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

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Manager
Joined: 17 Dec 2005
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A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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Updated on: 15 Sep 2014, 20:40
5
18
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Difficulty:

55% (hard)

Question Stats:

67% (02:00) correct 33% (02:20) wrong based on 449 sessions

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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

Originally posted by mrmikec on 16 Jun 2006, 22:35.
Last edited by Bunuel on 15 Sep 2014, 20:40, 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]

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06 Nov 2011, 22:21
5
2
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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##### General Discussion
Director
Joined: 06 May 2006
Posts: 748
Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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Updated on: 17 Jun 2006, 11:45
1
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

Originally posted by paddyboy on 16 Jun 2006, 23:54.
Last edited by paddyboy on 17 Jun 2006, 11:45, edited 2 times in total.
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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17 Jun 2006, 11:41
2
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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Location: San Francisco
Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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17 Jul 2009, 16:01
2
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]

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14 Jul 2013, 03:05
2
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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A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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12 Sep 2013, 01:36
2
Director
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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15 Sep 2014, 07:03
1
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]

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15 Sep 2014, 20:42
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]

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07 Jun 2015, 04:42
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]

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08 Jun 2015, 23:17
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.
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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13 Jun 2016, 22:37
mrmikec wrote:
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

1993: 1000

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

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

value of xy alone would not serve our purpose.

B is sufficient.
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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13 Jun 2016, 22:48
1
SonofAnarchy wrote:
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.

For anyone having the same query:

You have xy = 20

Number in 1995 = 1000 * (1 + x/100 + y/100 + xy/10000)

You can put x = 20/y and get:

Number in 1995 = 1000 * ( 1 + 20/100y + y/100 + 20/10000)

What next? We still have an unknown variable y. This is not an equation that will help you solve for y. You need to find the value of the highlighted expression. Also, even if this were an equation, it is likely to give you multiple values of y.

Another thing you can note is that xy = 20 means there are multiple values of x and y possible. e.g. x = 2, y = 10 or x = 4, y = 5 etc.
The value of the expression will change depending on which pair you take since x + y is different in each case.
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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07 Jan 2017, 17:15
How can this be done in 2 minutes?.. The factoring is super difficult. Is this a 600 Q?..
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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09 Jan 2017, 08:37
iliavko wrote:
How can this be done in 2 minutes?.. The factoring is super difficult. Is this a 600 Q?..

You have to do no factorisation here. Just realise that it is a question involving successive percentage changes. You can calculate that using the formula
x + y + xy/100

Stmtn 2 gives you this value.

Stmnt 1 gives you xy = 20 only. In this case, x + y can take different values (1+20 or 2+10 or 4+5 etc). So the expression given above will have different values.

Check this post for successive percentage changes: https://www.veritasprep.com/blog/2011/0 ... e-changes/
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Re: A manufacturer produced x percent more video cameras in 1994 than in  [#permalink]

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30 Mar 2019, 03:19
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Re: A manufacturer produced x percent more video cameras in 1994 than in   [#permalink] 30 Mar 2019, 03:19
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