# Lesson: Multiple Trips or Multiple Travelers

## Comment on Multiple Trips or Multiple Travelers

### For Ata I had the equation

For Ata I had the equation 100(t-1) and for Carl I got 75t. I set the both equal to each other because both are equal to d, the distance. I got t=4 plugged that back in the equations and got 300 for both of them. Is this another way to do it? ### That's perfect.

That's perfect.

Instead of setting Carl's travel time as 1 hour GREATER THAN Ata's travel time (which I did in my solution), you set Ata's travel time as 1 hour LESS THAN Carl's travel time. Perfectly valid approach.

### At 3:00 pm, a car has driven

At 3:00 pm, a car has driven 30 miles east. It will continue to drive east at 0.8 minutes per mile and then turn around and drive at 0.8 minutes per mile back to its original starting point. How far can it drive before turning around in order to arrive back to its original starting point by 3:40 pm? ### Good question!

Good question!

Let x = the distance the car drives east (until turning around)
So, x + 30 = distance the car travels to get to original starting point.

This means the TOTAL DISTANCE = x + (x + 30) = 2x + 30

We're told the TOTAL TRAVEL TIME (starting at 3:00 pm) is 40 minutes.

Let's rewrite the car's speed is miles per minute.
We're told the speed is 0.8 minutes per mile.
In other words, the speed is: 4/5 minutes per 1 mile.
Multiply both values by 5/4 to get an EQUIVALENT speed: 1 minute per 5/4 miles.
In other words, the car's SPEED = 5/4 miles per minute.

Now that we have our speed, distance and time, we can use the formula: DISTANCE = (SPEED)(TIME)

We get: 2x + 30 = (5/4)(40)
Simplify: 2x + 30 = 50
Solve: x = 10

So, the car must drive 10 more miles before turning around in order to arrive back to its original starting point by 3:40 pm

### At 3:00 pm car started to go

At 3:00 pm car started to go towards east and it traveled for 30miles at 5/4 miles/minute.
time taken by car to travel 30miles was 24 min ( (30) * (4/5) ). at 3:24 pm car completed 30miles. in next 16 min what will be the distance traveled by car is what the question asks, right?. so at constant rate car would go next 20miles ( (5/4) * (16) ).
whats the mistake in my understanding of the question? ### You have misread the question

You have misread the question. At 3:00pm the car is ALREADY 30 miles east of where it started.

Here's a crude diagram of the situation:

HOME.....>>........(CAR at 3pm)....x miles.....(STOP and head HOME)

We want to determine the value of x, which is the number of miles the car can travel east BEFORE turning around and arriving HOME at 3:40pm

### Oh sorry!, by 3:00pm car has

Oh sorry!, by 3:00pm car has DRIVEN 30miles. At3:00pm car is still moving towards east at 5/4 miles/minute. Question is asking that car has to reach back home by 3:40pm, so how far car can travel, from that point where it has already reached 30miles, isn't it?,

I approached the problem in following way,

let X be the total distance traveled by car (including 30miles for return)
car would travel 50 miles (5/4 *40), considering 40 minutes is its time.
car has moved some distance from point where it has started at 3:00pm, then returned
home, entire distance was 50 miles. Now X-30 will give how much distance it has moved east, right?. so I did 50-30 which gives 20 miles. still i am not getting what the mistake? please help me in understanding this clearly. ### You ALMOST have it.

You ALMOST have it.

Let Q = the location of the car at 3:00pm

The 20 miles you have as your answer includes the distance traveled east from point Q AND the the distance traveled east to get back to point Q (before continuing home)

So, we must divide 20 by 2 to get the correct answer of 10

### I approached this problem a

I approached this problem a little differently. I computed Ata's travel time as: d/100. We are told that Ata arrived an hour earlier. So, Carl's travel time with respect to Ata's is d/100 +1. Since d/75 and d/100 + 1 represent Ata's travel time, they must be equal.

d/75 = d/100 + 1 ---> d/75 = (d + 100)/100
100d = 75(d + 100) = 75d +7500 --> 100d - 75d = 7500
25d = 7500 --> d = 300 Nice work!

