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Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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Updated on: 29 Aug 2014, 14:42
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Suppose x, y, and z are positive integers such that xy + yz = 29 and xz + yz = 81. Which of the following variables has exactly one unique solution? (i) x (ii) y (iii) z A. none B. ii only C. iii only D. i and ii only E. ii and iii only
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Originally posted by plaverbach on 29 Aug 2014, 13:19.
Last edited by Bunuel on 29 Aug 2014, 14:42, edited 1 time in total.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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29 Aug 2014, 13:20
How can one solve in under 3 min?
Official ansower above: Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.
But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).
Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.
That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B.




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Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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30 Aug 2014, 08:06
plaverbach wrote: How can one solve in under 3 min?
Official ansower above: Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.
But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).
Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.
That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B. The way you have described should get you an answer in less than 2 mins. To make the calculation a bit faster, recognise that in the second equation \(z * (x+1) = 81\), 1 is already being added to x. So, it might be faster to substitute the value of x in terms of z. Doing that, the equation becomes: \(z * (29  z + 1) = 81\) \(30z  z^2 = 81\) \(z^2  30z + 81 = 0\) \((z  3) (z  27) = 0\) So, \(z = 3\) or \(z = 27\) Now, since z can take two values, without any further calculations, we know that x can take two values as well. Hence, B is correct. Hope that makes it clear.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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30 Aug 2014, 12:41
Here's another approach: Once we figure out that y=1, we use that to simplify the second equation to z(x+1) = 81. So what pairs of numbers make 81? 1*81, 3*27, or 9*9. Looking back at our first equation with y=1, we get x+z=29. So what do we know about x and z? They add up to 29, and when we add 1 to x, they multiply to yield 81. The only pair of factors that works is 3 and 27. So we could have x=2 and y=27, or x=26 and y=3.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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03 Sep 2014, 22:22
xy + yz = 29
y(x+z) = 1 * 29 (29 is prime number; cannot be factorized in any other way)
y = 1 .............. (1)
x+z = 29
x = 29z ........... (2)
xz + yz = 81
z(x+y) = 81
Placing values from (1) & (2)
z(29z+1) = 81
\(30z  z^2  81 = 0\)
This would give 2 values for z (27, 3) which in turn would also yield 2 values for x; only y has a distinct value
Answer = B



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Suppose x,y, and z are positive integers such that
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31 Jul 2015, 10:41
Suppose x,y, and z are positive integers such that xy+yz=29 and xz+yz=81 . Which of the following variables has exactly one unique solution? (i) x (ii) y (iii) z a)none b)ii only c)iii only d)i and ii only e)ii and iii only Please give a kudos and help me unlock the awesome Gmatclub test



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Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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31 Jul 2015, 10:47
Gmat1008 wrote: Suppose x,y, and z are positive integers such that xy+yz=29 and xz+yz=81 . Which of the following variables has exactly one unique solution? (i) x (ii) y (iii) z a)none b)ii only c)iii only d)i and ii only e)ii and iii only Please give a kudos and help me unlock the awesome Gmatclub test Please search for the question before you post. Topics merged.



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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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01 Jul 2016, 00:20
plaverbach wrote: Suppose x, y, and z are positive integers such that xy + yz = 29 and xz + yz = 81. Which of the following variables has exactly one unique solution?
(i) x (ii) y (iii) z
A. none B. ii only C. iii only D. i and ii only E. ii and iii only Responding to a pm: The question relies on your knowledge of factors. Work step by step through the given information. xy + yz = 29 (x + z)y = 29 (you need to immediately notice that 29 is prime) So the two factors are 1 and 29 only. (x+z) cannot be 1 so y MUST be 1 and (x + z) = 29 xz + yz = 81 (x + 1)z = 81 Factors of 81 > 1, 3, 9, 27, 81. The two factors cannot be 1 and 81 because (x+1) can be neither 1 nor 81 (since x + z is 29). The two factors cannot be 9 and 9 because if z is 9(then x would be 20), x cannot be 8. Then 3 and 27 must be the two factors. Now all we need to see is whether z can take both values 3 and 27. If z is 3, x is 26 and total they add up to 29  Works If z is 27, x is 2 and total again adds up to 29  Works So x and z can take 2 values. Only y has a single value. Answer (B)
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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24 Mar 2017, 12:20
Found this question in the Veritas prep tests. Like it, though it damaged my score.
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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31 Jul 2017, 16:44
Here is a solution.. Answer is II or B
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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17 Nov 2019, 17:08
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Re: Suppose x, y, and z are positive integers such that xy + yz = 29 and x
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