the wrong way before, and the person demonstrating the problem did not seem to explain or make it seem like it mattered as long as the total time was correct.

### Brent sir, i just had a

Brent sir, i just had a question. After i have completed all your videos, should i go for the egmat verbal course? do u recommend that? ### Sorry, but I've never

Sorry, but I've never reviewed that course. I can tell you that our videos lesson cover everything you need to know for all sections of the GMAT.

### Hi Brent, for the question

Hi Brent, for the question below I need your approach of this lesson to figure this question out.

https://gmatclub.com/forum/jerry-and-jim-run-a-race-of-2000-m-first-jerry-gives-jim-122757.html ### In the solution above did not

In the solution above did not understand why did you add 30 secs to Jerry's time in the first equation? In my head, 30 secs should be added to Jim's time as it would take 30 more secs to cover remaining d. In the second equation, 180 secs should be added to Jerry's time as it would take 180 secs for him to cover remaining d. But I am not getting the correct answer with this logic, where am I going wrong? ### GIVEN: Jerry gives Jim a

GIVEN: Jerry gives Jim a start of 200m and beats him by 30 seconds.

So, once Jerry crossed the finish line, 30 seconds elapsed until Jim crossed the finish line.
So, Jerry's travel time is LESS THAN Jim's travel time.
In fact, Jerry's travel time is 30 seconds LESS THAN Jim's travel time.
We might write: (Jerry's travel time) < (Jim's travel time)

Since Jerry's travel time is LESS THAN Jim's travel time, we need to add something to Jerry's travel time to make the two times equal.

Alternatively, since Jim's travel time is GREATER THAN Jerry's travel time, we could also make the times equal by SUBTRACTING 30 seconds from Jim's time.

KEY CONCEPT: If two expressions are not equal, then we must perform some operation on one value to make the values EQUAL.

Does that help?

Cheers,
Brent

### solving this question using

solving this question using ratios
a:c
100:75
4:3
distance is same
so time ratio
a:c
3:4
diff in ratio =1 and hours taken extra = 1
so total distance = 3 hrs *100 = 300 Nice!

### Hi Brent, could you explain

Hi Brent, could you explain why in the T=T equation we add 1 to Ata's travel time. Given that it takes Carl 1 hour longer than Ata to reach the destination, wouldn't we be adding "1" to Carl's time instead?

Or another way I see it is that we would subtract 1 from Atta's time to equal Carl's time. (such that T -1 = T)

Thanks! ### GIVEN: It takes Carl 1 hour

GIVEN: It takes Carl 1 hour longer than Ata to reach the destination.

When you're not sure what you need to do to create the necessary equation, try identifying some possible values that would satisfy the given information.

CASE 1:
Ata's travel time = 4 hours
and Carl's travel time = 5 hours
This satisfies the condition that "it takes Carl 1 hour longer than Ata to reach the destination"

CASE 2:
Ata's travel time = 20 hours
and Carl's travel time = 21 hours

CASE 3:
Ata's travel time = 6.5 hours
and Carl's travel time = 7.5 hours

For each case, Carl's travel time is 1 GREATER than Ata's travel time.
So, adding another hour to Carl's time will NOT make Carl's time EQUAL to Ata's time.

If we want to create an EQUATION, we need to add 1 hour to Ata's travel time.

That is: (Ata's travel time) + 1 = Carl's travel time
For CASE 1, we get: 4 + 1 = 5....perfect!
For CASE 2, we get: 20 + 1 = 21....perfect!
For CASE 3, we get: 6.5 + 1 = 7.5....perfect!

Cheers,
Brent

### Great comparison video thanks

Great comparison video thanks Brent.
One question regarding time of 1 hr earlier for Ata.
So A+1 = C in 1st formula make sense. However in 2nd & 3rd ways of calculation using distance and speed, A's time is A but C's time is A + 1 ? Shouldn't this be A - 1 to be equal if Ata's travel time = 4 hours and Carl's travel time = 5 hours?

So I calculate it as below is correct?

Ata's travel time = A -1
Carl's travel time = A

100 (A - 1) = 75A
100A -100 = 75A
25A = 100
A = 4
D = 75 * 4 = 300 ### Given: Carl's travel time is

Given: Carl's travel time is 1 hour more than Ata's travel time.
As long as your variables satisfy the given condition, then everything will work out fine.

In my solution, I let Ata's travel time = A, which means Carl's travel time = A+1
Notice that Carl's travel time (A+1) is 1 hour greater than Ata's travel time (A).

