If x, y and z are integers, is x odd?

Correct answer is C

Statement 1) for X to be odd, both y and z have to be odd.

We don't know whether y, z are odd or even. So insufficient.

2) x = y + z

One of y or z has to be odd and one has to be even for x to be odd. Not sufficient.

Combining the statements together.

yz = z + y

When z = y = 2 only satisfies the above equation. Hence, x is not odd.

So answer is C

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.



If x, y and z are integers, is x odd?

(1) yz = x

(2) x - y = z
In the original condition we have 3 variables (x,y,z) and we need 3 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), E is likely the answer. Using both 1) & 2) together, x=y+z=yz thus y=z=even --> x=even and the answer is no. However, if y=even, z=odd -->then x=odd, but out of scope(since yz=even) and also y=z=odd --> x=even, is also out of scope (since yz=odd). therefore, the only answer is y=z=even leading us to x=even. The answer is no, therefore the conditions sufficient. The answer is C.

If x, y and z are integers, is x odd?

(1) yz = x

(2) x - y = z

St 1 . yz = x

x could be even or odd .

yz = even * odd
or
yz= even * even

not sufficient

St 2 .
x= y +z
again x could be odd or even

So not sufficient

together
x=yz
or y+z = yz

The only number that satisfy the above condition is 2 .

Hence x even . Definite ans

C

(1) yz = x. Insufficient. x can be even or odd depending on the values of y and z
(2) x - y = z. x=y+z.Insufficient. x can be even or odd depending on the values of y and z

Using both, only one case exists, y,z=2,2 and x=4
Answer C

KAPLAN OFFICIAL SOLUTION:

This is a good example of a problem that does not seem to give us very much information in the stem or in the statements that is related to the question. However, by doing some algebra we can try to find a relationship.

First, let’s assess Statement 1. Statement 1 tells us nothing about whether x is even or odd, as we do not know anything about if y and z are odd. Therefore, statement 1 is not sufficient (and you can eliminate answer choices 1 & 4, or A & D).

Next, let’s assess Statement 2 (and remember—we are first assessing each statement on its own). Statement 2 runs into a similar problem as statement 1. We can isolate x here, to make the equation x = z + y, but without knowing anything about z and y, statement 2 is also insufficient (so you can eliminate answer choice 2 or B).

When considering the statements together, many test-takers will simply say that without knowing anything about y and z still, we cannot answer the question. However, we can do a little algebra here to gain some information. Since we know from statement 1 that x = yz and from statement 2 that x = y + z, we can say:

yz = y + z

yz – z = y

z(y-1) = y

Because y is an integer, we know that either y is even or y-1 is even. However, because an even times anything is an even, if y-1 is even then z(y-1) must equal an even number and y and y-1 cannot both be even. Therefore, the only way z(y-1) can be even is if z is even and y-1 is odd. Since we know that z is even and that y is even, we know that zy is also even. So, x must be even, because x = zy. The answer to our question, then, is ‘always no’ and the statements are sufficient together (answer choice 3 or C in Data Sufficiency).

just comebine and assume that yz is odd, then st2 cannot hold true => x cannot be an odd number.

I just want to add to the explanations above and clarify one thing: all three numbers could be 0 or all three could 2. So this question is also testing whether you know that zero is an even number.

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z=y=0 also satisfies the above equation. The answer remains C.

If x, y, and z are integers, is x odd?

(1) yz = x
Is x odd?
=> Is yz odd?
=> Are both y and z odd? (since the product of two integers is odd only if both of them are odd)
We have no idea!

Insufficient

(2) x - y = z
x = y + z
Is x odd?
=> Is y + z odd?
=> Is one of y, z even and the other odd?, i.e., is y even, z odd OR y odd, z even?
No idea!

Insufficient

Together,
x = yz = y + z

(i) y odd, z odd:
yz = odd and y+z = even
Impossible

(ii) y even, z even:
yz = even, y+z = even
Possible
So, x is even.

Is x odd?
NO

(iii) y even, z odd:
yz = even, y+z = odd
Impossible

(iv) y odd, z even
yz = even, y+z = odd
Impossible

Hence, we only get a NO

Sufficient

Correct Answer: C

Statement 1 says : x=yz

y=2,z=2 we get x=4 No
y=1,z=1 we get x= Yes


Statement 2 : x=z-y

z=4,y=2 we get x=2 NO
z=4,y=3 we get x=1 yes

On combining
We know that for y and z can be both even or one odd and one even
In both cases x=yz will always be even

hence we get a perfect NO. Therefore C