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## Comment on

Finding the Average Speed## Hi, I solved the Yoshi avg

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 dont understand why the d`s

## 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)

Start with 2d/1 x 80/5d

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

For more information on working with fractions, see https://www.gmatprepnow.com/module/gmat-arithmetic/video/1065 and https://www.gmatprepnow.com/module/gmat-arithmetic/video/1068

## The following problem is not

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

## 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

## Add a comment