Statement 1:
w/x = 1/z + 1/x
because w > x
w/x is >1
1/x < 1
so z=1
so equations becomes
w/x = 1 + 1/x
w=x+1
w and x are consecutive integers so their common divisor can only be 1
If y=1 then z becomes zero which could not be the case
so y is not a common divisor
Statement 2:
w-y-2=0 (factor out a w)
so w=y+2
hence w=x+1
w and x are consecutive integers so their common divisor can only be 1
If y=1 then z becomes zero which could not be the case
so y is not a common divisor
w, x, y, and z are integers
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Source: Beat The GMAT — Data Sufficiency |
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vishugogo
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Statement 1:
w/x = 1/z + 1/x
because w > x
w/x is >1
1/x < 1
so z=1
so equations becomes
w/x = 1 + 1/x
w=x+1
w and x are consecutive integers so their common divisor can only be 1
If y=1 then z becomes zero which could not be the case
so y is not a common divisor
Statement 2:
w-y-2=0 (factor out a w)
so w=y+2
hence w=x+1
w and x are consecutive integers so their common divisor can only be 1
If y=1 then z becomes zero which could not be the case
so y is not a common divisor
w/x = 1/z + 1/x
because w > x
w/x is >1
1/x < 1
so z=1
so equations becomes
w/x = 1 + 1/x
w=x+1
w and x are consecutive integers so their common divisor can only be 1
If y=1 then z becomes zero which could not be the case
so y is not a common divisor
Statement 2:
w-y-2=0 (factor out a w)
so w=y+2
hence w=x+1
w and x are consecutive integers so their common divisor can only be 1
If y=1 then z becomes zero which could not be the case
so y is not a common divisor
-
Matt@VeritasPrep
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Statement 1:
Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z
Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.
If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!
Statement 2 is much easier to work with:
w^2 - wy - 2w = 0
Since w is greater than 0, divide both sides by w.
w - y - 2 = 0
w - 2 = y
Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.
This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.
Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z
Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.
If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!
Statement 2 is much easier to work with:
w^2 - wy - 2w = 0
Since w is greater than 0, divide both sides by w.
w - y - 2 = 0
w - 2 = y
Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.
This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.
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GMATGuruNY
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A helpful rule to know:guerrero wrote:w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?
(1) w/x= z^-1+x^-1
(2) w^2-wy-2w=0
If x and z are positive integers, and x/z is an integer, then x/z is a FACTOR OF X.
Proof:
Let x/z = k, where k is an integer.
Thus:
x = k * z
The resulting equation indicates that k -- and thus x/z -- is a factor of x.
Since w>x>y>z>0, the least possible values are w=4, x=3, y=2, and z=1.
Statement 1: w/x = 1/z + 1/x
Multiplying each side by x, we get:
w = x/z + 1.
The resulting equation implies that w = (factor of x) + 1.
If x=3, its factors are 1 and 3.
Since w must be 1 more than either 1 or 3, and w>x, the only option is w = 3+1 = 4.
If x=6, its factors are 1, 2, 3 and 6.
Since w must be 1 more than 1, 2, 3, or 6, and w>x, the only option is w = 6+1 = 7.
The cases above illustrate that w and x are CONSECUTIVE INTEGERS.
Consecutive integers are COPRIMES: they share no factors other than 1.
Since y>1, and w and x share no factors other than 1, it is not possible that y is a factor of both w and x.
SUFFICIENT.
Statement 2: w²- wy- 2w = 0
w(w-y-2) = 0.
Since w≠0, it must be true that w-y-2 = 0.
Thus:
w = y+2.
If y=2, then w=4.
Since x must be an integer BETWEEN y and w, x=3.
The result:
w=4, x=3, y=2.
The case above indicates that w and x are CONSECUTIVE INTEGERS.
Since w and x are consecutive integers -- and thus share no factors other than 1 -- it is not possible that y is a factor of both w and x.
SUFFICIENT.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Anaira Mitch
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GMATGuruNY wrote:A helpful rule to know:guerrero wrote:w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?
(1) w/x= z^-1+x^-1
(2) w^2-wy-2w=0
If x and z are positive integers, and x/z is an integer, then x/z is a FACTOR OF X.
Proof:
Let x/z = k, where k is an integer.
