Magnus drives from City A to City B at 100 mph. Reaching

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Magnus drives from City A to City B at 100 mph. Reaching [#permalink]

15 Jul 2003, 09:28

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Magnus drives from City A to City B at 100 mph. Reaching City B he recalls that he has forgot to take a vallet and start driving back at 120 mph. Reaching City A he takes his vallet in no time (his wife throws the vallet into his car), and drives to City B again at 140 mph. What is his average speed for the entire, three-leg trip?

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15 Jul 2003, 09:58

I have erased your right solution leaving the answer only. You know the trick for an n-leg trip. Let us give other people to think this question over.

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15 Jul 2003, 10:05

rightoo sir :done

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15 Jul 2003, 10:39

Could you give the solution the N-leg Trip sums.
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15 Jul 2003, 10:40

Rather could you give the tip/ Reveal the trick-Master
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15 Jul 2003, 12:06

Power on your mind—if I simply post the formula, you will not understand the problem again. The question is not so difficult to ask for help immediately. So, give it a try.

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15 Jul 2003, 12:57

118 mph?
I solved the sum by assuming the distance =100.
300/(107/42)
=300*42/107

Is there a faster trick? I am sure there is.
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15 Jul 2003, 13:02

I also tried solving the sum by letting distance = x

I get a different answer? Could some tell me!!

s=x+x+x/(x/100)+(x/120)+(x/140)

s=3x*420/(107x)

Where am I going wrong?
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15 Jul 2003, 13:41

Who said you are going wrong!!!
Have confidence in what you do.

Your approach is perfectly fine.. You can generalize it for an N-round trip also.

Did you get that.. !!!

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15 Jul 2003, 13:56

With the X approach my answer doesnt come up to 118 mph. How come.

I am missing a Zero somewhere..cant figure out where..
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15 Jul 2003, 14:22

For a 2-legged race , the average speed is

2 x s1 x s2
---------------------
s1 x s2 + s2 x s1

For a 3-legged race
3 x s1 x s2 x s3
---------------------
s1 x s2 + s2 x s3 + s3 x s1

So now it's easy to find out for n-leeged race , right ?

In our example we have
s1 = 100, s2 = 120 a,d s3 = 140

Average speed =
3 x 100 x 120 x 140
-----------------------------------
100 x 120 + 120 x 140 + 140 x 100
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16 Jul 2003, 09:28

I have 118 mph as a solution, same formula as SATGATES

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16 Jul 2003, 10:37

The right formula is a so-called harmonical average: the number of members divided by the sum of their reciprocals. Brainless got the right approach.

S=3/(1/140+1/120+1/100)

As you can see, the lengh of the trip leg is not important here.

[#permalink] 16 Jul 2003, 10:37

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Magnus drives from City A to City B at 100 mph. Reaching

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Magnus drives from City A to City B at 100 mph. Reaching

Magnus drives from City A to City B at 100 mph. Reaching

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