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Math Expert V
Joined: 02 Sep 2009
Posts: 58402
If a, x, y, and z are integers greater than zero, are both x and y  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 72% (01:36) correct 28% (02:09) wrong based on 83 sessions

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If a, x, y, and z are integers greater than zero, are both x and y less than z?

(1) ax + xz + ay + yz < z^2 + az
(2) x < z

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Manager  Joined: 18 Feb 2016
Posts: 97
Re: If a, x, y, and z are integers greater than zero, are both x and y  [#permalink]

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1
2
We ca formulate the questions as follows:
x, y, and z > 0 [integers] and the question is: "x+y<z?"

(1) ax + xz + ay + yz < z^2 + az
x(a+z) + y(a+z) < z(z+a) [factor out x, y and z on both sides]
(a+z) (x+y) < z(z+a) [factor out (a+z) and then divide both sides by (a+z)
x+y < z --------------> sufficient

(2) x < z
the final decision depends on the value of y. ----------> not sifficient
Intern  Joined: 25 May 2015
Posts: 1
Re: If a, x, y, and z are integers greater than zero, are both x and y  [#permalink]

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deabas wrote:
We ca formulate the questions as follows:
x, y, and z > 0 [integers] and the question is: "x+y<z?"

(1) ax + xz + ay + yz < z^2 + az
x(a+z) + y(a+z) < z(z+a) [factor out x, y and z on both sides]
(a+z) (x+y) < z(z+a) [factor out (a+z) and then divide both sides by (a+z)
x+y < z --------------> sufficient

(2) x < z
the final decision depends on the value of y. ----------> not sifficient

In option (1) if we simplify the above expression we get,
ax + xz + ay + yz < z^2 + az
x(a+z) + y(a+z) < z(z+a)
(a+z) (x+y) < z(z+a)
(a+z) (x+y) - z(z+a)<0
(a+z) [(x+y) - z]<0

Either
(a+z)<0 & ((x+y)-z)>0
or
(a+z)>0 & ((x+y)-z)<0

=> (x+y)>z or (x+y)<z
Hence (A) is not sufficient
Math Expert V
Joined: 02 Aug 2009
Posts: 7971
Re: If a, x, y, and z are integers greater than zero, are both x and y  [#permalink]

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goelsarthak93 wrote:
deabas wrote:
We ca formulate the questions as follows:
x, y, and z > 0 [integers] and the question is: "x+y<z?"

(1) ax + xz + ay + yz < z^2 + az
x(a+z) + y(a+z) < z(z+a) [factor out x, y and z on both sides]
(a+z) (x+y) < z(z+a) [factor out (a+z) and then divide both sides by (a+z)
x+y < z --------------> sufficient

(2) x < z
the final decision depends on the value of y. ----------> not sifficient

In option (1) if we simplify the above expression we get,
ax + xz + ay + yz < z^2 + az
x(a+z) + y(a+z) < z(z+a)
(a+z) (x+y) < z(z+a)
(a+z) (x+y) - z(z+a)<0
(a+z) [(x+y) - z]<0

Either
(a+z)<0 & ((x+y)-z)>0
or
(a+z)>0 & ((x+y)-z)<0

=> (x+y)>z or (x+y)<z
Hence (A) is not sufficient

Hi,
Actually you have correctly stated that eliminating a+z on the two sides in normal circumstances would not be correct..
but here it works and A turns out to be sufficient..

the two cases are as observed by you -
Either
(a+z)<0 & ((x+y)-z)>0
or
(a+z)>0 & ((x+y)-z)<0

BUT it is given in Q that If a, x, y, and z are integers greater than zero
so a and z are >0 and therefore a+z>0..
SO ONLY one case is feasible -
(a+z)>0 & ((x+y)-z)<0
Statement I is therefore suff
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Location: India
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Re: If a, x, y, and z are integers greater than zero, are both x and y  [#permalink]

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Bunuel wrote:
If a, x, y, and z are integers greater than zero, are both x and y less than z?

(1) ax + xz + ay + yz < z^2 + az
(2) x < z

#1

ax + xz + ay + yz < z^2 + az
x(a+z)+y(z+a)<z(z+a)
x+y<z
sufficeint
z >x,a

#2
x<z, no info about y
insufficeint
IMO A Re: If a, x, y, and z are integers greater than zero, are both x and y   [#permalink] 01 Mar 2019, 04:50
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