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christoph
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Marc traveled from city A to city B covering as much 02 May 2005, 12:46
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Marc traveled from city A to city B covering as much distance in the second part as he did in the first part of this journey. His speed during the second part was twice as that of the speed during the first part of the journey. What is his average speed of journey during the entire travel?



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Marc traveled from city A to city B covering as much 02 May 2005, 13:50
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total distance= 2d
time A= d/speedA
time B= d/(2*speedA)

timeA+timeB= d/A+d/2A= d*(1/A+1/2A)=d*1/A*3/2=3/2*d/A
average= 2d/(3/2*d/A)

4/3 A :roll:
don't know if choices are actual numbers or functions...
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Marc traveled from city A to city B covering as much 03 May 2005, 05:33
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I agree . My method:
( d1 x V1 + d2 x ( 2 x V1 ) ) / 2 for d = 1
( V1 +2 V1 ) = 3 V1 / 2 => 1.333 V1
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Marc traveled from city A to city B covering as much 03 May 2005, 06:41
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OA:

The first part is 1/3rd of the total distance and the second part is 2/3rd of the total distance. He travels at s km/hr speed during the first half and 2s km/hr speed during the second half.

If 3 km is the total distance, then 1 km was traveled at s km/hr and 2 kms was traveled at 2s km/hr speed.

Hence average speed = total distance/total time=3/(1/s)+(2/2s)=3/(4/2s)=3s/2. This, however, = s+2s/2=3s/2 which is the arithmetic mean of the speeds of the two parts.
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Marc traveled from city A to city B covering as much 05 May 2005, 11:31
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do you agree with OA? why is it 1/3 and 2/3? it says "as much distance in the second part as he did in the first part of this journey". should not d1=d2?

thanks, guys
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Marc traveled from city A to city B covering as much 05 May 2005, 12:01
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july05
do you agree with OA? why is it 1/3 and 2/3? it says "as much distance in the second part as he did in the first part of this journey". should not d1=d2?

thanks, guys
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i dont agree with the OA for the same reason as you have mentioned. thats why i posted this question.
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Marc traveled from city A to city B covering as much 15 May 2005, 05:30
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Let the distance of the 2 parts be 'd' each.
Therefore total distance = 2d.

Now since the speed for the first part is half that of the 2nd part the time for the first part should be double that of the 2nd part.Let the time taken therfore be 2t and t.

Avg speed = Total dist/Total time=2d/3t.



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