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

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

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28 Aug 2003, 13:32
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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
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28 Aug 2003, 14:57
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
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29 Aug 2003, 00:46
Could you explain pls what exactly we need to find?
What is "the range"?
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Too much is not enough...

Manager
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29 Aug 2003, 12:38
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
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29 Aug 2003, 12:45
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?
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Sept 3rd

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29 Aug 2003, 13:03
mciatto, you are right. I thought it was a constant increase (in quantity) not in rate.
Manager
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29 Aug 2003, 14:21
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
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