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Algebra - Data Sufficiency; What is te value of y

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by GMATinsight » Tue Apr 12, 2016 5:21 am
lucas211 wrote:What is the value of y?

1) x^2 - y^2 = 5

2) x and y are each positive integers

Thanks in advance
Question : y = ?

Statement 1 : x^2 - y^2 = 5

i.e. (x+y)(x-y) = 5
Since we know nothing aboout x therefore y can't be calculated. Hence,
NOT SUFFICIENT


Statement 2 : x and y are each positive integers
But no concrete value of y can be calculated based on this information. Hence,
NOT SUFFICIENT

Combining the two statements

(x+y)(x-y) = 5 and both x and y arepositive integers therefore there sum and addition also much be Integers only

only (x+y)(x-y) = 1*5 is the possible solution cause (x+y) must be positive

i.e. x-y = 1 and x+y = 5 (cause x+y must be greater than x-y because of them being positive integers)

solving the two equations we get
x=3 and y=2 hence,
SUFFICIENT

Answer: Option C
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by [email protected] » Tue Apr 12, 2016 9:53 am
Hi lucas211,

This question can be solved by TESTing VALUES and a bit of 'brute force' math.

We're asked for the value of Y.

1) X^2 - Y^2 = 5

Since we know NOTHING about X and Y, those variables could be integers or non-integers.

IF...
Y = 1
X = root(6)
Then the answer to the question is 1.

IF...
Y = 0
X = root(5)
Then the answer to the question is 0.
Fact 1 is INSUFFICIENT

2) X and Y are POSITIVE INTEGERS.

From this Fact, we know that Y could be 1, 2, 3, 4, etc., but we don't know the exact value of Y.
Fact 2 is INSUFFICIENT

Combined, we know...
X^2 - Y^2 = 5
X and Y are POSITIVE INTEGERS.

Here, the options are significantly more limited. Since X and Y are INTEGERS, we're clearly dealing with PERFECT SQUARES in the first equation. I'm going to list out the first several positive perfect squares....
1
4
9
16
25
36
Etc.

From the equation in Fact 1, we know that the difference between the two perfect squares is 5. There is JUST ONE option that would fit: 9 and 4. Since Fact 2 tells us that the variables are POSITIVE integers, we know that X = 3 and Y = 2 is the ONLY option (those variables CANNOT be negative).
Combined, SUFFICIENT.

C

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by Matt@VeritasPrep » Tue Apr 12, 2016 6:26 pm
S1::

(x + y)(x - y) = 5

We could have 5 * 1 or (-5)*(-1), so this isn't sufficient.

S2::

We don't know anything about x and y, other than they're integers.

Now let's take them together.

Since x and y are both integers, (x + y) and (x - y) are both integers, so we have

(x + y) * (x - y) = integer * integer = 5

This means (x + y) and (x - y) are both factors of 5. Since x > 0 and y > 0, x + y > 0. 5 only has two factors (1 and 5), so we must have either

(x + y) = 5, (x - y) = 1
or
(x + y) = 1, (x - y) = 5

But in the second case, we'd have y = -2, which violates S2!

So our solution must be (x + y) = 5, (x - y) = 1, and we're set.
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by lucas211 » Wed Apr 13, 2016 12:28 am
Thanks a lot :-)
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by OptimusPrep » Wed Apr 13, 2016 7:56 pm
lucas211 wrote:Hello BTG

What is the value of y?

1) x^2 - y^2 = 5

2) x and y are each positive integers
Required: Value of y

Statement 1: x^2 - y^2 = 5
We have no clue about the value of x.
Insufficient

Statement 2:x and y are each positive integers
Again we do not know anything about the values of x and y
Insufficient

Combining both the statements:
(x-y)*(x+y) = 5
Since both x and y are positive integers, we have only two ways:
(x-y)*(x+y) = 5*1 or 1*5
Case 1:
x - y = 1
x + y = 5
On solving, we get x = 3, y = 2

Case 2:
x - y = 5
x + y = 1
On solving, we get x = 3, y = -2
Since both x and y are positive, case 2 cannot be considered.
Hence we have one solution
Sufficient.

Option C
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by Matt@VeritasPrep » Fri Apr 15, 2016 12:50 pm
lucas211 wrote:Thanks a lot :-)
No prob!

This is a neat problem, I think; I like GMAT problems like this, that make you think of algebra (or whatever) but actually respond better to some other approach.
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