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by sanjoy18 » Wed Sep 11, 2013 10:55 pm
(1): 5z = 2, z = 2/5; 5z = 10, z = 2... Not sufficient
(2): 3z = 2, z = 2/3; 3z = 6, z = 2... Not sufficient

Combining: Subtract the two: 2z = even - even = even

Subtracting again: 3z - 2z = even - even so z is even ... Ans. C
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by ganeshrkamath » Wed Sep 11, 2013 11:37 pm
vipulgoyal wrote:Is z even number ??
1. 5z is an even number
2. 3z is an even number

C
1. z = 2/5, z = 2
Not sufficient.

2. z = 2/3, z = 2
Not sufficient.

Both: 5z - 3z = 2z
Since 2z and 3z are both even numbers, z is an even number.

Choose C

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by Brent@GMATPrepNow » Thu Sep 12, 2013 7:12 am
I thought I'd point out that ganeshrkamath and sanjoy18 are both using versions of a nice divisibility rule that says:
If integers A and B are both divisible by k, then A+B is divisible by k, and A-B is divisible by k

In this question, k = 2 (that is, a number is even if it is divisible by 2)
So, if 5z is divisible by 2, and 3z is divisible by 2, then 5z-3z is divisible by 2
In other words, 2z is divisible by 2.

Furthermore, if 3z is divisible by 2, and 2z is divisible by 2, then 3z-2z is divisible by 2
In other words, z is divisible by 2.

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