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Re: If x and y are positive integers, what is the value of x?
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20 Apr 2017, 07:31

5

GMATPrepNow wrote:

If x and y are positive integers, what is the value of x?

1) 3x + 2y = 12 2) |9x² + 30xy + 25y²| = 21²

*kudos for all correct solutions

Hi, A good Q, easy to error..

Catch is x and y are POSITIVE integers

Let's see the statements... 1) 3x + 2y = 12 2y and 12 are even, so 3x will also be even.. Thus x is even... It can be 2 or 4.. Only x as 2 gives us an integer value of y .. X=2 and y=3.. Sufficient..

2) |9x² + 30xy + 25y²| = 21² \(|(3x+5y)^2=21^2.....3x+5y=21..\).. Again ONLY x as 2 & y as 3 satisfy the equation.. Sufficient

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20 Apr 2017, 08:03

1

x and y are positive integer statement 1st 3x+2y =12 , considering the question stem, it will result in only one solution i.e(2,3) statement 2nd will water down to 3x +5y =21 , considering question stem, it will result in only one solution (2,3)-> option D

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24 Apr 2017, 07:35

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Top Contributor

1

GMATPrepNow wrote:

If x and y are positive integers, what is the value of x?

1) 3x + 2y = 12 2) |9x² + 30xy + 25y²| = 21²

IMPORTANT: I created this question to highlight a common myth about Data Sufficiency questions as well as highlight a common mistake that students make.

Target question:What is the value of x?

Given: x and y are positive integers

Statement 1: 3x + 2y = 12 Some students will see this equation with 2 variables and automatically conclude that there are infinitely many solutions, in which case, statement 1 is not sufficient. Under most conditions, this conclusion would be correct. However, in this question, we have the given condition that x and y are positive integers, which severely limits the possible solutions. In fact, there is only ONE pair of positive integers that satisfy the equation: x = 2 and y = 3. So, we can be certain that x = 2 Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: |9x² + 30xy + 25y²| = 21² First factor the part inside the absolute value to get: |(3x + 5y)²| = 21² This means that EITHER (3x + 5y)² = 21² OR (3x + 5y)² = -(21²) We can quickly dismiss the second case, (3x + 5y)² = -(21²), since (3x + 5y)² must be greater than or equal to zero. So, it could never equal -(21²)

So, what about (3x + 5y)² = 21²? This means that either 3x + 5y = 21 or 3x + 5y = -21 If x and y are both positive, we know that 3x + 5y will be positive, which means there are no solutions to the equation 3x + 5y = -21 What about the equation 3x + 5y = 21? Under the restriction that x and y are POSITIVE INTEGERS, there is only ONE pair of positive integers that satisfy the equation: x = 2 and y = 3. So, once again, we can be certain that x = 2 Since we can answer the target question with certainty, statement 2 is SUFFICIENT

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04 Mar 2018, 20:21

hi Brent - thanks for posting this question and providing the explanation. when working with absolute values, I thought we would have to consider all possibilities as follows - please help me understand why this approach is incorrect (and since I ended up with 4 possibilities as shown below - I concluded that this statement was insufficient). I can see why case 2 and case 3 are invalid as RHS and LHS are not equal - but what about cases 1 and 4?:

case 1: 3x + 5y = 21

case 2: -(3x + 5y) = 21

and since we took the square root of both sides we were left with + / - 21 on the RHS, so we have two more cases:

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05 Mar 2018, 03:01

hudhudaa wrote:

hi Brent - thanks for posting this question and providing the explanation. when working with absolute values, I thought we would have to consider all possibilities as follows - please help me understand why this approach is incorrect (and since I ended up with 4 possibilities as shown below - I concluded that this statement was insufficient). I can see why case 2 and case 3 are invalid as RHS and LHS are not equal - but what about cases 1 and 4?:

case 1: 3x + 5y = 21

case 2: -(3x + 5y) = 21

and since we took the square root of both sides we were left with + / - 21 on the RHS, so we have two more cases:

case 3: 3x + 5y = -21

case 4: -(3x + 5y) = -21

Hi

You are right that when we are given (3x+5y)^2 = 21^2 then we have to consider 4 possibilities.

But here in this question we are also given that x/y are positive integers, so 3x+5y has to be a positive integer only, and it cannot be anything else. So we just go with the case 3x+5y = 21, and then we try for positive integer values of x/y (not even 0) which will satisfy the equation. We will see that the only possibility is x=2, y=3

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05 Mar 2018, 08:34

Top Contributor

hudhudaa wrote:

hi Brent - thanks for posting this question and providing the explanation. when working with absolute values, I thought we would have to consider all possibilities as follows - please help me understand why this approach is incorrect (and since I ended up with 4 possibilities as shown below - I concluded that this statement was insufficient). I can see why case 2 and case 3 are invalid as RHS and LHS are not equal - but what about cases 1 and 4?:

case 1: 3x + 5y = 21

case 2: -(3x + 5y) = 21

and since we took the square root of both sides we were left with + / - 21 on the RHS, so we have two more cases:

case 3: 3x + 5y = -21

case 4: -(3x + 5y) = -21

Good question.

You'll find that the solution to the case 1 equation is identical to the case 4 equation, and the solution to the case 2 equation is identical to the case 3 equation. The reason for this is that equations 1 and 4 are equivalent equations, and equations 2 and 3 are equivalent equations.

Take, for example, case 1: 3x + 5y = 21 If we multiply both sides of this equation by -1, we get -(3x + 5y) = -21 (the case 4 equation).