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9 at constant z

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9 at constant z

by sanju09 » Sat Mar 13, 2010 6:21 am
Three quantities x, y, and z are such that x y = k z, where k is a constant. When x is kept constant, y varies directly as z; when y is kept constant, x varies directly as z; and when z is kept constant, x varies inversely as y.

Initially, x was at 5 and x: y: z was 1: 3: 5. What is the value of x when y is 9 at constant z?
(A) 25/9
(B) 8
(C) 25/3
(D) 9
(E) 19/2
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by kstv » Sat Mar 13, 2010 7:40 am
x : y : z = 1 : 3 : 5
x = 5 y = 15 z = 25
xy = k z
k = 15*5 / 25 = 3

x = k z /y = 3 * 25 / 3 = 25/3 when z is kept constant, x varies inversely as y.
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IMO C[/spoiler]
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by gmatmachoman » Sat Mar 13, 2010 10:36 am
kstv wrote:x : y : z = 1 : 3 : 5
x = 5 y = 15 z = 25
xy = k z
k = 15*5 / 25 = 3

x = k z /y = 3 * 25 / 3 = 25/3 when z is kept constant, x varies inversely as y.
[spoiler]
IMO C[/spoiler]
@kstv Why did u take y = 3?? Is nt should y = 9??

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by Stuart@KaplanGMAT » Sat Mar 13, 2010 12:44 pm
sanju09 wrote:Three quantities x, y, and z are such that x y = k z, where k is a constant. When x is kept constant, y varies directly as z; when y is kept constant, x varies directly as z; and when z is kept constant, x varies inversely as y.

Initially, x was at 5 and x: y: z was 1: 3: 5. What is the value of x when y is 9 at constant z?
(A) 25/9
(B) 8
(C) 25/3
(D) 9
(E) 19/2
Good news - you'll never see this kind of wording on the actual GMAT - everything after the first sentence is redundant to the equation itself and is really just a distraction.

Let's try rephrasing this in a more GMATesque manner:
xy = kz, in which k is a constant. If the original value of x is 5, and if the original ratio of x:y:z is 1:3:5, then what's the value of x if y=9 and z remains constant?
Analyzing the problem, we think to ourselves: if the right side (i.e. kz) remains constant, then the left side must also remain constant. Therefore, x changes inversely to y. So, let's figure out the original value of y:

x:y = 1:3
x=5; therefore, y=15

Next, let's look at what's happened to y:

y started at 15 and has changed to 9. 9/15 = 3/5

Since y has been multiplied by 3/5, and since x changes inversely to y, we must multiply x by 5/3 to maintain equilibrium:

5 * 5/3 = 25/3... choose (C).
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by sanju09 » Mon Mar 15, 2010 2:04 am
Stuart Kovinsky wrote:
sanju09 wrote:Three quantities x, y, and z are such that x y = k z, where k is a constant. When x is kept constant, y varies directly as z; when y is kept constant, x varies directly as z; and when z is kept constant, x varies inversely as y.

Initially, x was at 5 and x: y: z was 1: 3: 5. What is the value of x when y is 9 at constant z?
(A) 25/9
(B) 8
(C) 25/3
(D) 9
(E) 19/2
Good news - you'll never see this kind of wording on the actual GMAT - everything after the first sentence is redundant to the equation itself and is really just a distraction.

Let's try rephrasing this in a more GMATesque manner:
xy = kz, in which k is a constant. If the original value of x is 5, and if the original ratio of x:y:z is 1:3:5, then what's the value of x if y=9 and z remains constant?
Analyzing the problem, we think to ourselves: if the right side (i.e. kz) remains constant, then the left side must also remain constant. Therefore, x changes inversely to y. So, let's figure out the original value of y:

x:y = 1:3
x=5; therefore, y=15

Next, let's look at what's happened to y:

y started at 15 and has changed to 9. 9/15 = 3/5

Since y has been multiplied by 3/5, and since x changes inversely to y, we must multiply x by 5/3 to maintain equilibrium:

5 * 5/3 = 25/3... choose (C).
Thank you Stuart for helping me write the wording the way GMAT do. Nice explanation as well.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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