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So, Even number is an Integer considered.

a+b means z cant be divisible by 3/5 both.Hence C.
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tonebeeze
Is z even?

(1) 5z is even
(2) 3z is even

z = integers, decimals, fractions, etc.
z = even ?

1) 5z is even
z = \(\frac{2}{5}\)
5z = 2, but z ≠ even.
z = 2
5z = 10, and z = even.
2 different answers. Insufficient.

2) 3z = even
z = \(\frac{2}{3}\)
3z = 2, but z ≠ even
z = 2, 3z = 6
z = even
2 different answers. Insufficient.

1+2)

5z = even and 3z = even
3 and 5 are prime => z cannot be a fraction as it has to satisfy both the equations.
=> z has to be even.
Sufficient. C is the answer.
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tonebeeze
Is z even?

(1) 5z is even
(2) 3z is even


(1) 5z is even
z=2, 5z is even yes
z=2/5 , 5z is even no
Insufficient.
(2) 3z is even
z=2, 3z is even yes
z=2/3 , 3z is even no
Insufficient.

Combining (1) and (2)
since even + even =even
Adding both (1) and (2)
5z+3z= even
or 8z=even
hence z could be odd or even.

hence answer should be E


Need an expert advice to analyse my logic.
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tonebeeze
Is z even?

(1) 5z is even
(2) 3z is even


(1) 5z is even
z=2, 5z is even yes
z=2/5 , 5z is even no
Insufficient.
(2) 3z is even
z=2, 3z is even yes
z=2/3 , 3z is even no
Insufficient.

Combining (1) and (2)
since even + even =even
Adding both (1) and (2)
5z+3z= even
or 8z=even
hence z could be odd or even.

hence answer should be E


Need an expert advice to analyse my logic.

Instead adding you should have subtracted 3z from 5z:

5z - 3z = even -even = even;

2z = even;

z = even/2 = integer.

If z is an integer, then for 3z to be even z must be even too.
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tonebeeze
Is z even?

(1) 5z is even
(2) 3z is even

#1
5z is even
z= 2 or 2/5
insufficient
#2
3z is even
z=2 or 2/3
insufficient
from 1 &2
z=2 or even integer only
IMO C
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tonebeeze
Is z even?

(1) 5z is even

z=2/5; 5z=2; z not even
z=2; 5z=10; z is even
Not Sufficient.

(2) 3z is even

z=2/3; 3z=2; z not even
z=2; 3z=6; z is even
Not Sufficient.

Combining both;
z can't be a fraction with denominator 3 or 5 because it has to satisfy both statements simultaneously.
Thus, z must be an even integer.
Sufficient.

Ans: "C"

So I chose C too, but I wasn't firm about whether z could not be a fraction. Is this strictly trial and error?
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tonebeeze
Is z even?

(1) 5z is even
(2) 3z is even
Remember: z is not given as integer

(1) 5z is even
Even numbers are of type 2, so 5z=2x.....\(z=\frac{2}{5}x\), where x is an integer.
If x= 1, then z=2/5, but if x=5, then z=2
Thus, 5z is even when z is even integer or when z has even numerator and 5 in denominator in its simplest form
Insuff

(2) 3z is even­
Even numbers are of type 2y, so 3z=2y.....\(z=\frac{2}{3}y\), where y is an integer.
If x= 1, then z=2/3, but if x=3, then z=2
Thus, 3z is even when z is even integer or when z has even numerator and 3 in denominator in its simplest form
Insuff

Combined
z is even integer is common in the two statements.
Thus, sufficient

C
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