### For the first method of

For the first method of solving the problem you stated Ata's time+1=Carl's time. Why not the other way round ie Ata's time = Carl's travel time +1 ### We're told that Ata arrived

We're told that Ata arrived in Townville 1 hour earlier than Carl.
So, Ata spent LESS TIME driving.
In other words, Ata's travel time is 1 hour LESS THAN Carl's driving time.

At the moment, (Ata's travel time) ≠ (Carl's driving time)
We want to somehow create an EQUATION.
How do we do this?

Since Ata's travel time is 1 hour LESS THAN Carl's driving time, we can make their travel times equal by adding 1 hour to the smaller travel time.
In other words, if add 1 hour to Ata's travel time (the lesser value), we can create the equation:
(Ata's travel time) + 1 = (Carl's driving time)

In your proposed equation, you are taking the bigger value (Carl's driving time) and making it even bigger. This approach won't create an equation.

Does that help?

Cheers,
Brent

### Is my mistake taking time and

Is my mistake taking time and travel time to be the same. Because at 4.16 in the video Carl's time was increased by 1 ### At 4:16 in the video, we are

At 4:16 in the video, we are equating the DISTANCES (not the times, as we did in the first solution).

Cheers,
Brent

### Can you solve the first GrE

Can you solve the first GrE reinforcement activities question using another equation besides travel distance ### You bet.

You bet.
Here's my full solution: https://greprepclub.com/forum/running-on-a-10-mile-loop-in-the-same-dire...

Cheers,
Brent

Please kindly solve the second also in terms of rate only. Happy to help.

Here's my full solution: https://greprepclub.com/forum/bill-and-ted-each-competed-in-a-240-mile-b...

Cheers,
Brent

### I got the equation right.but

I got the equation right.but i lost it when i simplified 240/t+5 as 240+5t/t i multiplied 5 by t and it's denominator 1 by t also so i could combine them. Is this wrong? ### It's hard to tell what you're

It's hard to tell what you're doing if you don't add spaces or brackets.

If you're talking about taking the equation 240/t + 5 = 240/(t - 4), and first combining the terms on the left side, then here's what we need to do:

Take: 240/t + 5 = 240/(t - 4)
Rewrite 5 as 5t/5 to get: 240/t + 5t/t = 240/(t - 4)
Combine terms on left side to get: (240 + 5t)/t = 240/(t - 4)

Now that we have two fractions equaling each other, we can . . .
Cross multiply to get: (240 + 5t)(t - 4) = (240)(t)
Use FOIL to expand: 240t - 960 + 5t² - 20t = 240t
Subtract 240t from both sides: -960 + 5t² - 20t = 0
Rearrange: 5t² - 20t - 960 = 0
Divide both sides by 5 to get: t² - 4t - 192 = 0
Factor to get: (t - 16)(t + 12) = 0
So, EITHER t = 16 OR t = -12
...etc

For more on dealing with fractions in equations, watch https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

Cheers,
Brent

### Thanks this is exactly what i

Thanks this is exactly what i did i just made a minor mistake when i cross multiplied. i would love to see this question solved equating distances also ### I tried solving equating

I tried solving equating distances using t=bill's time and t-4= Ted's time. B= TED'S RATE AND B-5=BILL'S RATE. I GOT B=15 BUT Bill's rate =B-5 and since b is 15,15-5=10.I know this is incorrect.i just wanted to see if i can solve it with a different variable assignation. Can you correct me? ### In order to determine what

In order to determine what went wrong with your solution, I need to see your full solution.
I can tell you that you assigned the variables correctly.

### With all steps being the same

With all steps being the same.
Bill distance=Ted's distance
(B-5)T=B(T-4)
BT-5T=BT-4B**subtract BT from both sides = -5T=-4B***divide both sides by -5 will give T=4/5*B(4B/5)

It's said that Bill traveled 240 miles So,(Bill's speed)(Bill's travel time)=240
(B-5)T=240;replacing t with 4B/5
(B-5)4B/5=240;on expansion we get
4B^2-20B/5=240; multiplying both sides by 5 gives 4B^2-20B=1200
4B^2-20B-1200=0 dividing through by 4 gives B^2-5B-300=0
Factor (B+15)(B-20)=0 B=15
Bill's rate is given as B-5 so 15-5=10. ### So close!!