Thus:
x = k * z
The resulting equation indicates that k -- and thus x/z -- is a factor of x.
Since w>x>y>z>0, the least possible values are w=4, x=3, y=2, and z=1.
Statement 1: w/x = 1/z + 1/x
Multiplying each side by x, we get:
w = x/z + 1.
The resulting equation implies that w = (factor of x) + 1.
If x=3, its factors are 1 and 3.
Since w must be 1 more than either 1 or 3, and w>x, the only option is w = 3+1 = 4.
If x=6, its factors are 1, 2, 3 and 6.
Since w must be 1 more than 1, 2, 3, or 6, and w>x, the only option is w = 6+1 = 7.
The cases above illustrate that w and x are CONSECUTIVE INTEGERS.
Consecutive integers are COPRIMES: they share no factors other than 1.
Since y>1, and w and x share no factors other than 1, it is not possible that y is a factor of both w and x.
SUFFICIENT.
Statement 2: w²- wy- 2w = 0
w(w-y-2) = 0.
Since w≠0, it must be true that w-y-2 = 0.
Thus:
w = y+2.
If y=2, then w=4.
Since x must be an integer BETWEEN y and w, x=3.
The result:
w=4, x=3, y=2.
The case above indicates that w and x are CONSECUTIVE INTEGERS.
Since w and x are consecutive integers -- and thus share no factors other than 1 -- it is not possible that y is a factor of both w and x.
SUFFICIENT.
The correct answer is D.
Great Solution Mitch. Easily Understandable.
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hazelnut01
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Dear @Matt@VeritasPrep, Could you help to elaborate the statement highlighted in red? I could not image the number.Matt@VeritasPrep wrote:Statement 1:
Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z
Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.
If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!
Statement 2 is much easier to work with:
w^2 - wy - 2w = 0
Since w is greater than 0, divide both sides by w.
w - y - 2 = 0
w - 2 = y
Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.
This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.
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DavidG@VeritasPrep
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Matt was invoking the following axiom: two consecutive integers cannot have any divisors in common aside from 1.ziyuenlau wrote:Dear @Matt@VeritasPrep, Could you help to elaborate the statement highlighted in red? I could not image the number.Matt@VeritasPrep wrote:Statement 1:
Let's simplify first.
w/x = 1/x + 1/z
w/x = (x+z)/xz
w = (x+z)/z
wz = (x+z)
wz - z = x
z(w-1) = x
(w-1) = x/z
Since w > x and w and x are integers, (w - 1) cannot be less than x. If z is an integer, x/z will be less than x UNLESS z = 1. So (w - 1) = x, and z = 1.
If x = (w - 1), then x and w are consecutive integers, and hence only have ONE common divisor: the number 1 itself. Since z = 1, and y > z, y CANNOT be a common divisor of x and w. SUFFICIENT!
Statement 2 is much easier to work with:
w^2 - wy - 2w = 0
Since w is greater than 0, divide both sides by w.
w - y - 2 = 0
w - 2 = y
Since w and y are integers, and w > x > y, x must therefore be (w - 1). At this point we reach the same conclusion that we did in Statement 1, and this is statement is also SUFFICIENT.
This is a pretty sophisticated question, and it strikes me as something you'd only see if you're scoring Q49+ on the actual exam.
To illustrate, take two random consecutive numbers, 15 and 16
Factors of 15: 1, 3, 5, 15
Factors of 16: 1, 2, 4, 8, 16
Notice, they have no factors in common, aside from 1. This makes sense. If 15 has 3 as a factor, the next number that will have 3 as a factor is 18. If 15 has 5 as a factor, the next number that will have 5 as a factor is 20. So we're continually leaping over that 16.
As a consequence, if I asked you if variable 'y' was a common divisor of 15 and 16, you'd know that the only way this could be true would be if y = 1. right? Similarly, if y is not equal to 1, then y can't be a common divisor of 15 and 16.
Same idea here. If x = w - 1, then x and w are consecutive integers. (In my scenario, w =16 and x = 15.) If z = 1 and y > z, then we know that y is NOT equal to 1. If y is NOT equal to 1, and x and w are consecutive integers, then y cannot be a divisor of those numbers, as two consecutive numbers have no factors in common, aside from 1.
