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x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 05:40
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x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\) A. 12 B. 6 C. 0 D. 3 E. 4
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Re: x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 05:53
trankimphuong wrote: x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\)
A. 12 B. 6 C. 0 D. 3 E. 4 \(z^2  y^2  x^2=(x+2)^2(x+1)^2x^2=x^2+4x+4x^22x1x^2=2xx^2+3=x(2x)+3\) let us see if the choices can fit in this equation.. A. 12.........\(x(2x)+3=12.....x(2x)=15 = (3*5).. x = 5\) YES B. 6.........\(x(2x)+3=6.....x(2x)=9\) x(2x)={x(x2)} so a multiple of two numbers with difference 2. can we write 9 as product of two such numbers..NO C. 0.........\(x(2x)+3=0.....x(2x)=3 = (3*1).. x = 3\) YES D. 3.........\(x(2x)+3=3.....x(2x)=0 = (2*0).. x = 2\) YES E. 4.........\(x(2x)+3=4.....x(2x)=1 = (1*(1)).. x = 1\) YES B
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 05:57
trankimphuong wrote: x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\)
A. 12 B. 6 C. 0 D. 3 E. 4 Another method to solve is going by process of elimination(substituting values for x,y,z s.t x<y<z) A. 12 is possible for consecutive integers 3,2, and 1 C. 0 is possible for consecutive integers 1,0, and 1 D. 3 is possible for consecutive integers 0,1, and 2 E. 4 is possible for consecutive integers 1,2, and 3 Therefore, Option B(6) cannot be the value of \(z^2  y^2  x^2\)
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x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 12:25
Let x be first integer y = x + 1 z = x + 2
So z^2  y^2  x^2 = x^2 +4x + 4 (x^2 +2x + 1)  x^2
This would be = 2x + 3  x^2 2x  x^2 X (2  x ) + 3
Now if we put in values for x
X = 0 then the equation = 3 possible X = 1 then the equation = 4 possible X = 2 then the equation = 3 possible X= 3 then the equation = 0 possible X = 4 then the equation =  5 possible X = 5 then the equation = 12 possible
Answer choice B
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Re: x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 12:51
chetan2u wrote: trankimphuong wrote: x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\)
A. 12 B. 6 C. 0 D. 3 E. 4 \(z^2  y^2  x^2=(x+2)^2(x+1)^2x^2=x^2+4x+4x^22x1x^2=2xx^2+3=x(2x)+3\) let us see if the choices can fit in this equation.. A. 12.........\(x(2x)+3=12.....x(2x)=15 = (3*5).. x = 5\) YES B. 6.........\(x(2x)+3=6.....x(2x)=9\) x(2x)={x(x2)} so a multiple of two numbers with difference 2. can we write 9 as product of two such numbers..NO C. 0.........\(x(2x)+3=0.....x(2x)=3 = (3*1).. x = 3\) YES D. 3.........\(x(2x)+3=3.....x(2x)=0 = (2*0).. x = 2\) YES E. 4.........\(x(2x)+3=4.....x(2x)=1 = (1*(1)).. x = 1\) YES B hello chetan2u i didnt get how did you get from this \(z^2  y^2  x^2\) you got this \((x+2)^2(x+1)^2x^2\) I didnt get why only x`s i see whereas there is also \(z\) and \(y\) ? and where did you get +2 and +1 ? ;? pls explain thanks!



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Re: x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 13:04
dave13 wrote: chetan2u wrote: trankimphuong wrote: x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\)
A. 12 B. 6 C. 0 D. 3 E. 4 \(z^2  y^2  x^2=(x+2)^2(x+1)^2x^2=x^2+4x+4x^22x1x^2=2xx^2+3=x(2x)+3\) let us see if the choices can fit in this equation.. A. 12.........\(x(2x)+3=12.....x(2x)=15 = (3*5).. x = 5\) YES B. 6.........\(x(2x)+3=6.....x(2x)=9\) x(2x)={x(x2)} so a multiple of two numbers with difference 2. can we write 9 as product of two such numbers..NO C. 0.........\(x(2x)+3=0.....x(2x)=3 = (3*1).. x = 3\) YES D. 3.........\(x(2x)+3=3.....x(2x)=0 = (2*0).. x = 2\) YES E. 4.........\(x(2x)+3=4.....x(2x)=1 = (1*(1)).. x = 1\) YES B hello chetan2u i didnt get how did you get from this \(z^2  y^2  x^2\) you got this \((x+2)^2(x+1)^2x^2\) I didnt get why only x`s i see whereas there is also \(z\) and \(y\) ? and where did you get +2 and +1 ? ;? pls explain thanks! Hi dave I’ll give my self the liberty to answer your question. It says x,y,z are consecutive integers and (x<y<z) If x is the least of these 3 consecutive integers Then y = x + 1 And z = x + 2 He just changed the variables in a way to make it simpler to deal with. I hope this clarifies your doubt. Posted from my mobile device



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x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 13:06
pushpitkc wrote: trankimphuong wrote: x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\)
A. 12 B. 6 C. 0 D. 3 E. 4 Another method to solve is going by process of elimination(substituting values for x,y,z s.t x<y<z) A. 12 is possible for consecutive integers 3,2, and 1 C. 0 is possible for consecutive integers 1,0, and 1 D. 3 is possible for consecutive integers 0,1, and 2 E. 4 is possible for consecutive integers 1,2, and 3 Therefore, Option B(6) cannot be the value of \(z^2  y^2  x^2\) hey pushpitkcWhy do you pick negative numbers ? when squared negative number turns positive. thanks may be you know Salsanousi



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Re: x,y,z are consecutive integers (x<y<z). Which of the following
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02 Oct 2018, 23:10
dave13 wrote: pushpitkc wrote: trankimphuong wrote: x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of \(z^2  y^2  x^2?\)
A. 12 B. 6 C. 0 D. 3 E. 4 Another method to solve is going by process of elimination(substituting values for x,y,z s.t x<y<z) A. 12 is possible for consecutive integers 3,2, and 1 C. 0 is possible for consecutive integers 1,0, and 1 D. 3 is possible for consecutive integers 0,1, and 2 E. 4 is possible for consecutive integers 1,2, and 3 Therefore, Option B(6) cannot be the value of \(z^2  y^2  x^2\) hey pushpitkcWhy do you pick negative numbers ? when squared negative number turns positive. thanks may be you know SalsanousiHey dave13The reason I used negative numbers is that I needed a negative answer. That was possible only if the magnitude of x was greater than y and y<z. Hope that clears your confusion
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Re: x,y,z are consecutive integers (x<y<z). Which of the following &nbs
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