Given a and b are integers, is b even?

A)3a+4b is even
B)3a+5b is even



statement 1)

odd+odd=even and even+even=even

if 3a=odd then a=odd
if 4b=odd then b cannot be even or odd

if 3a=even then a=even
if 4b=even then b=even

b is even. sufficient

statement 2)

same logic applies here

if 3a=even then a=even
if 5b=even then b=even

if 3a=odd then a=odd
if 5b=odd then b=odd

b can be either even or odd. insufficient

A

pls confirm

Shah,
No. Why would say A is sufficient.

All that the first statement tells you is that a must be sufficient.

whatever the value of b, odd or even, 4b is even.

3a+4b is even => 3a MUST be even => a must be even.

A is clearly INSUFFICIENT.

Will post the OA after a few more replies.

I’d go with C
1. 3a + 4b = Even
In this case, since 4 is even, it must be that
Even + Even = Even
“a” here must be even but “b” is unknown. Because it is being multiplied by 4, “b” is hidden.
2. 3a + 5b = Even
In this case, it can be both cases
Odd + Odd = Even
Even + Even = Even
Thus, using the same logic as S1, “b” is hidden.
When we combine, we can write them as following, using 2k and 2m as the even number
3a + 4b = 2k
3a + 5b = 2m
Adding the two, we get
6a + 9b = 2k+2m
The term 6a will be even, no matter what, thus the term 9b must be even as well, thus b must be even
C.

IMO C

1)3a+4b is even
even +even=even
odd+odd =even

but since "4b" has 4, it is even and so is 3a. but b can be odd or even
not SUFF

2)3a+5b is even,
again, 5b can be even or odd. and so b can be even or odd.
NOT SUFF

combining, here is something you must notice. From 1, we know that 3a is even.
so in 2, 5b must be even
so b must be even
SUFF
hence C

answer is C

Thanks. The OA was given as E, but am sure it is C.

Just wanted to make sure I did not miss anything.