# Lesson: Finding the Average Speed

## Comment on Finding the Average Speed

### Hi, I solved the Yoshi avg

Hi, I solved the Yoshi avg speed in another almost similar way: as the question says, the speed Yoshi traveled from Twnsville to Villageton was 20mile/hour

This means the distance between the two cities is 20 mile. as though, the total distance is 20*2=40. for total time I just added 1 hour plus 1/4 of the hour on the way back as he was 4 times faster.

Then using the formula avg speed=total dist/total time

=40/(5/4)= 160/5= 32

What do you think?

### Great approach, Zoser.

Great approach, Zoser.

Since we aren't told the distance between Townville and Villageton, you assigned an arbitrary value of 20 miles to the distance and performed the necessary calculations.

Another useful value you could have assigned to the distance between Townville and Villageton is 80 miles.

In this case, the time TO Villageton = 80/20 = 4 hours, and the time back TO Townville = 80/80 = 1 hour.

So average speed = (total distance)/(total time)
= (80 + 80)/(4 +1)
= 160/5
= 32 miles per hour

Thanks a lot.

### I believe you're referring to

I believe you're referring to the expression 2d x 80/5d

NOTE: This is equivalent to 2d/1 x 80/5d

There are a couple of ways to deal with this.

APPROACH #1: Multiply the fractions.
We get: 2d/1 x 80/5d = 160d/5d
Divide to and bottom by d to get: 160/5
Divide to and bottom by 5 to get: 32/1 (which is equal to 32)

APPROACH #2: Simplify the fractions BEFORE multiplying (see 4:35 of https://www.gmatprepnow.com/module/gmat-arithmetic/video/1068)

Take 2d and 5d and divide both by d to get 2 and 5

So, we have: 2d/1 x 80/5d = 2/1 x 80/5
= 160/5
= 32

### The following problem is not

The following problem is not registering for me. In other words I struggle with remembering the approach.

I see you have solved it and it seems longer to me

https://gmatclub.com/forum/bill-travels-from-point-a-to-point-b-at-a-constant-speed-of-238295.html

Will appreciate any help to remember the steps

### I believe that, in order to

I believe that, in order to master this question type, it's very useful to start with a word equation (as I often do in my solutions).

Once you have a valid word equation, you can replace words with algebraic expressions and equations.

Keep at it!

Cheers,
Brent

### Thoughts on picking numbers

Thoughts on picking numbers here for distance? I chose 80 as the distance because it is the LCM of 20 and 80. Using this approach I was quickly able to get to (80+80)/5 = 32. Will this method typically be OK?

### You're referring to the

You're referring to the question that starts at 3:35 in the above video.

Your approach is perfect! Choosing "nice" values is often a great strategy for these kinds of questions.

Cheers,
Brent

### Hi Brent, is there any

Hi Brent, is there any logical explanation as to why the "Ds" cancel out in the the second example. With this I mean, how are we able to obtain the solution given that the question does not give us any values for distance and/or time of travel.

I completely understand the steps taken in your solution, but what I am unclear about is the logic/theory behind this approach.

Thanks!

### Great question!

Great question!

To understand why the distance is irrelevant, let's see what happens if we assign some values to the distance from Townville to Villageton.
-------------------------------------------------------

CASE A: Let's say the distance = 80 miles
So, the TOTAL distance = 160 miles

Now let's calculate TOTAL travel time.

Time for 1st leg of trip = distance/speed = 80/20 = 4 hours
Time for return leg of trip = 80/80 = 1 hours
So, TOTAL travel time = 4 + 1 = 5 hours

Average speed = 160/5 = 32 mph

-------------------------------------------------------
CASE B: Let's say the distance = 40 miles
So, the TOTAL distance = 80 miles

Time for 1st leg of trip = distance/speed = 40/20 = 2 hours
Time for return leg of trip = 40/80 = 0.5 hours
So, TOTAL travel time = 2 + 0.5 = 2.5 hours

Average speed = 80/2.5 = 32 mph

-------------------------------------------------------
CASE C: Let's say the distance = 240 miles
So, the TOTAL distance = 480 miles

Time for 1st leg of trip = distance/speed = 240/20 = 12 hours
Time for return leg of trip = 240/80 = 3 hours
So, TOTAL travel time = 12 + 3 = 15 hours

Average speed = 480/15 = 32 mph

-----------------------------
So, no matter what distance we use, the average speed is always the same.

Also, notice that, when we HALVE the distance (as we did in CASE B), the travel times are HALVED as well.
Likewise, when we TRIPLE the distance (as we did in CASE C), the travel times are TRIPLED as well.

So, when it comes to the average speed, the HALVING or the TRIPLING cancel out.

That is: Average speed = (HALVED distance)/(HALVED time) = distance/time
And: That is: Average speed = (TRIPLED distance)/(TRIPLED time) = distance/time

Does that help?

Cheers,
Brent

### Wow, awesome explanation!

Wow, awesome explanation! Thanks a lot Brent :)

### Can average speed questions

Can average speed questions be solved using weighted averages?

For example in the first example :

40 miles per hour was at 2/5ths the time and 60 miles per hour was at 3/5ths the time so

2/5*40 + 3/5*60 gives 52.

Is this a valid approach?

### Very nice! I never thought

Very nice! I never thought about using Weighted Averages before.

