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If x and y are positive numbers and z = xy^2, a 50 percent

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Manager
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Joined: 03 Sep 2006
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If x and y are positive numbers and z = xy^2, a 50 percent  [#permalink]

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New post 22 Mar 2007, 07:26
00:00
A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 8 sessions

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If x and y are positive numbers and z = xy^2, a 50 percent increase in x and 20 percent decrease in y would result in which of the following changes in z?
(A) A decrease of 4%
(B) A decrease of 14%
(C) An increase of 4%
(D) An increase of 20%
(E) An increase of 30%

P.S> I cannot get an answer from the answer list

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Manager
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Re: PS: z = xy^2 increase-decrease. Please, explain.  [#permalink]

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New post 22 Mar 2007, 07:47
Let z1 = xy^2

then z2= (3x/2)(4y/5)^2
change in z= (24xy^2)/25 - xy^2

= -xy^2/25
In terms of % = ((-xy^2/25)/xy^2) *100 ....4 % decrease
Hence A
Manager
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New post 22 Mar 2007, 08:08
1
Thanx a lot.

I think I've overstudied today (Singapore time now 23.08) =)

Of course:
z1 = xy^2
z2 = 1.5x * (0.8y)^2 = 1.5x * 0.64y^2 = 0.96xy^2

Hence, diff. is 4%
Senior Manager
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New post 22 Mar 2007, 22:21
My answer is A too :)

z = xy^2

50% increase in x => 150x/100
20% decrease in y=> 80y/100:

z = (150x/100) * (80y * 80y / 10000)

z = 96/100 xy^2 => 96% of xy^2 => 4% decrease :)
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New post 22 Mar 2007, 23:13
A.

z = 1.5x*(0.8y)^2
z = 1.5*0.64xy^2
z = 0.96xy^2

:)
Senior Manager
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New post 23 Mar 2007, 01:52
A answer also.

Can be solved by assuming number values.

x = 10
y = 10
z = 10 * (10)^2
= 1000


x = 15
y = 8
z = 15 * 64
= 960

Decrease = 40/1000 * 100
= 4%
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Re: If x and y are positive numbers and z = xy^2, a 50 percent  [#permalink]

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New post 06 Apr 2017, 15:50
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Re: If x and y are positive numbers and z = xy^2, a 50 percent &nbs [#permalink] 06 Apr 2017, 15:50
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