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A manufacturer produced x percent more video cameras in 1994 than in

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A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 16 Jun 2006, 21:35
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A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more in 1995 than in 1994. If the manufacturer produced 1,000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995?

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

Last edited by Bunuel on 15 Sep 2014, 19:40, edited 3 times in total.
Edited the question.
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 16 Jun 2006, 22:54
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.
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 17 Jun 2006, 10:41
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paddyboy wrote:
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.
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 17 Jul 2009, 15:01
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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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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 06 Nov 2011, 21:21
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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 than in [#permalink] New post 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 than in [#permalink] New post 11 Sep 2013, 11:24
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 12 Sep 2013, 00:36
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 15 Sep 2014, 06:03
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I got E because of typo made in the question.

Original question :: A manufacturer produced x percent more video cameras in 1994 than in 1993
Typo :: A manufacturer produced x percent ____ video cameras in 1994 than in 1993

Bunuel please edit the question stem, it must be "x percent more"
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 15 Sep 2014, 19:42
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PiyushK wrote:
I got E because of typo made in the question.

Original question :: A manufacturer produced x percent more video cameras in 1994 than in 1993
Typo :: A manufacturer produced x percent ____ video cameras in 1994 than in 1993

Bunuel please edit the question stem, it must be "x percent more"


Thank you! Edited the typo.
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 07 Jun 2015, 03:42
93 –10^3
94 – 10^3(1+x/100)
95- 10^3(1+y/100)(1+x/100)

To calculate: 10^3 ( 1 + 1/10^2(x + y +xy/10^2))

Clearly B is sufficient
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Re: A manufacturer produced x percent more video cameras in 1994 than in [#permalink] New post 08 Jun 2015, 22:17
superman wrote:
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



hey,
i have a query.
we substituted the value of "xy" in the equation and are left with two unknown variable x and y.
Cant we substitute the value x=20/y also in the equation and make it a single variable equation?
From that we can find the value of x and y separately.
Re: A manufacturer produced x percent more video cameras in 1994 than in   [#permalink] 08 Jun 2015, 22:17
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