In your solution, you let Ata's travel time = A-1, which means Carl's travel time = A
Notice that Carl's travel time (A) is 1 hour greater than Ata's travel time (A-1).

So, your solution is perfectly fine, although I would argue that it's somewhat confusing to use the variable A to represent Carl's travel time.
Nevertheless, as you can see by your solution, it's perfectly valid.

### Thanks Brent for confirmation

Thanks Brent for confirmation.
One last bit to clarify is carry on using Ata's travel time = A-1, I'm getting -300 as below?

d/100 -1 = d/75
d-100/ 100 = d/75
100d = 75d - 7500
25d = -7500
d = -300 ### This is where your variable

This is where your variable assignments get you in trouble/

We know that: Ata's travel time + 1 = Carl's travel time
So we get: d/100 + 1 = d/75

### Get it thanks Brent and didn

Get it thanks Brent and didn't expect assignment of time +/-1 hr in time could cause this type of issue in different calculatin.
So it's good to stick to Ata's travel time + 1 = Carl's travel time whenever it's 1 hr earlier in work rate issue then.
Thanks Brent

### A woman is planning a trip

A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

15 minutes
25 minutes
1 hour 15 minutes
1 hour 40 minutes
4 hours 30 minutes ### Hi Jalaj,

Hi Jalaj,

In my opinion, this is not a GMAT-worthy question. The main reason is that it really just has one brute-force (aka time-consuming) solution.
When answering questions on the various GMAT forums, I post to my website only those questions I believe to be GMAT-worthy. The above question would not "make the grade."

Cheers,
Brent

Okay. Thanks!

### Hi Brent, this question is

Hi Brent, this question is wordy and tricky, need your help?

At Consolidated Foundries, for a resolution to become policy, a quorum of at least half the 20 directors must pass the resolution by at least a two-thirds majority. At a meeting of the board of directors, did resolution X pass or fail?

(1) Ten directors voted for the resolution.

(2) Seven directors voted against the resolution. ### Hi Jalaj,

Hi Jalaj,

This question has nothing to do with time/distance/rate questions.

Here's my full solution: https://gmatclub.com/forum/at-consolidated-foundries-for-a-resolution-to...

Cheers,
Brent

### Oh my bad. Will make sure of

Oh my bad. Will make sure of it the next time.

On the question though doesn't the line "A quorum of at least half the 20 directors must pass the resolution..." means that there are 20 directors? ### The wording of the question

The wording of the question isn't very good, but I believe there are 20 directors, but we don't know how many of those 20 directors actually attended the meeting to vote.

### Exactly my thinking!

Exactly my thinking!

Was about to post until I saw this :-)

### Hi Brent,

Hi Brent,

Need your help in understanding the below question:

https://gmatclub.com/forum/jacks-watch-gains-10-seconds-every-hour-jose-s-watch-loses-15-seconds-275413.html ### Hi Brent,

Hi Brent,

Thank you for the video, I've learned a lot from you.

I have a question in regards to this question.

On the first method we used, why do we add 1 on Ata's travel time.

Since Ata's arrived an hour earlier than Carl's, shouldn't it be Ata's Time= Carl's time - 1 instead.

For example, if the if it takes 5 hours for Carl travels from Towsnville to Villageton, then it will only take 4 hours for Ata's travel time.

But if we say Ata's travel time + 1 equals to Carl travel time, does that make it invalid because of Ata's time is (5+1) then Carl's travel time would be 5, which means Ata's travel an hour later compare to Carl.

Could you please explain? Thank you ### Good question!

Good question!

I'll start by saying that the following example of yours is incorrect: "For example, if the if it takes 5 hours for Carl travels from Towsnville to Villageton, then it will only take 4 hours for Ata's travel time"

Rather than think about travel times, let's first think of things in terms of ARRIVAL times.
If Carl ARRIVED at 5:00pm, then that means Ata ARRIVED at 4:00pm (1 hour earlier)
So, Carl's travel TIME = 5 hours, and Ata's travel TIME = 4 hours.

In fact, Ata's travel time will always be 1 hour less than Carl's travel time (since Ata's trip ends 1 hour earlier than Carl's trip).

As such, we need to add 1 hour to Ata's travel time.

Does that help?

Cheers,
Brent

### Hey Brent,

Hey Brent,

https://gmatclub.com/forum/al-and-ben-are-drivers-for-sd-trucking-company-one-snowy-day-ben-lef-242857.html

How can you assume that the distance the 2 drivers covered is the same? You wrote: Al rate * time = Ben rate * time +3

Thanks for your clarification, maybe I just missed some info.