So close!!
Everything is perfect to this point: (B + 15)(B - 20) = 0
This tells us that EITHER B = -15 OR B = 20
Since B cannot be negative, B must equal 20
This means Bill's speed = B - 5 = 20 - 5 = 15

Cheers,
Brent

### Well my factoring produced B

Well my factoring produced B+15 and B-20 i can't take B=20 because of the negative sign. I was thinking there was an error in the calculation that produced a negative sign for 20 and positive for 15.we can't just pick a number and change it's sign to suit our calculation or can we?

### Oh I'm supposed to say B+15=0

Oh I'm supposed to say B+15=0 which is -15
B-20=0 WHICH IS 20. JEEZ THANKS BRENT ### Exactly.

Exactly.
If (B + 15)(B - 20) = 0, then we want values of B that satisfy the equation.

If we think B = 15 is a solution, we can verify this by plugging B = 15 into the equation.
We get: (15 + 15)(15 - 20) = 0 (doesn't work)

For more on solving quadratic equations, watch: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

### https://greprepclub.com/forum

https://greprepclub.com/forum/bill-spends-two-days-driving-from-point-a-to-point-b-11887.html

I assumed that the speed on day 1 is y+5 and speed on day 2 is y. Then i used it in distance = speed * time
680 = (2y+5) * 18
y is then 16.4
what is wrong here? For example, if Bill drives 6 mph on Day 1, and he drives 1 mph on Day 2, we can't then conclude that, over the 2-day period, his average speed was 7 mph.

Also, you are not taking into consideration the TIME that Bill spent driving EACH day.

Cheers,
Brent

### Hi,

Hi,

How do you do p262 #15.

Thank you!! ### I'm happy to help!

I'm happy to help!

ASIDE: In the future, please post the entire question. There are 3 editions of the GRE Official Guide.

A group can charter a particular aircraft at a fixed total cost. If 36 people charter the aircraft rather than 40 people, then the cost per person is greater by \$12.

(a) What is the fixed total cost to charter the aircraft?
(b) What is the cost per person if 40 people charter the aircraft?

(a) Let T = the fixed total cost to charter the aircraft
With 40 people sharing the cost, the cost per person = T/40
With 36 people sharing the cost, the cost per person = T/36

GIVEN: The per-person cost with 36 people is \$15 MORE THAN the per-person cost with 40 people.
In other words: (cost per person with 36 people) = (cost per person with 40 people) + 15
Substitute values to get: (T/36) = (T/40) + 15 [Solve for T]

Since 360 is a multiple of 36 and 40, we can eliminate the fractions by multiplying both sides of the equation by 360.
When we do this, we get: 10T = 9T + 5400
Solve: T = \$5400

----------------------------------
(b) With 40 people sharing the cost, the cost per person = T/40
Now that we know T = \$5400, the cost per person = \$5400/40 = \$135

Cheers,
Brent

### Hi,

Hi,
Very nice techniques. I came up with a solution without a lot of workings. Because there's a difference of one complete hour total distance needs to be the LCM of 100 and 75, which is 300. ### That's a great observation!

That's a great observation!
Unfortunately, it was only a coincidence.

For example, if Ata's speed were 5 km per hour, and Carl's speed were 3 km per hour, the distance would be 7.5 km.
7.5 is not the LCM of 3 and 5

Similarly, if Ata arrived in Villageton TWO hours before Carl, the distance would be 600 km.
600 is not the LCM of 75 and 100

Cheers,
Brent

### You are right, I can't think

You are right, I can't think of any logic for Ata's 5 km and Carl's 3km per hour speed scenario, they have a difference of one hour but the distance is half of 5 and 3's LCM.

Whilst on the other hand, if we take 100 and 75km speeds for the 1-hour difference it's the LCM, for the 2-hour difference it's the second common multiple which is 600 and for the 3-hour difference the third common multiple which is 900.

Thank you for such an amazing series though, it'helping remove the brain rust. ### Great observation!

Great observation!
I'm sure there's a mathematical reason why it doesn't work for 5 kmh and 3 kmh, but I don't see it immediately.

Cheers,
Brent

### Hi need help with question 4

Hi need help with question 4 -
https://greprepclub.com/forum/on-a-partly-cloudy-day-derek-decides-to-walk-back-from-work-12332.html

This statement makes no sense to me
"Since s is an integer and the av. speed is 2.8
hence the s and s+1 should be 2 and 3"

Isnt average of 2 and 3, 2.5. 2.8 is not an average of 2 and 3. ### https://greprepclub.com/forum

https://greprepclub.com/forum/on-a-partly-cloudy-day-derek-decides-to-wa...

