If a and b are positive integers such that a < b, is b even?
(1) b/2 - a/2 is an integer.
(2) 3b/4 - a/2 is an integer.
OA B
Experts please tell me why A is not sufficient
Manhattan
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The answer is B
Statement 1 - It states (b-a)/2 is integer which means that b-a is even and b and a can be odd together or even together
(7-1) both odd
(6-2) both even
Insufficient
Statement 2 - It states that (3b-2a)/4 is an integer which means that 3b-2a is even.
a can be odd. But 2a is even and if 3b-2a needs to be an even integer, 3b needs to be even. Therefore b is even.
Sufficient.
Regards
Anup
Statement 1 - It states (b-a)/2 is integer which means that b-a is even and b and a can be odd together or even together
(7-1) both odd
(6-2) both even
Insufficient
Statement 2 - It states that (3b-2a)/4 is an integer which means that 3b-2a is even.
a can be odd. But 2a is even and if 3b-2a needs to be an even integer, 3b needs to be even. Therefore b is even.
Sufficient.
Regards
Anup
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Target question: Is b even?das.ashmita wrote:If a and b are positive integers such that a < b, is b even?
(1) b/2 - a/2 is an integer.
(2) 3b/4 - a/2 is an integer.
Statement 1: b/2 - a/2 is an integer
We can combine the fractions to get (b-a)/2 is an integer.
If (b-a)/2 is an integer, then b-a must be even.
So, statement 1 is really just telling us that b-a is even. There are several pairs of values that satisfy this condition. Here are two:
case a: a=3 and b=5, in which case b is not even
case b: a=2 and b=6, in which case b is even
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: 3b/4 - a/2 is an integer
We can combine the fractions to get (3b-2a)/4 is an integer.
If (3b-2a)/4 is an integer, then 3b-2a must be divisible by 4.
If 3b-2a is divisible by 4, then 3b-2a must be even
Well, we know that 2a will be even for all integer values of a
So, if 3b - 2a is even then 3b must also be even.
If 3b is even, then b must be even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer = B
Cheers,
Brent