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# Algebra Help? A bit confused...

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During his morning commute to work in rush hour traffic, Justin's average speed was 30 mi/h. During his afternoon commute back home along the same route, his average speed was 60 mi/h. What was his average speed for the entire round trip?

ATTEMPT:

Okay, so I figured that you would have the same distance right?

So all you need to do is take an average of 30, and 60???

Which is 45???

Could somebody check my work?

Thank you!

\(tommarvoloriddle\)

Nov 9, 2019
edited by tommarvoloriddle  Nov 9, 2019

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Not quite, Tom......

The average rate   =  2 * product of the rates          2 * 30 * 60            3600

____________________  =  ___________  =   _____  =  40 mph

sum of the rates                      30 + 60                90

P.S.....I can prove this if you want.....

Nov 9, 2019
edited by CPhill  Nov 9, 2019
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T-T

Where have I gone wrong?

Was it an assumption?

tommarvoloriddle  Nov 9, 2019
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Can you prove it?

I got a teensy bit confused...

IRRELEVANT SORRY I GET DISTRACTED!

tommarvoloriddle  Nov 9, 2019
edited by tommarvoloriddle  Nov 9, 2019
edited by tommarvoloriddle  Nov 9, 2019
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OK  ....call  the distance one way,  D

So....the round trip  =  2D

The time one way  is    D /30

The time back  is   D /60

So.....the average   rate   =     Total Distance                 2D

____________  =   ____________

Total Time                 D/30 + D/60

If  you simplify this.....you should get  40 mph

Nov 9, 2019
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Ohhhh...

Thank you, CPhill!

tommarvoloriddle  Nov 9, 2019
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Actually, I believe it should be 40mph. If we let the distance be 60 miles, than it takes Justin 2 hrs to work, and 1 hr back home. That sums up to a total of 3 hrs for a distance of 2*60 = 120 miles. Dividing 120 by 3 gives us an answer of 40mph.

Nov 10, 2019