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If x = u^2 – v^2 and y = 2uv

Expert replies
by gmatdriller » Thu Sep 24, 2015 3:40 pm
If x = u^2 - v^2, y = 2uv, and z = u^2 + v^2 , and if x = 11, what is the value of z
(1) y = 60
(2) u = 6

statement (2) is easier- sufficient

For statement (1), I expressed (u^2 - v^2) as a difference of two squares => (u+v)(u-v)
Recognizing that u^2 - v^2 can be expressed as a difference of two squares
x = (u+v)(u-v); x^2 = (u+v)^2(u-v)^2
x^2 = (u^2 + v^2 + 2uv)(u^2 + v^2 - 2uv)
x^2 = (z + 60)(z - 60)... z = u^2 + v^2, given
11^2 = z^2 - 60^2..... x =11, given
Z^2 = 11^2 + 60^2...BUT we know that because z is expressed as a sum of squares, it CANNOT be -VE
So, the value of z takes the +VE component of Z^2
Statement (2) is also SUFFICIENT

Has anyone got a shorter, more elegant approach for statement (1)?
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Source: — Data Sufficiency |

by GMATGuruNY » Thu Sep 24, 2015 4:26 pm
gmatdriller wrote:If x = u^2 - v^2, y = 2uv, and z = u^2 + v^2 , and if x = 11, what is the value of z?

(1) y = 60
(2) u = 6
The values in the problem -- 2, 11, 60 and 6 -- imply that x, u, v, y and z are all INTEGERS.

Question stem, rephrased:
What is the value of u² + v²?

Since x = 11 and x = u² - v², we get:
11 = u² - v².
Implication:
u and v are PERFECT SQUARES with a difference of 11.
Make a list of perfect squares:
1, 4, 9, 16, 25, 36, 59...
Only the values in red yield the required difference of 11, implying that u²=36 and that v²=25.
Thus, it is almost guaranteed that u=±6 and that v=±5.

Statement 1: y=60
Since y=2uv and y=60, we get:
60=2uv
30=uv.
Here, either u=6 and v=5 or u=-6 and v=-5, in accordance with the red values above.
No other factor combination for uv=30 will satisfy the constraint that u² - v² = 11.
Thus, u² + v² = (±6)² + (±5)² = 36+25=61.
SUFFICIENT.

Statement 2: u=6
If u=6, then v=±5, in accordance with the red values above.
Thus, u² + v² = (6)² + (±5)² = 36+25=61.
SUFFICIENT.

The correct answer is D.
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by Matt@VeritasPrep » Fri Sep 25, 2015 1:21 am
I don't like the integer assumption, so let's do it another way. We have 11 = u² - v² to start.

S1 tells us that uv = 30, so u = 30/v. Plugging this in, we have

11 = (30/v)² - v², or

11 = 900/v² - v², or

11v² = 900 - v�, or

v� + 11v² - 900 = 0, or

(v² + 36)(v² - 25) = 0

So v² = 25, and v = 5 or v = -5. From 2uv = 60, this gives {v = 5, u = 6} or {v = -5, u = -6}, and in either case u² + v² = 61; SUFFICIENT.
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by GMATGuruNY » Fri Sep 25, 2015 3:17 pm
I don't like the integer assumption, so let's do it another way. We have 11 = u² - v² to start.

S1 tells us that uv = 30, so u = 30/v. Plugging this in, we have

11 = (30/v)² - v², or

11 = 900/v² - v², or

11v² = 900 - v�, or

v� + 11v² - 900 = 0, or

(v² + 36)(v² - 25) = 0

So v² = 25, and v = 5 or v = -5. From 2uv = 60, this gives {v = 5, u = 6} or {v = -5, u = -6}, and in either case u² + v² = 61; SUFFICIENT.
That all of the unknowns are integers is not so much an assumption as a conclusion based upon the way numbers -- and DS problems -- work.
Since the two statements cannot contradict each other, the value in statement 2 (u=6) must also satisfy statement 1.
In statement 1, u=6 implies that v=5.
It is not possible for uv = 30 and u² - v² = 11 to have both an integer solution (u=6, v=5) and a non-integer solution.
Implication:
All of the unknowns in the problem MUST be integers.

While the problem is designed to make us evaluate statement 1 algebraically -- an approach that would probably prove too time-consuming for many students -- a savvy test-taker can take advantage of the GMAT's penchant for pretty numbers.
Given uv = integer and u² - v² = integer, it is virtually guaranteed on the GMAT that u and v will be integers.
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by Samuel117 » Wed May 18, 2016 7:53 am
Matt@VeritasPrep wrote:I don't like the integer assumption, so let's do it another way. We have 11 = u² - v² to start.

S1 tells us that uv = 30, so u = 30/v. Plugging this in, we have

11 = (30/v)² - v², or

11 = 900/v² - v², or

11v² = 900 - v�, or

v� + 11v² - 900 = 0, or

(v² + 36)(v² - 25) = 0

So v² = 25, and v = 5 or v = -5. From 2uv = 60, this gives {v = 5, u = 6} or {v = -5, u = -6}, and in either case u² + v² = 61; SUFFICIENT.
Hi Matt, how do you go from v� + 11v² - 900 = 0, to (v² + 36)(v² - 25) = 0 ? How to do this fast without having to use the quadratic formula? thank you very much
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by 800_or_bust » Wed May 18, 2016 9:48 am
Samuel117 wrote:
Matt@VeritasPrep wrote:I don't like the integer assumption, so let's do it another way. We have 11 = u² - v² to start.

S1 tells us that uv = 30, so u = 30/v. Plugging this in, we have

11 = (30/v)² - v², or

11 = 900/v² - v², or

11v² = 900 - v�, or

v� + 11v² - 900 = 0, or

(v² + 36)(v² - 25) = 0

So v² = 25, and v = 5 or v = -5. From 2uv = 60, this gives {v = 5, u = 6} or {v = -5, u = -6}, and in either case u² + v² = 61; SUFFICIENT.
Hi Matt, how do you go from v� + 11v² - 900 = 0, to (v² + 36)(v² - 25) = 0 ? How to do this fast without having to use the quadratic formula? thank you very much
Yeah, that would be a challenge to factor.
800 or bust!
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