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If a and b are integers, is b even? (1) 3a + 4b is even (2) 3a + 5b is even

(1) 3a + 4b is even --> if \(a\) is even, then it's not necessary for \(b\) to be even, may be even or odd. Not sufficient. (2) 3a + 5b is even --> if \(a\) is even, then \(b\) is even too, but if \(a\) is odd, then \(b\) is odd too. Not sufficient.

IMO C. Let me try to explain my approach for this one. Addition of 2 numbers will be even when both of them are either even or odd. Now lets check option 1. 3a + 4b is even. 3a can be either even or odd depending value of a. 4b will be even irrespective b being even or odd. Hence, 3a should be even if the sum has to be even. But, we cannot confirm whether b is odd or even with this statement. Now lets check option 2. 3a + 5b is even. 3a can be either even or odd depending value of a. 5b can be either even or odd depending value of a. Hence, a and b can be either even or odd. This also not sufficient to conclude whether b is even or off. Combing these 2 statements. Both 3a + 4b and 3a + 5b are even. Then as per first statement 3a should be even. Then 4b and 5b should be even if the sum has to be even in both the cases. Then b should be even. Hence, C is the answer. Please let us know the OA and explanation if I am wrong or ambiguous.
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If a and b are integers, is b even? (1) 3a + 4b is even (2) 3a + 5b is even

If a sum is even, then both numbers are even or both numbers are odd.

Statement 1: 4b is even, and 3a + 4b is even, which means 3a is even and hence it tells us a is even. This is insufficient.

Statement 2: 3a + 5b is even, this means that either both a and b are even or both a and b are odd, since either way the sum will be even. But that is insufficient too.

Combining both statements, we know that a is even, which means for the second statement to be valid, b also has to be even.

If a and b are integers, is b even? (1) 3a + 4b is even (2) 3a + 5b is even

(1) 3a + 4b is even --> if \(a\) is even, then it's not necessary for \(b\) to be even, may be even or odd. Not sufficient. (2) 3a + 5b is even --> if \(a\) is even, then \(b\) is even too, but if \(a\) is odd, then \(b\) is odd too. Not sufficient.

If a and b are integers, is b even? (1) 3a + 4b is even (2) 3a + 5b is even

(1) 3a + 4b is even --> if \(a\) is even, then it's not necessary for \(b\) to be even, may be even or odd. Not sufficient. (2) 3a + 5b is even --> if \(a\) is even, then \(b\) is even too, but if \(a\) is odd, then \(b\) is odd too. Not sufficient.

Hi Just a small doubt, can we consider b to be zero. Is zero treated as an even integer. Thx

Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

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Re: If a and b are integers, is b even? [#permalink]

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