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29 Dec 2014, 08:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
76% (01:11) correct
24%
(01:08)
wrong
based on 187
sessions
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Tough and Tricky questions: Word Problems.
Is z an odd integer?
(1) z is divisible by 3.
(2) The square root of z is an integer divisible by 3.
Kudos for a correct solution.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
29 Dec 2014, 10:50
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We can solve this one easily by picking numbes.
Statement 1: if we could fine one odd and one even integer, both divisible by 3, we could show that this statement is insufficient. 3 and 6 are such numbers. To prove it by using number properties, we can just write 3as 1*3 and 6 as 2*3. As 3 is odd, multiplying it with odd and even numbers will have odd and even results. Insufficient.
Statement 2: With the same approach, we can show that the second statement is also insufficient. We just need to square 3 and 6. As 3is odd, it's square,9, must also be odd. As 6 can be written as 2*3, it's square is \((2*3)^2=2^2*3^3=36\), it must be even. Therefore statement 2is also insufficient.
Both taken together: as we see from the factorization, the numbers picked for statement 2 (9 and 36) already fulfill both statements and still do not allow a clear conclusion whether z is odd or even. Answer E.
Statement 1: if we could fine one odd and one even integer, both divisible by 3, we could show that this statement is insufficient. 3 and 6 are such numbers. To prove it by using number properties, we can just write 3as 1*3 and 6 as 2*3. As 3 is odd, multiplying it with odd and even numbers will have odd and even results. Insufficient.
Statement 2: With the same approach, we can show that the second statement is also insufficient. We just need to square 3 and 6. As 3is odd, it's square,9, must also be odd. As 6 can be written as 2*3, it's square is \((2*3)^2=2^2*3^3=36\), it must be even. Therefore statement 2is also insufficient.
Both taken together: as we see from the factorization, the numbers picked for statement 2 (9 and 36) already fulfill both statements and still do not allow a clear conclusion whether z is odd or even. Answer E.
08 Jan 2015, 03:26
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St (1) : 3 has both odd and even multiples, so z can be odd or even..... Insufficient.
St (2): Z can be 9 or 36, according to this statement.....Insufficient.
Combining both statements, still 9 AND 36 satisfy both statements......Answer is E.
St (2): Z can be 9 or 36, according to this statement.....Insufficient.
Combining both statements, still 9 AND 36 satisfy both statements......Answer is E.
