if w, x, y, z are integers such that w/x and y/z are integers, is w/x + y/z odd?

Question

1) wx + yz is odd
2) wz + xy is odd


Choice A: Statement 1 is sufficient by itself
Choice B: Statement 2 is sufficient by itself
Choise C: Need both statements together to find the answer
Choice D: Either statement is sufficient by itself
Choice E: Neither statement.

Please select the best choice for this question and explain why you picked it.

Answer 1

Well, we seem to have several incorrect explanations so far.

Firstly, the first one alone is not sufficient. Let x = w = 2, y=z=1. Then wx+yz = 4+1 = 5 is odd, but w/x + y/z = 2 is even.

However, the second one alone is sufficient. Note that w/x + y/z = (wz + xy)/xz. If wz + xy is odd, then x and z must both be odd to make that an integer; therefore the quotient must be odd.

So the answer is B.

Answer 2

Remember this:

for the sum of two terms to be odd, one and only one of the terms must be odd (2 + 3 = 5)

for the product of two factors to be odd, both factors must be odd (7 * 5 = 35)

in a division, if the dividend is odd, the divisor must be odd and the quotient will be odd (35 / 7 = 5) <--Stephen uses this factoid

if the product of two numbers is odd, their quotient will be odd also (15 * 5 = 75; 15 / 5 = 3)

but if the product of two numbers is even, their quotient could be even or odd (42 * 6 = 252; 42 / 6 = 7)(8 * 4 = 32; 8 / 4 = 2)

Armed with this, we can see that statement 1 is useless. In statement 1, either wx or yz must be odd, but not both. Let's say that wx is odd and yz even. Then w/x is odd, but we can't say anything about y/z. (Kellenraid is wrong in his third line. If wx is odd, both w and x must be odd, not what he says.)

Stephen's method of dismissing statement 1 is the best, just giving a counter-example.

Answer 3

Using ONLY Statement 1 we can know ONE of the follow posibilities is true.
1. wx is even & yz is odd & w is divisible by 2x & x is even & y is odd / z is odd / (w/x) is even / (y/z) is odd
2. wx is even & yz is odd & w is even & x is even & y is odd & z is odd / (w/x) is odd / (y/z) is odd
3. wx is even & yz is odd & w is even & x is odd & y is odd & z is odd / (w/x) is even / (y/z) is odd
4. wx is odd & yz is even & w is odd & x is odd & y is divisible by 2z & z is even / (w/x) is odd / (y/z) is even
5. wx is odd & yz is even & w is odd & x is odd & y is even & z is even / (w/x) is odd / (y/z) is odd
6. wx is odd & yz is even & w is odd & x is odd & y is even & z is odd / (w/x) is odd / (y/z) is even

Using only statement 1 we can’t determine if w/x + y/z is odd because in posibilities 1, 3, and 6 it would result in an odd result and in 2, 4, and 5 it would result in an even result.

Using ONLY statement 2 we can know ONE of the following posibilities is true.

7. wz is even & xy is odd & w is even & z is odd & x is odd & y is odd & (w/x) is even / (y/z) is odd
8. wz is odd & xy is even & w is odd & z is odd & x is odd & y is even & (w/x) is odd / (y/z) is even

The other possiblities were eliminated because x must be a factor of w and z must be a factor of y.

Both of the posiblities listed require w/x + y/z to be odd, therefore we don’t need statement 1 at all.

The answer is B

Answer 4

The choice is D.
Take choice 1 to begin: to be true, then wx is odd or even
if wx is odd, then either w is odd or x is odd
Since the above must be true, then take w to be odd, then x must be even; that means that y and z are either both odd or both even
Go back to w/x + y/z, by the premise above, then since w is odd, x even, then w/x is odd and y/z must be even.
Since both 1 and 2 are going to be odd, then the same reasoning will apply to make the sum w/x + y/z to always be odd whether you use 1 or 2; hence, choice D

Answer 5

E is right. Statement 1 has no relevance. Statement 2 is not sufficient as sum of two integers need not be odd. We need some more inf about these integers.

Answer 6

w/x integer => w = mx (m is some integer)

y/z integer => y = nz (n is some integer)

wx + yz = m^2 * x + n^2 * z is odd (read m squared x ...)
=> m^2 * x is odd or n^2 * z is odd
=> (x & m are odd) OR (z & n are odd) [ONLY one set is ODD]

w/x + y/z = m + n = Odd (only one of them id odd)

So stmt 1 is sufficient (CHOICE A)