Data Sufficiency

Is z even?
(1) 5z is even. (2) 3z is even.

Ans C

I know this is a basic question but can you please explain in detail
Appreciate it

Thanks

Assuming z is an integer

(1) 5z is even, which means that z must be even, since 5 is odd, odd * odd = odd, and odd * even = even

(2) 3z is even, which means that z must be even, since 3 is odd, odd * odd = odd and odd * even = even

So, it should be (D), not (C)

Is z even?
(1) 5z is even. (2) 3z is even.

Ans:

Statement 1:

Break 5z as 5*z. Now since it is even their must be a 2 somewhere in z because it cant be in the 5. So, z is an even number.

Statement 2:

The same logic applies. 3 is a prime and odd number and cant have 2 as a factor. So even this is Sufficient.

Both are SUFFICIENT!!

Assuming z is an integer

Ah, but that's one of the most important things to know about Data Sufficiency - you cannot assume anything!

Your goal in these situations should be to get the less-popular answer to the overall question - to challenge yourself to find the unique case that most won't find.

So for statement 1, it should be pretty easy to get the answer "YES", z is even. Because if z is 2, 5z = 10 and you've satisfied the statement. But is there a case out there in which 5z could be even but z could be not even? Well, since the question doesn't rule out a fraction, z = 2/5 also works. 2/5 * 5 = 2, which is even, but 2/5 itself is not an even number so we get the answer "NO".

Similarly for statement 2, it's pretty easy to get "NO" if you call z an even integer, because then 3z would definitely be even and the statement would be satisfied. But z = 2/3 also works for the statement so we could get z as something other than an even number - we can still get "NO".

Now, taken together those denominators of 5 and 3 don't work for us anymore. 2/3 * 3 is even, but 2/3 * 5 is not, so we can't use 2/3 (or 2/5 for that matter) anymore when both statements are required. That's why the answer is C - when both statements are present then z must be an even number.


The big takeaways here are:

1) Don't assume anything that you're not given (or that you can't logically infer from what is given)

2) Because of the above, make your goal to use all the available types of numbers in order to get the "other" answer to the overall question.

Thanks Brian.

It was silly to assume it as an interger!!

This is awesome Brian thanks ...

Brian@VeritasPrep wrote:

Assuming z is an integer

Ah, but that's one of the most important things to know about Data Sufficiency - you cannot assume anything!

Your goal in these situations should be to get the less-popular answer to the overall question - to challenge yourself to find the unique case that most won't find.

So for statement 1, it should be pretty easy to get the answer "YES", z is even. Because if z is 2, 5z = 10 and you've satisfied the statement. But is there a case out there in which 5z could be even but z could be not even? Well, since the question doesn't rule out a fraction, z = 2/5 also works. 2/5 * 5 = 2, which is even, but 2/5 itself is not an even number so we get the answer "NO".

Similarly for statement 2, it's pretty easy to get "NO" if you call z an even integer, because then 3z would definitely be even and the statement would be satisfied. But z = 2/3 also works for the statement so we could get z as something other than an even number - we can still get "NO".

Now, taken together those denominators of 5 and 3 don't work for us anymore. 2/3 * 3 is even, but 2/3 * 5 is not, so we can't use 2/3 (or 2/5 for that matter) anymore when both statements are required. That's why the answer is C - when both statements are present then z must be an even number.


The big takeaways here are:

1) Don't assume anything that you're not given (or that you can't logically infer from what is given)

2) Because of the above, make your goal to use all the available types of numbers in order to get the "other" answer to the overall question.

Brian@VeritasPrep wrote:

Assuming z is an integer

Ah, but that's one of the most important things to know about Data Sufficiency - you cannot assume anything!

Your goal in these situations should be to get the less-popular answer to the overall question - to challenge yourself to find the unique case that most won't find.

So for statement 1, it should be pretty easy to get the answer "YES", z is even. Because if z is 2, 5z = 10 and you've satisfied the statement. But is there a case out there in which 5z could be even but z could be not even? Well, since the question doesn't rule out a fraction, z = 2/5 also works. 2/5 * 5 = 2, which is even, but 2/5 itself is not an even number so we get the answer "NO".

Similarly for statement 2, it's pretty easy to get "NO" if you call z an even integer, because then 3z would definitely be even and the statement would be satisfied. But z = 2/3 also works for the statement so we could get z as something other than an even number - we can still get "NO".

Now, taken together those denominators of 5 and 3 don't work for us anymore. 2/3 * 3 is even, but 2/3 * 5 is not, so we can't use 2/3 (or 2/5 for that matter) anymore when both statements are required. That's why the answer is C - when both statements are present then z must be an even number.


The big takeaways here are:

1) Don't assume anything that you're not given (or that you can't logically infer from what is given)

2) Because of the above, make your goal to use all the available types of numbers in order to get the "other" answer to the overall question.

Why 2/5 is not en even number?

it is 0.4 and is divisible by 2...Could you elaborate this part pls?

Thnk you

Ozlemg wrote: Why 2/5 is not en even number?

it is 0.4 and is divisible by 2...Could you elaborate this part pls?

Thnk you

Parity is a property of integers only.

I am not clear about how we can combine them and say the answer is statement C is the answer.

Is it
5z-3z= 2z
=> 2z=2
=> z= 2/2 which yields 1 and we can say Z is not even and its sufficient

Please explain
Thanks