1) \(4y\) will always be even. Then we have \(5x - even = even\). For this to be the case, \(5x\) must be even. Since 5 can't be even, then x must be even. Thus the product \(xy\) will be even. Sufficient.

2) \(6x\) will always be even. Then we have \(even + 7y = even\). Thus \(7y\) is even, and \(y\) is even, and \(xy\) is even. Sufficient.

Answer: D

Target question: Is xy even?

Statement 1: 5x - 4y is even
Let's test all 4 cases
case a: x is even and y is even: In this case 5x-4y is EVEN
case b: x is even and y is odd: In this case 5x-4y is EVEN
case c: x is odd and y is even: In this case 5x-4y is ODD
case d: x is odd and y is odd: In this case 5x-4y is ODD
So, cases a and b are both possible.
In both cases the product xy is even
So, xy must be even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Aside: If anyone is interested, we have a video (below) on testing possible cases for these question types

Statement 2: 6x+7y is even
Let's test all 4 cases
case a: x is even and y is even: In this case 6x+7y is EVEN
case b: x is even and y is odd: In this case 6x+7y is ODD
case c: x is odd and y is even: In this case 6x+7y is EVEN
case d: x is odd and y is odd: In this case 6x+7y is ODD
So, cases a and c are both possible.
In both cases the product xy is even
So, xy must be even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D
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For xy to be even, either x or y or both must be even.

(1) 5x - 4y is even
E-E=E
O-O=E

Since 4y is even, 5x has to be even. 5 is not even, hence x is even.

xy will be even.

(2) 6x + 7y is even
E+E= E
O+O=E

Since 6x is even, 7y has to be even. And because 7 is not even y has to be even.

xy will be even.

Both statements are sufficient.

D is the answer
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D. in either case x and y has to be even.

(1) 5x - 4y is even: x has to be even. suff.
(2) 6x + 7y is even: y has to be even. suff.

agree with D.

1) 5x - 4y is even -> x must be multiple of 2
5*2k-4y = 2*(any number) even
sufficient
2)6x + 7y is even[/
y should multiple of 2

even
sufficient

In my opinion, this type of problem is super easy if you really understand what you're being asked, and then re-frame the question to be more straightforward. It's more about the question than the math.

Given: (x,y) > 0, AND are integers

Original Question: Is xy even?

We know that any number multiplied by an even number is even, so the question is really asking: Do we know if either "X" or "Y" (or both, or neither) is even?

Reframing the question this way is better, in my opinion.



A.) 5x - 4y is even.

For 5x - 4y to be even, both 5x and 4y have to be of the same parity. ("Parity" is the word that describes whether a number is odd or even)
In other words, odd + odd, or even + even, produce even results.



5x - 4y

We know off the bat that "4y" is even, because 4 is even. 4 times anything would be even. So we have ( 5x - EVEN NUMBER = EVEN NUMBER. ) Therefore, we know that "5x" is also even, because 4y (even) and 5x have to be the same parity. (even minus even = even)

In order to make 5x even, "x" has to be even. That's because 5 is odd.

Knowing that "x" is even is enough to answer the question: Do we know if either "X" or "Y" (or both, or neither) is even?

SUFFICIENT.


B.) 6x + 7y is even

Same drill here. For 6x + 7y to be even, both terms have to be the same parity. Since 6 is even, we know that 6x is even. Since 6x is even, we know that 7y is even. Since 7y is even, we know that y is even (since 7 is odd).

Knowing that "y" is even is enough to answer the question: Do we know if either "X" or "Y" (or both, or neither) is even?

SUFFICIENT

The answer is D

\(S1: 5x - 4y = even => 5x = even+4y => 5x = even => x = even => xy = even. Suff\)

\(S2: 6x+7y = even => 7y = even - 6x =>7y = even => y = even => xy = even. Suff\)

Both S1 and S2 are individually sufficient. Final answer = D
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If x and y are positive integers, is the product xy even
(1) 5x - 4y is even
(2) 6x + 7y is even

1. for 5x-4y = Even, x must be even & y may or may not be even. Since x is even, we can conclude xy is even. SUFFICIENT.
Even – Even = Even
Even – Odd = Odd
Odd – Even = Odd
Odd – Odd = Odd

2. for 6x + 7y is even, y must be even. Since y is even, we can conclude xy is even. SUFFICIENT.
Even + Even = Even
Odd + Even = Odd
Even + Odd = Odd
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Statement 1. 5x – 4y is even i.e. 5x – 4y = 2k
Difference of two numbers is even if either both of them are odd or both of them are even.
Here, 4y is always even. So, 5x is also even.
Since 5x is even, we can say that x is even.
Now, xy will be even because the product of an even no with any positive integer is always even. Hence, Sufficient.
Statement 2. 6x + 7y is even.
6x is eve n because it is divisible by 2. So, 7y has to be even.
For 7y to be even, y has to be even.
We know the product of an even no with a positive integer is even.
Hence, xy is even. Hence, Sufficient.

Can this problem be broken out algebraically rather than testing values?

e.g.

For Statement 1 I did:
(1) 5x - 4y is even

4x+1x - 4y = E
4x -4y +1x = E
4(x-y) + 1x = E; X MUST BE EVEN; if X is even XY = Even

so sufficient;

(2) 6x + 7y is even
6x + 6y + y = Even
6(x+y) + y = Even; Y must be even as E + E = 0 and 6*(X+Y) = E

so sufficient;

Answer = D

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