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A math problem
Posted: Sat, 01 Nov 2014 12:17:38
by Chasbo00
You are driving to work and the traffic sucks. You have dirven exactly half way to work and averaged only 20 miles per hour. How fast must your average speed be during the second half of the way to work to average 40 miles per hour for the entire trip to work?
A. 80 miles per hour
B. 50 miles per hour
C. 60 miles per hour
D. Faster than the speed of light
Re: A math problem
Posted: Sat, 01 Nov 2014 12:32:52
by dorminWS
D.
Re: A math problem
Posted: Sat, 01 Nov 2014 14:54:05
by Chasbo00
dorminWS wrote:D.
Anyone else?
Re: A math problem
Posted: Sat, 01 Nov 2014 15:33:46
by HighExpert
C would be my answer.
Re: A math problem
Posted: Sat, 01 Nov 2014 16:20:14
by Snakester
It would have to be "C" .
Re: A math problem
Posted: Sat, 01 Nov 2014 18:29:02
by Palladin
How far is it to work?
Hell - I didn't want to go to work today anyway. I'll never make it on time...

Re: A math problem
Posted: Sat, 01 Nov 2014 18:45:15
by OakRidgeStars
I'm usually stuck behind the driver going 20MPH wishing I could be driving 60.

Re: A math problem
Posted: Sat, 01 Nov 2014 18:55:22
by Monkey
By process of elimination, it has to be D. A, B & C aren't fast enough to support every Virginia driver's twice-daily quest to qualify for NASCAR...
Re: A math problem
Posted: Sat, 01 Nov 2014 19:26:15
by Chasbo00
D is the correct answer.
It's not possible to double your average speed after going half way as you have used up all the time just going half way. To make the math simple, use 40 miles as the total distance to work (it could be any distance). To average 40 miles per hour would take one hour, but you have already used 1 hour just going half way (20 miles at 20 miles per hour). So, D is the only answer as you will have to bend time by going faster than the speed of light.
Remember what you were told about one of the multiple choice answers being ridiculous and you could discount it - that's not always the case.
For those more mathematically inclined, you need to use the harmonic mean and not a standard arithmetic mean when averaging rates.
Re: A math problem
Posted: Sat, 01 Nov 2014 22:24:56
by AlanM
Miss Bright could do it with ease.
There once was a lady named Bright
who could travel much faster than light.
She left one day in an Einstein way,
and returned the previous night.
Re: A math problem
Posted: Sat, 01 Nov 2014 23:47:59
by HighExpert
Is this some of that new math? You know, 2+6=5.
Re: A math problem
Posted: Sun, 02 Nov 2014 00:07:57
by Chasbo00
HighExpert wrote:Is this some of that new math? You know, 2+6=5.
Nope, it's the same old real math applied to some simple physics.
Re: A math problem
Posted: Sun, 02 Nov 2014 09:48:36
by Radiac
Answer is C.
Re: A math problem
Posted: Sun, 02 Nov 2014 10:36:30
by Radiac
Oops. Guess I should have cheated and looked for the answer first. Ah well.
Re: A math problem
Posted: Mon, 03 Nov 2014 09:26:09
by dorminWS
Chasbo00 wrote: D is the correct answer.
It's not possible to double your average speed after going half way as you have used up all the time just going half way.
Exactly. No math involved, really. Just a little mental elbow grease.
Chasbo00 wrote:
Remember what you were told about one of the multiple choice answers being ridiculous and you could discount it - that's not always the case.
In this case, three answers were nonsense, and the one that was right SOUNDED ridiculous. I reckon questions like this are the REASON for the general rule about one of the multiple choice answers being ridiculous and you can discount it. Misdirection.
Chasbo00 wrote:
For those more mathematically inclined, you need to use the harmonic mean and not a standard arithmetic mean when averaging rates.
Harmonic mean....... Now THERE's a term I haven't seen or heard since Hector was a pup.
Re: A math problem
Posted: Mon, 03 Nov 2014 10:54:01
by MarcSpaz
Man, I was expecting something like .1*.1. Your example sucks, because no one in my area averages 20mph on the way to work. its simply not fathomable.
Re: A math problem
Posted: Mon, 03 Nov 2014 13:30:16
by wittmeba
Chasbo00 wrote:D is the correct answer.
It's not possible to double your average speed after going half way as you have used up all the time just going half way. To make the math simple, use 40 miles as the total distance to work (it could be any distance). To average 40 miles per hour would take one hour, but you have already used 1 hour just going half way (20 miles at 20 miles per hour). So, D is the only answer as you will have to bend time by going faster than the speed of light.
Remember what you were told about one of the multiple choice answers being ridiculous and you could discount it - that's not always the case.
For those more mathematically inclined, you need to use the harmonic mean and not a standard arithmetic mean when averaging rates.
I know you posted this to open a discussion

... so here goes.
I've seen this before. I don't know the answer, however, I don't believe your answer fits the problem. There is nothing to say you have used all your time. I could drive half way to to my mailbox and average 20 MPH without using 1 hour but still obtain a MPH value (distance and time).
If work is more than 40 miles I would agree. Still thinking about this, however. Do we have all the pieces of the equation?
Re: A math problem
Posted: Mon, 03 Nov 2014 14:03:07
by Chasbo00
wittmeba wrote:
I've seen this before. I don't know the answer, however, I don't believe your answer fits the problem. There is nothing to say you have used all your time. I could drive half way to to my mailbox and average 20 MPH without using 1 hour but still obtain a MPH value (distance and time).
The one hour was associated with my example of using a total distance of 40 miles and driving half way (20 miles) at 20 miles per hour. In this example, it takes 1 hour to drive half way. If you are going to average 40 miles per hour for the entire trip, this too would take one hour, but you have already taken an hour to go half way.
I only used 40 miles total distance as an example because it made the math easy. The total distance to work can be any value.
wittmeba wrote:
If work is more than 40 miles I would agree. Still thinking about this, however. Do we have all the pieces of the equation?
The actual distance and speed do not matter, you can use any values. Once you have gone half way, it's not possible to double your average speed for the entire trip without somehow going back in time.
Re: A math problem
Posted: Mon, 03 Nov 2014 14:24:41
by MarcSpaz
I was wondering who was going to catch that. No one did. You should have waiting longer.
Plus, its not possible to exceed the speed of light. No one seemed to catch that either.
[ Post made via Mobile Device ] 
Re: A math problem
Posted: Mon, 03 Nov 2014 14:38:51
by dorminWS
MarcSpaz wrote:I was wondering who was going to catch that. No one did. You should have waiting longer.
Plus, its not possible to exceed the speed of light. No one seemed to catch that either.
[ Post made via Mobile Device ] 
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
What are you talking about that nobody "caught"?
I was just about to post this:
.....................................
To put it another way, once you have gone half the distance at half the rate of speed you wanted to achieve, you have used all the time you had available to cover the whole distance at twice the rate of speed you actually achieved. So no matter how fast you cover the last half of the distance, you will still be on the road for more time than you would have travelled to cover the whole distance at the target rate of speed. You don't NEED any numbers to get to the answer. Plug in any numbers you want that fit those parameters; it doesn’t matter. The answer will be the same whether you were going 10 miles or a 1,000.