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A manufacturer produced x percent video cameras in 1994 than

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Manager
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A manufacturer produced x percent video cameras in 1994 than [#permalink]  16 Jun 2006, 21:35
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35% (medium)

Question Stats:

59% (02:20) correct 40% (01:36) wrong based on 55 sessions
A manufacturer produced x percent video cameras in 1994 than in 1993 and y percent more in 1995 than in 1994. If he produced 1000 video cameras in 1993, how many did he produce in 1995?

(1) xy = 20
(2) x + y + xy/100 = 9.2
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Sep 2013, 00:29, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
Director
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B.

Take,
m = 2003 production
n = 2004 production
p = 2005 production

n = (100 + x)*m/100 -- 1
p = (100 + y)*n/100 -- 2

Substituting 1 in 2,
=> p = (100 + x)(100 + y)*m/(100*100)
=> p = (1 + x/100 + y/100 + xy/10000)*m
=> p = (100 + x + y + xy/100)*m/100

#1. xy = 20

Insufficient; as we get
p = (100 + x + y + 20/100)*m/100 => still have 2 unknowns

#2. x + y + xy/100 = 9.2

Sufficient; as we get
p = (100 + 9.2)*m/100 => no more unknowns

Last edited by paddyboy on 17 Jun 2006, 10:45, edited 2 times in total.
Manager
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B.

Take,
m = 2003 production
n = 2004 production
p = 2005 production

n = (100 + x)*m/100
p = (100 + y)*n/100

=> p = (100 + x)(100 + y)*mn/(100*100)
=> p = (1 + x/100 + y/100 + xy/10000)*mn
=> p = (100 + x + y + xy/100)*mn/100

#1. Insufficient
#2. Sufficient

I'm sorry I don't understand the rationale the answer. where does the extra N come from?
Director
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Typos galore!

Edited my original post. Hope its better now
Manager
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1
KUDOS
B.

Take,
m = 2003 production
n = 2004 production
p = 2005 production

n = (100 + x)*m/100
p = (100 + y)*n/100

=> p = (100 + x)(100 + y)*mn/(100*100)
=> p = (1 + x/100 + y/100 + xy/10000)*mn
=> p = (100 + x + y + xy/100)*mn/100

#1. Insufficient
#2. Sufficient

I got the following:

1993 = m
1994 = n
1995 = p

1994 = ((100+x)/100)m
1995 = ((100+y)/100)n

sub N

1995 = ((100+x)/100)((100+y)/100)m

= ((10000+100x+100y+xy)/10000)m

= ((1+x/100+y/100+xy/10000))m

C1. xy = 20
INSUFFICENT

C2. x+y+xy/100= 9.2
rewritten as

x/100+y/100+xy/10000 = 9.2(1/100)=9.2/100

Thus
subing condition2 into the underlined statement gets me

( 1+ 0.092)m = ??

What happens to m? We just assume its 100.
Manager
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M = 1000 (2003 production)

DOH!!

Thanks, is my algebra correct?
VP
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Schools: Duke (Fuqua) - Class of 2012
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A manufacturer produced x percent more video cameras in 1994 [#permalink]  17 Jul 2009, 13:47
Expert's post
A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more video cameras 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
_________________

Last edited by asimov on 18 Jul 2009, 06:41, edited 1 time in total.
Current Student
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Re: video cameras [#permalink]  17 Jul 2009, 15:01
This is tricky but simple

Lets the number of increase in camera be m for 1993 to 1994

==> m = 1000 * x/100 = 10x

Lets the number of increase in camera be n for 1994 to 1995

==> n = (1000 + 10x) * y/100 = (100 + x) * y/10 = 10y + xy/10

Now the cameras in 1995 = 1000 + a + b = 1000 + 10x + 10y + xy/10 = 10(100+x+y+xy/100)

Now using stmt 2 we can solve the above equation hence Stmt 2 is sufficient and hence answer is B

Note from stmt 1 we wont know values of x and y
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A manufacturer produced x percent video cameras in 1994 than [#permalink]  06 Nov 2011, 05:22
A manufacturer produced x percent video cameras in 1994 than in 1993 and y percent more in 1995 than in 1994. If he produced 1000 video cameras in 1993, how many did he produce in 1995?

(1) xy = 20
(2) x + y + xy/100 = 9.2
Director
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Re: DS Question [#permalink]  06 Nov 2011, 21:21
The question has one critical word missing. Am assuming as follows:
A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more in 1995 than in 1994. If he produced 1000 video cameras in 1993, how many did he produce in 1995?

Video cameras produced in 1995 = 1000 * (1+x/100) * (1+y/100)
= 1000 * (1 + x/100 + y/100 + xy/10000) cameras
= 1000 + 1000/100 (x + y + xy/100) cameras
= 1000 + 10 (x + y + xy/100) cameras

Using statement (1), we know xy but not know x and y separately. Insufficient.
Using statement (2), we know x + y + xy/100 and so can calculate the expression above. Sufficient.

(B) it is.
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Re: A manufacturer produced x percent more video cameras in 1994 [#permalink]  14 Jul 2013, 02:05
1993
1000

1994
1000( 1+\frac{x}{100})

1995
1000(1+\frac{x}{100)}(1+\frac{y}{100})

solve for 1995 and take \frac{1}{100} common you will get statement 2 sufficient!
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Re: A manufacturer produced x percent more video cameras in 1994 [#permalink]  11 Sep 2013, 11:24
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Re: A manufacturer produced x percent video cameras in 1994 than [#permalink]  12 Sep 2013, 00:36
Expert's post
Re: A manufacturer produced x percent video cameras in 1994 than   [#permalink] 12 Sep 2013, 00:36
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