Cheers,
Brent

### Bill travels from point A to

Bill travels from point A to point B at a constant speed of 120 kilometers per hour. Upon reaching point B, he immediately heads back to point A. On his return trip to point A, Bill spends half the time traveling 20 kilometers per hour, and half the time traveling 60 kilometers per hour. What is Bill’s average speed, in kilometers per hour, for the entire roundtrip?

A) 48
B) 60
C) 66 2/3
D) 72
E) 80

A-B round trip so dividing one way trip by 2 to get distance divided into 4 parts.

4D
--------------------------------
D D D D
---- + --- + ---- + ------
120 120 20 60

After solving the average speed i got was 48kmph

Kindly help!

Thanks.

You made a mistake by dividing the TOTAL distance into 4 parts.

The question reads "On his return trip, Bill spends half the TIME traveling 20 kmh, and half the TIME traveling 60 kmh."

Notice that the question talks about half the TIME, not half the DISTANCE.
If the question had read "Bill spends half the DISTANCE traveling 20 kmh... etc", then your solution would be correct.

ASIDE: You may be asking yourself "Isn't half the TIME the same as half the DISTANCE?"
The answer is "no, they are not the same"
Here's why:

Let's say it took me 2 hours to drive from point A to point B.
Let's say I spent half the TIME driving at 1000 mph and half the TIME driving at 1 mph.
Since it's a 2-hour trip, I must have spent 1 hour driving at 1000 mph and 1 hour driving at 1 mph.
So, in the first hour I drove 1000 km, and in the second hour I drove 1 km.
As you can see, this is not the same as driving 1000 mph for half the DISTANCE and 1 mph for half the DISTANCE.

Does that help?

Cheers,
Brent

### Thanks Brent,,

Thanks Brent,,

I'm still unsure about my approach, please validate this approach if its correct.

since, the distance is not half and time is half so...

D
----------------------
1/2 x d/20 + 1/2 x d/60

= after solving i get answer as 60kmph. Is it a correct approach? If yes, can you explain me in detail how i got this answer? Please explain explaining my solution in detail.

Thanks !

### Hi Rohan,

Hi Rohan,

I'm not entirely sure what you are doing in your solution.
You have the EXPRESSION D/(1/2 x d/20 + 1/2 x d/60), but that is not an equation.

So, I'm not sure what you mean when you say "after solving" (since there is not EQUATION to solve).

Can you tell me (in detail :-), what each part of your solution represents?

Cheers,
Brent

### Hi Brent,

Hi Brent,

I am struggling with this question: https://gmatclub.com/forum/bill-travels-from-point-a-to-point-b-at-a-constant-speed-of-238295.html

My approach was the following:
120 km/h --> 2t
60 km/h --> t
20 km/h --> t

Distance = 240t + 60t + 20t
D= 320t
Rate= 320t/4t =80

I am guessing the approach is wrong because the time traveled from A to B does not equal the time travel from B to A? Could I use this approach if the time were equal? And how could I correct my approach?

Thank you so much!
BR Pia

The error occurs here:
"My approach was the following:
120 km/h --> 2t
60 km/h --> t
20 km/h --> t"

We know the time spent driving 60 kmh = time spent driving 20 kmh
So, it's perfectly to let each time = t hours

HOWEVER, we cannot say that 2t = the time spent driving 120 kmh
That is, we cannot say that the time spent driving from A to B = the time spent driving from B to A.

To understand why this is, consider the same question but with different speeds:

REVISED QUESTION: Bill travels the 120 km from point A to point B at a constant speed of 12,000,000 kilometers per hour. Upon reaching point B, he immediately heads back to point A. On his return trip to point A, Bill spends half the time traveling 0.0000000002 kilometers per hour, and half the time traveling 0.0000000006 kilometers per hour. What is Bill’s average speed, in kilometers per hour, for the entire roundtrip?

We can see that the trip from A to B will take less than 1 second.
However, the trip from B to A will take days and days.

So, it's incorrect to say that the time spent driving from A to B = the time spent driving from B to A.

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

https://gmatclub.com/forum/reiko-drove-from-point-a-to-point-b-at-a-constant-speed-and-138183.html

Explanation for Ans A

UFFICIENT: Say the trip is d miles long in each direction, so that the round-trip distance is 2d miles. According to this statement, Reiko took (2d miles)/(80 miles/hour) = d/40 hours to drive the entire round trip.

Reiko could not have driven from B to A in zero time, so it must have taken her less than d/40 hours to drive from A to B. Therefore, her speed on the trip from A to B must have been (d miles)/(LESS than d/40 hours) = 1/(LESS than 1/40 hours) = GREATER than 40 miles per hour.

(Note: when dividing by a "less than" number, flip the sign to "greater than." If we had been dividing by a "greater than" number, we would have flipped the sign to "less than.")

Alternatively, try the above with real numbers: say the trip is 80 miles long in each direction, so that the round-trip distance is 160 miles. According to this statement, Reiko took (160 miles / 80 miles/hour) = 2 hours to drive the entire round trip.

Reiko could not have driven from B to A in zero time, so it must have taken her less than 2 hours to drive from A to B. Therefore, her speed on the trip from A to B must have been (80 miles)/(LESS than 2 hours) = (40 miles)/(LESS than 1 hour) = GREATER than 40 miles per hour.

Doubt:

If 80km in less than 2, how can they assume 40km will be less than 1hr.

Could you please me understand the gap in my reasoning?

Thanks

### I'm not crazy about that

I'm not crazy about that solution.
I just posted my full solution here: https://gmatclub.com/forum/reiko-drove-from-point-a-to-point-b-at-a-cons...

Take a look, and see what you think.