If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
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x an even integer
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j_shreyans
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GMATGuruNY
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Statement 1: 3x is even.If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
It's possible that x = 2, which is an even integer.
It's possible that x = 2/3, which is not an even integer.
INSUFFICIENT.
Statement 2: 5x is even.
It's possible that x = 2, which is an even integer.
It's possible that x = 2/5, which is not an even integer.
INSUFFICIENT.
Statements 1 and 2 combined:
5x-3x = even - even = even.
Since 5x-3x = 2x, 2x is even.
3x-2x = even - even = even.
Since 3x-2x=x, x is even.
SUFFICIENT.
The correct answer is C.
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j_shreyans
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Hi Guru ,Statements 1 and 2 combined:
5x-3x = even - even = even.
Since 5x-3x = 2x, 2x is even.
3x-2x = even - even = even.
Since 3x-2x=x, x is even.
SUFFICIENT.
Thanks for your reply, but i am unable to understand the combined part..
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GMATGuruNY
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When one even integer is subtracted from another even integer, the result is an even integer.j_shreyans wrote:Hi Guru ,Statements 1 and 2 combined:
5x-3x = even - even = even.
Since 5x-3x = 2x, 2x is even.
3x-2x = even - even = even.
Since 3x-2x=x, x is even.
SUFFICIENT.
Thanks for your reply, but i am unable to understand the combined part..
In math terms:
even - even = even.
Statement 1: 3x is even
Statement 2: 5x is even
Since 5x is even and 3x is even, their difference -- 2x -- is also even.
Since 3x is even and 2x is even, their difference -- x -- is also even.
Thus, the two statements combined imply that x is an even integer.
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[email protected]
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Hi j_shreyans,
Mitch's explanation shows the value of knowing your Number Property rules. In the event that you don't spot that pattern (when it's there), you can still "discover" the pattern by TESTing VALUES. Here's how....
Once you've determined that Fact 1 is INSUFFICIENT and Fact 2 is INSUFFICIENT, you combine Facts....
3X is an EVEN integer
5X is an EVEN integer
We're told that X is a POSITIVE number. Let's list out values for each Fact and see if a pattern emerges....
3X = EVEN integer.....X COULD be 2/3, 4/3, 6/3 = 2, 8/3, 10/3, 12/3 = 4, etc.
5X = EVEN integer.....X COULD BE 2/5, 4/5, 6/5, 8/5, 10/5 = 2, 12/5, 14/5, 16/5, 18/5, 20/5 = 4, etc.
Since the denominators will always be 3 (for the first group) and 5 (for the second group), most values of X will not fit both Facts. However, the ones that DO fit both Facts (that we can immediately see) are 2 and 4. If we increase the lists, that pattern will continue....6, 8, 10, etc. This provides the proof that X is ALWAYS going to be an EVEN integer, so the answer to the question is ALWAYS YES.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich[/b]
Mitch's explanation shows the value of knowing your Number Property rules. In the event that you don't spot that pattern (when it's there), you can still "discover" the pattern by TESTing VALUES. Here's how....
Once you've determined that Fact 1 is INSUFFICIENT and Fact 2 is INSUFFICIENT, you combine Facts....
3X is an EVEN integer
5X is an EVEN integer
We're told that X is a POSITIVE number. Let's list out values for each Fact and see if a pattern emerges....
3X = EVEN integer.....X COULD be 2/3, 4/3, 6/3 = 2, 8/3, 10/3, 12/3 = 4, etc.
5X = EVEN integer.....X COULD BE 2/5, 4/5, 6/5, 8/5, 10/5 = 2, 12/5, 14/5, 16/5, 18/5, 20/5 = 4, etc.
Since the denominators will always be 3 (for the first group) and 5 (for the second group), most values of X will not fit both Facts. However, the ones that DO fit both Facts (that we can immediately see) are 2 and 4. If we increase the lists, that pattern will continue....6, 8, 10, etc. This provides the proof that X is ALWAYS going to be an EVEN integer, so the answer to the question is ALWAYS YES.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich[/b]
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Brent@GMATPrepNow
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Target question: Is x an even integer?j_shreyans wrote:If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
Statement 1: 3x is an even integer
There are two possible cases that satisfy this condition:
Case a: x is an even integer
Case b: x is a fraction with 3 in the denominator and an even number in the numerator. For example, x COULD equal 2/3 or 4/3 or 8/3, or 10/3, etc). In this case x is NOT an even integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: 5x is an even integer
There are two possible cases that satisfy this condition:
Case a: x is an even integer
Case b: x is a fraction with 5 in the denominator and an even number in the numerator. For example, x COULD equal 2/5 or 4/5 or 6/5, or 8/5, etc). In this case x is NOT an even integer
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that x is EITHER an even integer OR a fraction with 3 in the denominator
Statement 2 tells us that x is EITHER an even integer OR a fraction with 5 in the denominator
Since both statements are true, it must be the case that x is an even integer
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent
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Matt@VeritasPrep
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Here's an algebraic approach.
S1 tells us that x = m/3, where m is an even integer
S2 tells us that x = n/5, where n is an even integer
Clearly neither statement alone is sufficient. Together, we have x = m/3 and x = n/5, so we know that m/3 = n/5, or 3n = 5m. This tells us that m is a multiple of 3 and that n is a multiple of 5.
Hence x = m/3 = (multiple of 3)/3 = integer and x = n/5 = (multiple of 5)/5 = integer, and we know that x must be an integer.
S1 tells us that x = m/3, where m is an even integer
S2 tells us that x = n/5, where n is an even integer
Clearly neither statement alone is sufficient. Together, we have x = m/3 and x = n/5, so we know that m/3 = n/5, or 3n = 5m. This tells us that m is a multiple of 3 and that n is a multiple of 5.
Hence x = m/3 = (multiple of 3)/3 = integer and x = n/5 = (multiple of 5)/5 = integer, and we know that x must be an integer.
