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655-705 (Hard)|   Algebra|      
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Bunuel
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If x, y, and z are positive real numbers such that x(y + z) = 152, y(z 05 Dec 2019, 01:30
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Difficulty:

55% (hard)

Question Stats:

71% (03:13) correct 29% (03:14) wrong based on 231 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

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D
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rajatchopra1994
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If x, y, and z are positive real numbers such that x(y + z) = 152, y(z 05 Dec 2019, 02:34
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Bunuel
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

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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 Please give kudos, if you find my explanation good enough :please
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Ansh777
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If x, y, and z are positive real numbers such that x(y + z) = 152, y(z 05 Dec 2019, 03:56
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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)

Adding A+B+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

Answer= D
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If x, y, and z are positive real numbers such that x(y + z) = 152, y(z 05 Dec 2019, 05:50
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Bunuel
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

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

Ans (D)
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If x, y, and z are positive real numbers such that x(y + z) = 152, y(z 29 May 2024, 14:33
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Maybe not the best way but applied some intuition

152= 19*8
162 = 18*9
170 = 17*10

Clearly we can identify 8;9;10 are the numbers can be easily verified too

And we can directly multiply

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If x, y, and z are positive real numbers such that x(y + z) = 152, y(z 16 Aug 2025, 09:00
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