Great question!!

Many students (incorrectly) believe that finding one's average speed is as easy as just finding the average of the different speeds traveled.

I cover the flaw in that reasoning from 0:20 to 1:30 of the following video: https://www.greenlighttestprep.com/module/gre-word-problems/video/914

The big takeaway is this:
If someone travels x miles per hour for part of a trip, and then travels y miles per hour for the remainder of the trip, then the AVERAGE speed will be BETWEEN x and y miles per hour.

So for example, if Derek travels 2 mph for a while and then travels 3 mph for a while, then his average speed will be BETWEEN 2 and 3 mph.

Cheers,
Brent

### Hi Brent ,

Hi Brent ,
Could you please tell me why did you consider the distances as different as he is going from work to home days may be different but, distance is same.

Thanks ### In the question, Derek is

In the question, Derek is only walking in one direction (from work to home)
For part of distance he walks at a speed of s mph (when sunny) and for part of that distance, he walks at a speed of s+1 mph (when cloudy)

So, we can write: (distance traveled while sunny) + (distance traveled while cloudy) = TOTAL distance traveled

Does that help?

Cheers,
Brent

### Hello,

Hello,
In the second exercise, why do you add 4 instead of -4 .
Ted completed 4 hours sooner so it should not be -4?

Exercise:
Ted completed the race 4 hours sooner than Bill did
(Bill's travel time) = (Ted's travel time) + 4

Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Let B = Bill's travel speed
So, B + 5 = Ted's average speed

Ted completed the race 4 hours sooner than Bill did
(Bill's travel time) = (Ted's travel time) + 4 Given: Ted completed the race 4 hours sooner than Bill did
So, for example, it could be the case that the race began at noon, and Ted completed the race at 3pm and Bill completed the race at 7pm.

In this case, Ted's travel time = 3 hours
And Bill's travel time = 7 hours

Notice that Ted's travel time is FOUR HOURS LESS THAN Bill's travel time
In order to make their travel times EQUAL, we must add 4 hours to Ted's travel time.
That is: (Bill's travel time) = (Ted's travel time) + 4

Does that help?

Cheers,
Brent

### Running on a 10-mile loop in

Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles per hour and Rob ran at a constant rate of 6 miles per hour. If they began running at the same point on the loop, how many hours later did Sue complete exactly 1 more lap than Rob?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Could we solve this as follows :
8t ( sue speed/ per hour ) = 6t ( Robsue speed/ per hour) +10
2t= 10
t = 5

10t = 8t + 10
t = 5 In fact, it's exactly how I MEANT to create that equation, but I made the mistake of letting Sue's rate = 10 mph and Rob's rate = 8 mph.
Not sure where I came up with those numbers :-)

I have edited my response accordingly.

Cheers and thanks!

### On a partly cloudy day, Derek

On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of "s" miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

I reviewed your answer and had a difficulty to understand as below part ( how did you identify Dreck total traveled miles?) :

If Derek's average speed is 2.8 mph, then let's say that he traveled a total of 28 miles. Since there is no information that tells us the distance that Derek walked, we can assume that the distance is irrelevant (otherwise would not be able to answer the question).

So, instead of assigning some variable to the distance walked, I just chose a "nice" value that would work with the given information,

Here's my solution to the same problem, but this time I don't assign a nice value to the distance: https://greprepclub.com/forum/on-a-partly-cloudy-day-derek-decides-to-wa...

Cheers,
Brent

### Hi Brent

Hi Brent
For the quwstion ata and carl

I had approached in this way

Da=100*t1

Dc=75*t2

t2-t1=1

D/75 - D/100 =1
100D - 75D = 7500
25D=7500
D=300
Thanks ### Perfect approach!

Perfect approach!

### Hi Brent,

Hi Brent,

https://greprepclub.com/forum/al-and-ben-are-drivers-for-sd-trucking-company-one-snowy-da-17770.html
For this problem I set it up the same way you did, however i set t as Ben's travel time so I had: 40(t+3) +20(t) = 240 and solved for t, is it wrong to start with t for Ben, given that we know that the retrieval of data happened after 11 AM?