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# Confused Questions

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Intern
Joined: 06 Jan 2010
Posts: 20

Kudos [?]: 2 [0], given: 0

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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?

Kudos [?]: 2 [0], given: 0

Manager
Joined: 26 Nov 2009
Posts: 162

Kudos [?]: 62 [0], given: 5

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23 Jan 2010, 10:47

Kudos [?]: 62 [0], given: 5

Manager
Joined: 27 Apr 2008
Posts: 191

Kudos [?]: 94 [0], given: 1

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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.

Kudos [?]: 94 [0], given: 1

Manager
Joined: 27 Apr 2008
Posts: 191

Kudos [?]: 94 [0], given: 1

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24 Jan 2010, 18:55
Your answer to question 2) should be right. Plug in to check:
$$(\frac{1 truck}{5 hours})(\frac{15}{16}hours)+(\frac{1 truck}{3 hours})(\frac{15}{16}hours) = \frac{8}{16}truck$$

Kudos [?]: 94 [0], given: 1

Intern
Joined: 06 Jan 2010
Posts: 20

Kudos [?]: 2 [0], given: 0

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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

Kudos [?]: 2 [0], given: 0

Intern
Joined: 06 Jan 2010
Posts: 20

Kudos [?]: 2 [0], given: 0

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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!

Kudos [?]: 2 [0], given: 0

Manager
Joined: 02 Oct 2009
Posts: 193

Kudos [?]: 22 [0], given: 4

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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...

Kudos [?]: 22 [0], given: 4

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