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

 It is currently 01 Jul 2015, 17:21

### 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

# Q. In 1970 the production of mobile phones was .8 million.

Author Message
TAGS:
Intern
Joined: 28 Aug 2003
Posts: 1
Location: india
Followers: 0

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

Q. In 1970 the production of mobile phones was .8 million. [#permalink]  28 Aug 2003, 12:32
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Q. In 1970 the production of mobile phones was .8 million. During
the 11 year period from 1970 to 1980 the production of mobiles
increased at a constant rate every year. If a total of 14.3 million
mobiles were produced during the period then what is the range of
the mobile phone production for the 11 year period.

I apologise for no answer selections, but can anyone solve?

My method

1970=.8
1971=.8x
1972=.8x^2
1973=.8x^3
....
1980=.8x^10

The sum of these will equal 14.3, but what next?
What we need to find is (.8x^10) - .8

Many thanks
Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

bufff... lets see, heres my reasoning:
lets call S=0.8+0.8*x+...+0.8*x^10
then S=0.8*((x^11)-1)/(x-1)=14,8
rearrenging we have
S=0.8*[[x*((x^10)-1)/(x-1)]+1]=14,8
thus 0.8*((x^10)-1)=14*(x-1)/x=14*(1-1/x)

we know that x>1, so the range asked is 0-14
Manager
Joined: 28 Feb 2003
Posts: 147
Location: Kiev
Followers: 1

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

Could you explain pls what exactly we need to find?
What is "the range"?
_________________

Too much is not enough...

Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay
Followers: 1

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

The range is 1 (1.8 million - 0.8 million)

We have 14.3 in 11 years so 14.3 / 11 = 1.3 this is the middle number
1.3-0.8 = 0.5 so the increase per year is 0.1 million
The range will be the gratest minus the least so 1.8 - 0.8 = 1

hope is OK
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

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

I don't think that is right.

In your case, tha rate of increase from .8 to .9 is not the same (it is greater) as the increase from .9 to 1.0.

Therefore, the increase can't be a constant number, but rather a constant rate.

Not that I know how to solve this, but I don't think that is right...unless of course the problem is worded incorrectly.

Nimit, can you double check the wording, or perhaps provide answer choices?
_________________

Sept 3rd

Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay
Followers: 1

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

mciatto, you are right. I thought it was a constant increase (in quantity) not in rate.
Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

i think the only way to solve this problem in 2 min and without calculator is to treat it as an arithmetic progression, rather than as a geometric progression... the answer provided by MBA04 would then be ok... although in my opinion the question is misleading
Similar topics Replies Last post
Similar
Topics:
2 A certain phone manufacturer ships its products in crates. A crate con 7 11 Feb 2015, 05:45
2 The table above shows quantities of the various mobile phone models so 4 29 Dec 2014, 07:52
1 A mobile phone is available for $39000 cash or$17000 as dow 3 19 Aug 2014, 19:34
If x is the product of the positive integers from 1 to 8, 1 01 Jun 2008, 12:56
1 If n is the product of the integers from 1 to 8, inclusive, 4 22 Aug 2007, 10:22
Display posts from previous: Sort by