Cheers,

Philipp ### Good catch!!!

Good catch!!!
You're totally right; it should be (Ben's travel distance) + (Al's travel distance) = 240 miles
Not (Ben's travel distance) = (Al's travel distance)

I have edited my response here: https://gmatclub.com/forum/al-and-ben-are-drivers-for-sd-trucking-compan...

Thanks for the heads up, Philippi!

Cheers,
Brent

### Hi Brent, can you please

Hi Brent, can you please share your solution to Jim and Jerry race question?

https://gmatclub.com/forum/jerry-and-jim-run-a-race-of-2000-m-first-jerry-gives-jim-122757.html

Thanks!
Kashaf ### Sure thing. Here's my full ### Hello Sir,

Hello Sir,

I come across your question on GMAT Club, and I am trying to solve it with "Distance" instead; however, I got the wrong answer. Could you please tell me where I did incorrectly?

Let x be the time the boat moves up the current
Let 10-x be the time the boat in still water

Rate= 16 --> Against Current
Rate= 20 --> In still water

Since it's round trip, the distance is the same.

Distance of 1st trip = Distance of 2nd trip
16(x) = 20(10-x)
x= 5.55-->"This is for the time

Distance= 16 x 5.5 = 88?

However, it's wrong:(

https://gmatclub.com/forum/the-current-in-a-river-is-4-mph-a-boat-can-travel-20-mph-in-still-wat-212278.html Be careful. During the trip up and down the river, the boat is either travelling WITH the current or AGAINST the current.
That is, the boat never travels in still water during the trip.

GIVEN: The current in a river is 4 mph. A boat can travel 20 mph in still water.
So, when the boat is travelling UPSTREAM, its speed is 16 mph (20 - 4 = 16).
And, when the boat is travelling DOWNSTREAM, its speed is 24 mph (20 + 4 = 24).

Cheers,
Brent

### Hi Mr. Brent,

Hi Mr. Brent,

I hope you have a great week so far.

I ran into this question while I am practicing. Could you please enlighten me?

https://gmatclub.com/forum/a-boat-traveled-upstream-90-miles-at-an-average-speed-of-100767.html

Here is what I did:
Distance travel upstream + Distance travels downstream = 180

Let t be the time travels downstream
Let (t + 1/2) travels upstream

Distance = Rate x time
Therefore:
Distance travels upstream = (v-3)(t + 1/2) = vt +1/2v-3t-3/2
Distance travels downstream = (V+3)(T) = Tv + 3t

Then I combined these two distances together and set it equal to 180.

(vt + 1/2v - 3t - 3/2) + tv +3t = 180
I simplified everything, but I can't get rid of the variable V to get t.

Could you please tell me what I did wrong? Thanks You've set up everything correctly, David.
However, since you decided to use TWO variables in your solution, then we must find TWO equations in order to solve the question.

In your solution, you use the fact that the total distance traveled equals 180 miles to create the equation (vt + 1/2v - 3t - 3/2) + tv + 3t = 180.

We need to use another piece of information to write a second equation.
Well, we also know that the boat traveled the SAME distance in each direction.

We get: (distance traveled upstream) = (distance traveled Downstream)
Substitute to get: vt + 0.5v - 3t - 1.5 = vt + 3t
Subtract vt from both sides: 0.5v - 3t - 1.5 = 3t
Subtract 3t from both sides: 0.5v - 6t - 1.5 = 0
Multiply both sides by 2 to get: v - 12t - 3 = 0
Rearrange to get: v = 12t + 3
This is our second equation (in addition to the equation you solved earlier)

Now take your first equation: (vt + 1/2v - 3t - 3/2) + tv + 3t = 180
Simplify: 2vt + 0.5v - 1.5 = 180
Replace v with 12t + 3 to get: 2t(12t + 3) + 0.5(12t + 3) - 1.5 = 180
Expand: 24t² + 6t + 6t + 1.5 - 1.5 = 180
Simplify: 24t² + 12t = 180
Subtract 180 from both sides: 24t² + 12t - 180 = 0
Divide both sides by 12 to get: 2t² + t - 15 = 0
Factor: (2t - 5)(t + 3) = 0
If 2t - 5 = 0, then t = 2.5
If t + 3 = 0, then t = -1

Since the time must be positive, the correct answer is t = 2.5

Cheers,
Brent

Hi Brent,