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 Q45  V27 GMAT 2: 640  Q47  V31
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Re: Confused Questions [#permalink]
Your answer to question 2) should be right. Plug in to check:
\((\frac{1 truck}{5 hours})(\frac{15}{16}hours)+(\frac{1 truck}{3 hours})(\frac{15}{16}hours) = \frac{8}{16}truck\)
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Re: Confused Questions [#permalink]
mrblack wrote:
For question 1), is the answer 3 mins?

Let X represent the average time to travel for each day (excluding the 30 mins extra on the first day). Thus we have the equation:

Average time = \(\frac{(X+30)+9X}{10}=30\)

X=27

Therefore, the holdup increased the average time by 3 mins.


What's the meaning of "first outward journey"? In my opinion, this is a go-back journey. so it needs totally 20 days.
my equation is:Delay 30/(10+10)=1.5 m
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Re: Confused Questions [#permalink]
mrblack wrote:
Your answer to question 2) should be right. Plug in to check:
\((\frac{1 truck}{5 hours})(\frac{15}{16}hours)+(\frac{1 truck}{3 hours})(\frac{15}{16}hours) = \frac{8}{16}truck\)


I really agree with you!
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Re: Confused Questions [#permalink]
mrblack is on the money...!

clarity lies in the assumption that 30 mins is average time; like saying your car does 30mpg on an average...BUT... so going back on the question again... for the first phase 30mins of hold is added... and that is rest 9 days average holds steady... but for the first day it adds 30 mins of additional time...



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