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Math Expert V
Joined: 02 Sep 2009
Posts: 61274
If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 78% (03:18) correct 22% (02:01) wrong based on 18 sessions

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If x, y, and z are positive real numbers such that $$x(y + z) = 152$$, $$y(z + x) = 162$$, and $$z(x + y) = 170$$, then $$xyz$$ is

A. 672
B. 688
C. 704
D. 720
E. 750

Are You Up For the Challenge: 700 Level Questions

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Senior Manager  G
Joined: 16 Feb 2015
Posts: 355
Location: United States
Re: If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

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1
1
Bunuel wrote:
If x, y, and z are positive real numbers such that $$x(y + z) = 152$$, $$y(z + x) = 162$$, and $$z(x + y) = 170$$, then $$xyz$$ is

A. 672
B. 688
C. 704
D. 720
E. 750

Are You Up For the Challenge: 700 Level Questions

Explanation:

opening all the brackets & then Add.
2(xy+yz+zx)=484
xy+yz+zx=242
y(x+z) + zx = 242
zx=242-162 = 80

Option D is only multiple of 80.
x,y,z are 8,9,10

IMO-D Please give kudos, if you find my explanation good enough Manager  S
Joined: 03 Nov 2019
Posts: 54
Re: If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

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x(y+z)=152
xy+xz=152 ...........(A)

y(z+x)=162
yz+yx=162 ............(B)

z(x+y)=170
zx+zy=170 .............(C)

2(xy+yz+zx)=484
xy+yz+zx=242 ........(D)

Substituting eq A in D, B in D and C in D respectively we get
yz=90
zx=80
xy=72

Multiplying all these
(xyz)^2=90*80*72 =9*10*8*10*8*9= (8*9*10)^2
xyz=720

VP  P
Joined: 24 Nov 2016
Posts: 1204
Location: United States
Re: If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

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Bunuel wrote:
If x, y, and z are positive real numbers such that $$x(y + z) = 152$$, $$y(z + x) = 162$$, and $$z(x + y) = 170$$, then $$xyz$$ is

A. 672
B. 688
C. 704
D. 720
E. 750

Are You Up For the Challenge: 700 Level Questions

$$x(y + z) = 152…xy+xz=152$$
$$y(z + x) = 162…yz+yx=162$$
$$z(x + y) = 170…zx+zy=170$$
$$zx+zy=170-(xy+xz=152)…zy-xy=18$$
$$zy-xy=18+(yz+yx=162)…2zy=180…zy=90$$
$$zy=90…yz+yx=162…yx=162-90=72$$
$$yx=72…xy+xz=152…xz=152-72=80$$
$$zy*yx*xz=90*72*80…x^2y^2z^2=72*72*100…xyz=720$$

Ans (D) Re: If x, y, and z are positive real numbers such that x(y + z) = 152, y(z   [#permalink] 05 Dec 2019, 04:50
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# If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  