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A anufacturer produced x percent more video cameras in 1994

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A anufacturer produced x percent more video cameras in 1994 [#permalink] New post 24 Jan 2008, 15:03
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A anufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more videocameras in 1995 than in 1994. if the manufacturer produced 1000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995?

1) xy=20
2) x+y+(xy/100)=9.2
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Re: DS: video cameras [#permalink] New post 24 Jan 2008, 15:11
netcaesar wrote:
A anufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more videocameras in 1995 than in 1994. if the manufacturer produced 1000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995?

1) xy=20
2) x+y+(xy/100)=9.2


1) insufficient -- see below
2) sufficient 1000*(1+x/100)(1+y/100) = 1000*(1+(x+y)/100 + xy/(100*100)) = 1000*(1+[x+y+(xy/100)]/100)=1000*(1+9.2/100)=1,092. //obviously xy is not sufficient ... -- try 4% and 5% and then 1% and 10%.
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Re: DS: video cameras [#permalink] New post 24 Jan 2008, 15:19
Should be C.

Coz we need to use both 1 and 2 to determine that X and Y are 4 and 5.

Otherwise, as per 1, X and Y can be 1 & 20 or 2 &10 or 4 &5.

Combining 1 and 2, X & Y are 4 & 5.

Whats OA ?
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Last edited by suntaurian on 26 Jan 2008, 14:44, edited 1 time in total.
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Re: DS: video cameras [#permalink] New post 24 Jan 2008, 23:49
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B

N=1000*\frac{(100+x)}{100}*\frac{(100+y)}{100}=10*(100+x+y+\frac{xy}{100})
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Re: DS: video cameras [#permalink] New post 25 Jan 2008, 07:11
Why do you not need xy = 20 to solve if we don't know whether the solution is (5,4) or (10,1)?
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Re: DS: video cameras [#permalink] New post 25 Jan 2008, 07:43
sonibubu wrote:
Why do you not need xy = 20 to solve if we don't know whether the solution is (5,4) or (10,1)?


because condition 2 is sufficient, it gives all the info you need to perform calculations.
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Re: DS: video cameras [#permalink] New post 25 Jan 2008, 08:33
OA is B.


But, is there any faster way to solve it?
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Re: DS: video cameras [#permalink] New post 07 Jul 2008, 11:41
walker wrote:
x+y+(xy/100)=9.2


I like walker's solution. It makes sense.

We start out with 1000.

x = the % increase from 1993 to 1994, and y = the % increase from 1994 to 1995.

if you add x% + 7% that would be the increases of the base (on 1000) not counting the extra you get from y% time the increased number for the prior year.

Like if you had 10% increase from '93 to '94, that'd be 1100 total. Then the next year if you had a 20% increase, you'd have 220. But you could view this as

x% = 10% and y=20%, so x + y = 30%. That means you have 300 increase from '93 to '95, but you have to account for the y% of the increase to '94 numbers. This is (10 * 20) /100 * 1000, or 20 / 100 = .2 * 1000 = 20. So you have 320. This is the same as

10% increase '93 to '94 = 100 increase. '94 to '95 is 20% of the '94 number (1100) so an increase of 220. Total is 1320 produced in '95.

Very nice walker to see how that works. You're always on top of things like this. Kudos!
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Re: DS: video cameras [#permalink] New post 07 Jul 2008, 13:44
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netcaesar wrote:
A anufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more videocameras in 1995 than in 1994. if the manufacturer produced 1000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995?

1) xy=20
2) x+y+(xy/100)=9.2


B.

It gives you the variables if you calculate this out.
Re: DS: video cameras   [#permalink] 07 Jul 2008, 13:44
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