9. If X, Y and Z are positive integers, is X greater than Z – Y?
(1) X – Z – Y > 0.
2) z^2 = x^2+y^2
(1) X – Z – Y > 0.
2) z^2 = x^2+y^2
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
gud prob
Source: Beat The GMAT — Data Sufficiency |
x – y – z > 0namitrajiv wrote:hI stop@800,
Can you kindly prove, hows b insufficient ,
x > z + y
y is positive
so if x is greater than z+y, it will certainly be greater than z-y as we are subtracting something [2y] positive.
so
B is sufficient
Example:
100 > 50
100 will always be greater than 50-x
where x is positive integer
Hope this helps!!!
no actually I wanted to knw hw B is insufficient,
i.e z^2 = x^2+y^2 ,
it seems to me this is sufficient to prove that
x>z-y
e.g z= 13, y = 12, x =5
z= 13 , y = 12, x = 5
I cant find a case where the condition is insufficient
i.e z^2 = x^2+y^2 ,
it seems to me this is sufficient to prove that
x>z-y
e.g z= 13, y = 12, x =5
z= 13 , y = 12, x = 5
I cant find a case where the condition is insufficient
z^2 = x^2+y^2stop@800 wrote:36, 77, 85
85^2 = 36^2 + 77^2
36>85-77
77>85-36
x>z-y
So B suff with D as answer
I agree with A but your example made me unsure
GMAT/MBA Expert
-
Ian Stewart
- GMAT Instructor
- Posts: 2623
- Location: Montreal
- Thanked: 1090 times
- GMAT Score:780
stop@800 explained well above why the first statement is sufficient.
The second statement is also sufficient. If we know that x, y and z are positive, and that the Pythagorean relationship holds:
x^2 + y^2 = z^2
then we know that we can make a right angled triangle with sides x, y and z, where z is the hypotenuse. And in any triangle, the sum of two sides is always larger than the third side:
x + y > z
x > z - y
So the answer should be D.
The second statement is also sufficient. If we know that x, y and z are positive, and that the Pythagorean relationship holds:
x^2 + y^2 = z^2
then we know that we can make a right angled triangle with sides x, y and z, where z is the hypotenuse. And in any triangle, the sum of two sides is always larger than the third side:
x + y > z
x > z - y
So the answer should be D.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
-
cubicle_bound_misfit
- Master | Next Rank: 500 Posts
- Posts: 246
- Location: Texaco Gas Station
- Thanked: 7 times
Hi Ian,
in GMATLAND, ain't 0 a positive integer?
Also, the question stem does not say X Y Z are all different.
how can phythagoras be used then?
in GMATLAND, ain't 0 a positive integer?
Also, the question stem does not say X Y Z are all different.
how can phythagoras be used then?
Cubicle Bound Misfit
GMAT/MBA Expert
-
Ian Stewart
- GMAT Instructor
- Posts: 2623
- Location: Montreal
- Thanked: 1090 times
- GMAT Score:780
No, 0 is not a positive integer. It is an integer, but it isn't positive, and it isn't negative.cubicle_bound_misfit wrote:Hi Ian,
in GMATLAND, ain't 0 a positive integer?
Also, the question stem does not say X Y Z are all different.
how can phythagoras be used then?
If you have two lines of length x and y, and you connect them at right angles, the hypotenuse z will automatically satisfy
x^2 + y^2 = z^2
That's Pythagoras. It doesn't matter if x and y are different or equal. So you can make a triangle with sides x, y and z, and x+y must be greater than z.
(and while it's not especially important, in this question, x actually must be different from y if the second statement is true, because we know x, y and z are all integers -- if x were an integer and equal to y, then z would need to be equal to root(2)*x, which wouldn't be an integer).
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
