# Lesson: Multiple Trips or Multiple Travelers

## Comment on Multiple Trips or Multiple Travelers

(travel time at FASTER speed) = (5/7)(travel time at SLOWER speed)

time = distance/speed

Let x = Jolene's SLOWER (actual) speed
So, x + 10 = Jolene's FASTER speed

Substitute values into word equation to get: 150/(x + 10) = (5/7)(150/x)
Simplify to get: 150/(x + 10) = 750/7x
Cross multiply to get: (750)(x + 10) = (7x)(150)
Expand: 750x + 7500 = 1050x
Subtract 750x from both sides: 7500 = 300x
So: x = 7500/300 = 25

So, Jolene's speed was 25 mph.

QUESTION: How many minutes did the trip take ?
time = distance/ speed
So, her time = 150/25 = 6 HOURS
6 HOURS = 360 MINUTES

Cheers, Brent

### Hi Brent,

Hi Brent,

Alex and James start running each other from two towns 40 kilometers apart. Alex runs at a constant speed of 9km/h while James runs at a constant speed of 11km/h. At the same time, Bill starts cycling towards James from the same starting point as Alex at a constant speed of 15km/h. Once he reaches James, he turns around and starts cycling towards Alex. Upon reaching Alex, he turns again and starts cycling towards James. This process is repeated until Alex and James cross each other. In this process, how much distance does Bill cycle ?
A.18km, B.22km, C.30km,D.40km, E.80km

### Tricky!

Tricky!

This is a shrinking gap question.
9 kmh + 11 kmh = 20 kmh
So the gap shrinks at a rate of 20 kilometers per hour

time = distance/speed

Alex and James are originally separated by 40 km
time = 40/20 = 2
So, it'll take 2 HOURS for the two runners to meet.

During those 2 HOURS, Bill cycles at a rate of 15 km.
Distance = (speed)(time) = (15)(2) = 30
So, Bill's total travel distance = 30 km

### Hi Brent,

Hi Brent,

Mike and Bryan walk up a moving escalator. The escalator moves at a constant speed. Mike takes five steps for every four of Bryan's. Mike gets to the top of the escalator after having taken 30 steps, while Bryan takes only 27 steps to reach the top. If the escalator was turned off, how many steps would they have each to take to reach the top?
A.37, B.42, C.54, D.57, E.60

### Hi Brent I have the same

Hi Brent I have the same question as Joelle2020

### My apologies to Joelle2020. I

My apologies to Joelle2020. I don't know how I missed that question! Here's my belated solution:

Let's say Mike's speed is 5 steps PER MINUTE
So, Bryan's speed is 4 steps PER MINUTE

Travel time = (total steps)/(speed in steps per minute)

So, if Mike takes 30 steps, then his travel time is 6 minutes (30/5 = 6)
Likewise, if Bryan takes 27 steps, then his travel time is 6.75 minutes (27/4 = 6.75)

Let x = escalator's speed in steps PER MINUTE

So, 5 + x = Mike's overall speed (in steps PER MINUTE)
And, 4 + x = Bryan's overall speed (in steps PER MINUTE)

Since Mike and Bryan each traveled the same distance, we can write:
(Mike's distance) = (Bryan's distance)
Distance = (speed)(time)
So, we get: (5 + x)(6) = (4 + x)(6.75)
Expand: 30 + 6x = 27 + 6.75x
Solve: x = 4

So, the escalator travels at a speed of 4 steps per minute.

Important: Notice that the escalator travels at the SAME SPEED that Bryan travels (4 steps per minute)
So, when Bryan takes 27 steps to reach the top, the escalator did half the work.
So if the escalator were turned off, Brian would need to take twice as many steps (i.e., 54 steps)

-------------------
ANOTHER APPROACH
After completing the above solution, I realize that there's a much faster approach.

Given: Mike takes five steps for every four of Bryan's.
So, at the moment when Mike reaches the top, he has taken 30 steps, and Bryan has taken 24 steps.
30 - 24 = 6
So, at the moment Mike reaches the top, Bryan has 6 steps remaining.
However, to get to the top, Bryan takes only 3 more steps (since he takes a total of 27 steps)

This tells us that the escalator is travelling at the same speed that Bryan is travelling.
For example, in the time it takes Bryan to take one step, the escalator travels one step.
So, when Bryan takes 27 steps to reach the top, the escalator did half the work.
So if the escalator were turned off, Brian would need to take twice as many steps (i.e., 54 steps)

Cheers,
Brent

### Had to say that it is really

Had to say that it is really annoying to solve the equation so I just plug in C to the equation to get the answer.

https://gmatclub.com/forum/a-bus-trip-of-450-miles-would-have-taken-1-hour-less-if-the-average-203850.html

### Question link: https:/

Great move! Most students don't bother to test answer choices, but there are tons of GMAT questions that can be solved this way. In fact, there are many times when testing the answers choices is the fastest approach.

### Where is the solution of this

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

### There are lots of solutions

There are lots of solutions in that link.
If you're looking for my solution, it's here on page 2: https://gmatclub.com/forum/jerry-and-jim-run-a-race-of-2000-m-first-jerr...

### Hi Brent, are you able to

Hi Brent, are you able to elaborate on the use of the alligation rule?

I'm referring to one of the solutions posted for this question: https://gmatclub.com/forum/a-certain-road-trip-was-completed-in-two-parts-interrupted-by-an-hour-233301.html

It looks like an interesting technique to use on questions, but I don't know its full context for use.
I was also confused because they put the average of 55 and 63 as 60 instead of 59, so that would change the ratio they get at the end.

Thank you!

### I'm not a fan of the

I'm not a fan of the alligation method. At least I wasn't a fan when it was first explained to me. I don't even remember how it works, but I recall finding it limiting because it only works in certain situations.
I can tell you that all GMAT questions can be solved using methods other than alligation.

### Thanks Brent for all the

Thanks Brent for all the brilliant golden nuggests video. Great innvoative teaching and thiinking out of the box tips. Loving it !!!

### Thanks Kimberly. That's nice

Thanks Kimberly. That's nice of you to say.

### Hi Brent, could you help

Hi Brent, could you help below question?

There are 12 A machines can print 850,000 in 8 hrs. There are 8 B machines can print 720,000 in 8 hrs. If 4 A machines and 3 B machines work simultaneously for 12 hrs, how many can both print?

### Given: 12 A machines can

Given: 12 A machines can print 850,000 in 8 hrs
So, in 4 hours, 12 A machines can print 425,000 (half the time means half the quantity)
So, in 12 hours, 12 A machines can print 1,275,000 (triple the time means triple half the quantity)
So, in 12 hours, 4 A machines can print 425,000 (dividing the number of machines by 3 means dividing the quantity by 3)

Given: 8 B machines can print 720,000 in 8 hrs.
So, 1 B machine can print 90,000 in 8 hrs (dividing the number of machines by 8 means dividing the quantity by 8)
So, 3 B machines can print 270,000 in 8 hrs (tripling the number of machines means tripling the quantity by 8)
So, 3 B machines can print 135,000 in 4 hrs (halving the work time means halving the quantity)
So, 3 B machines can print 540,000 in 12 hrs (tripling the work time means tripling the quantity)

If 4 A machines and 3 B machines work simultaneously for 12 hrs, they can print 425,000 + 540,000, which is 965,000

### Great thanks Brent.

Great thanks Brent.