stupid math question I can't figure out

[TheWart]

Here is the question:

A cyclist bikes x distance at 10 miles per hour and returns over the same path at 8 miles per hour. What is the cyclist's average rate for the round trip in miles per hour?






The given answer is 8.9

How do you get that? I saw the question and was like, uhhh, of course it is 9mph....but that's wrong!

 

[blinky8225]

The return trip time changes. It takes longer when he goes back 8 miles per hour. There for he spends more time at 8 miles per hour, so the rate should slightly lean toward 8 from 9.

 

[TheWart]

Originally posted by: blinky8225
The return trip time changes. It takes longer when he goes back 8 miles per hour. There for he spends more time at 8 miles per hour, so the rate should slightly lean toward 8 from 9.

Right, he is going 1.25 times as slow, but I don't see how the length of the trip matters in this case, cause he is going two speeds: 10mph and 8mph, so how could the avg not be 9?

sorry if I am being bone headed, lol

 

[Cancer12]

He spends longer time going 8 miles an hour

 

[rbV5]

Just do the problem. He'll do 20 miles in 2.25 hours.

 

[JohnCU]

duh nevermind

 

[blinky8225]

Originally posted by: TheWart

Originally posted by: blinky8225
The return trip time changes. It takes longer when he goes back 8 miles per hour. There for he spends more time at 8 miles per hour, so the rate should slightly lean toward 8 from 9.

Right, he is going 1.25 times as slow, but I don't see how the length of the trip matters in this case, cause he is going two speeds: 10mph and 8mph, so how could the avg not be 9?

sorry if I am being bone headed, lol

From the equation, Distance (mi) = rate (mi/h) * time (h):

10 * t1 = x --> t1 = x / 10
8 * t2 = x --> t2 = x / 8

Average speed is (Total Distance) / (Total Time) = (2x) / (t1 + t2) = (2x) / (x/10 + x/8)

= (2) / (1/10 + 1/8) = (16) / (8/10 + 1) = (160) / (18) = (80/9) = 8.88888888888888888888888888..., which rounds to 8.9. Yay math!

No integrals required, JohnCU. This looks like a common SAT problem that tries to trick you into putting 9.

 

[Chiropteran]

Originally posted by: Cancer12
He spends longer time going 8 miles an hour

Good thing it wasn't an English grammar question.

 

[JohnCU]

Originally posted by: blinky8225

From the equation, Distance (mi) = rate (mi/h) * time (h):

10 * t1 = x
8 * t2 = x

Average speed is (Total Distance) / Time = (2x) / (t1 + t2) = (2x) / (x/10 + x/8)

= (2) / (1/10 + 1/8) = (16) / (8/10 + 1) = (160) / (18) = (80/9) = 8.88888888888888888888888888..., which rounds to 8.9. Yay math!

No integrals required, JohnCU. This looks like a common SAT problem that tries to trick you into putting 9.

yeah i realized that after i wrote it out 😱

 

[summit]

assume the trip is some fixed number. 80 miles for easy math. it takes him 8 hrs there and 10 hrs back. thats a total of 18 hrs. 160 miles divided by 18 hrs is ~8.9 miles per hour.

holla back young'n.

 

[TheWart]

ahhh, ok.

I feel dumb now. Thanks everyone!

 

[OCGuy]

The answer is: copy off the Asian next to you

 

[Taejin]

Originally posted by: Ocguy31
The answer is: copy off the Asian next to you

:roll:

 

[blinky8225]

Originally posted by: JohnCU

Originally posted by: blinky8225

From the equation, Distance (mi) = rate (mi/h) * time (h):

10 * t1 = x
8 * t2 = x

Average speed is (Total Distance) / Time = (2x) / (t1 + t2) = (2x) / (x/10 + x/8)

= (2) / (1/10 + 1/8) = (16) / (8/10 + 1) = (160) / (18) = (80/9) = 8.88888888888888888888888888..., which rounds to 8.9. Yay math!

No integrals required, JohnCU. This looks like a common SAT problem that tries to trick you into putting 9.

yeah i realized that after i wrote it out 😱

No worries, it happens to all of us after taking Calculus 3, Differential Equations, and whatever other high level math classes--we forget the easy stuff and make things harder for ourselves.

 

[jagec]

The answer is "it depends on how long he waits at the other end."😛

If the cyclist were to bike for one hour @ 10MPH and then one hour @ 8MPH, the average would be 9 MPH.

But this is constant distance, not constant time. Put it this way: Let's say you go 10,000MPH for one hour, and then walk back at 1 MPH. You have travelled 20,000 miles total, and taken 10,001 hours to do so. The average speed, therefore, is just below 2MPH...much closer to the slower speed than the faster.

 

[magreen]

How cool am I that I did it in my head and got the right answer (80/9) ?