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If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)

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If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)  [#permalink]

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New post 12 Jan 2015, 09:06
1
7
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

71% (03:05) correct 29% (02:57) wrong based on 129 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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Re: If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)  [#permalink]

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New post 12 Jan 2015, 11:00
3
xy+xz = 152 --------------------------------------1)
yz+yx = 162---------------------------------------2)
xz+zy = 170---------------------------------------3)

subtracting 1) from 2) we have, yz-xz = 10--------------4)

now by adding 4) to 3) we have

2yz =180
or yz=90.

since yz =90, therefore xyz must be a multiply of 90. now, out of given options only option D is a multiple of 90, hence answer must be D.
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Re: If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)  [#permalink]

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New post 13 Jan 2015, 01:23
2
xy+xz = 152 --------------------------------------1)
yz+yx = 162---------------------------------------2)
xz+zy = 170---------------------------------------3)

Re-writing equation 3 as follows:

xz+zy = 162 + 8

xz+zy = yz+yx + 8

xz = yx + 8 ............... (4)

Adding (1) & (4)

2xz = 160

xz = 80

xyz has to be multiple of 80, only 720 fits in

Answer = D
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Re: If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)  [#permalink]

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New post 22 Jul 2015, 21:29
what about term "real number" meaning that xyz can be any from answer choices?
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If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)  [#permalink]

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New post 02 Nov 2015, 08:20
1
damn, I solved this in a more difficult and time-consuming way (3:54)!!!
maybe because I did not see how else to approach this problem:
xy+xz=152
xy+yz=162
xz+yz=170

if we add each 2 we get:
2xy+xz+yz=152+162
2xy+z(x+y)=314 -> we know z(x+y)=170, which means that 2xy=144 or xy = 72

xy+xz=152
xz+yz=170
2xz + y(x+z)=322 -> we know y(x+z) = 162, which means that 2xz = 160 or xz=80

xy+yz=162
xz+yz=170
2yz+x(y+z)=332 -> we know x(y+z) = 152, which means that 2yz = 180 or yz = 90

now we have xy*xz*yz = x^2*y^2*z^2 - ok, if we know the value of this, we can find xyz:
which means that xyz = sqrt(72*80*90)
prime factorization of 72 - 2x36=2*3*12=2*3*2*6=2*3*2*2*3
prime factorization of 80 = 10*8 = 2*2*2*2*5
prime factorization of 90 = 9*10 = 3*3*2*5

sqrt (2*3*2*2*3*2*2*2*2*5*3*3*2*5)

well, we have 8 of 2's, 4 of 3's, and 2 of 5's. thus, xyz=2^4*3^2*5 = 16*5*9 = 80*9 = 720.


I did not make assumptions that the number is a multiple of other number, since the question mentions real numbers, not integers!!!
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Re: If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)  [#permalink]

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Re: If x,y ,and z are positive real numbers such that x(y+z) = 152, y(z+x)   [#permalink] 12 Nov 2018, 19:41
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