GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 20 Jan 2020, 14:44

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60515
If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

### Show Tags

05 Dec 2019, 01:30
00:00

Difficulty:

35% (medium)

Question Stats:

75% (03:19) correct 25% (02:01) wrong based on 16 sessions

### HideShow timer Statistics

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

_________________
Senior Manager
Joined: 16 Feb 2015
Posts: 252
Location: United States
Concentration: Finance, Operations
Re: If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

### Show Tags

05 Dec 2019, 02:34
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
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]

### Show Tags

05 Dec 2019, 03:56
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
Joined: 24 Nov 2016
Posts: 1076
Location: United States
Re: If x, y, and z are positive real numbers such that x(y + z) = 152, y(z  [#permalink]

### Show Tags

05 Dec 2019, 05:50
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, 05:50
Display posts from previous: Sort by