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Confused Questions

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Joined: 06 Jan 2010
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Confused Questions [#permalink]

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New post 23 Jan 2010, 06:12
Hi All, I get confused on the following 2 questions.

1. If the average return journey time over 10 days is 30 minutes, but on the first outward journey a hold up added 30 minutes to that journey, by how much did the delay increase the daily average?
( my answer is 1.5 minutes)

2. If it takes one person 5 hours to load a truck while another person can complete the task in 3 hours, how ling it takes them to half fill the truck if they work together at the same rate?
(my answer is x/5+x/3=1/2, x=15/16 hour)

Unfortunately, none of my answers is correct! what's the problem?

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Re: Confused Questions [#permalink]

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New post 23 Jan 2010, 10:47
please post the answer choices

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Re: Confused Questions [#permalink]

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New post 24 Jan 2010, 18:47
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.

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Re: Confused Questions [#permalink]

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New post 24 Jan 2010, 18:55
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]

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New post 27 Jan 2010, 07:08
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]

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New post 27 Jan 2010, 07:12
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]

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New post 27 Jan 2010, 10:50
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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Re: Confused Questions   [#permalink] 27 Jan 2010, 10:50
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