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Ah, but that's one of the most important things to know about Data Sufficiency - you cannot assume anything!Assuming z is an integer
Brian@VeritasPrep wrote:Ah, but that's one of the most important things to know about Data Sufficiency - you cannot assume anything!Assuming z is an integer
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:Ah, but that's one of the most important things to know about Data Sufficiency - you cannot assume anything!Assuming z is an integer
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.
Parity is a property of integers only.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
For statement 1, there are 2 possible scenarios to consider.adity_13 wrote: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
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