Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Aug 2016, 01:25

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Confused Questions

Author Message
TAGS:

### Hide Tags

Intern
Joined: 06 Jan 2010
Posts: 20
Followers: 0

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

### Show Tags

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?
Manager
Joined: 26 Nov 2009
Posts: 175
Followers: 3

Kudos [?]: 56 [0], given: 5

### Show Tags

23 Jan 2010, 10:47
Manager
Joined: 27 Apr 2008
Posts: 190
Followers: 1

Kudos [?]: 65 [0], given: 1

### Show Tags

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.
Manager
Joined: 27 Apr 2008
Posts: 190
Followers: 1

Kudos [?]: 65 [0], given: 1

### Show Tags

24 Jan 2010, 18:55
Your answer to question 2) should be right. Plug in to check:
$$(\frac{1 truck}{5 hours})(\frac{15}{16}hours)+(\frac{1 truck}{3 hours})(\frac{15}{16}hours) = \frac{8}{16}truck$$
Intern
Joined: 06 Jan 2010
Posts: 20
Followers: 0

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

### Show Tags

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
Intern
Joined: 06 Jan 2010
Posts: 20
Followers: 0

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

### Show Tags

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!
Manager
Joined: 02 Oct 2009
Posts: 197
Followers: 1

Kudos [?]: 18 [0], given: 4

### Show Tags

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...
Re: Confused Questions   [#permalink] 27 Jan 2010, 10:50
Similar topics Replies Last post
Similar
Topics:
confusion with |x| 2 31 Oct 2013, 07:34
Confusing Percentages 0 21 Jul 2013, 21:50
2 Absolute Value Confusion 7 23 May 2010, 20:35
Very confused PS 6 09 Feb 2009, 03:20
1 SET theory ... BIG CONFUSION 7 17 Jul 2008, 18:43
Display posts from previous: Sort by