Problem Solving

abc

Answer "A"

for ab^2/c to be even, ab^2 must be even
=>either of a or b must be even
II is irrelevant as -ve sign won't effect the odd/even nature
III is wrong as c won't have any effect on the result

mathbyvemuri wrote:Answer "A"

for ab^2/c to be even, ab^2 must be even
=>either of a or b must be even
II is irrelevant as -ve sign won't effect the odd/even nature
III is wrong as c won't have any effect on the result

6/2 and 6/3 will have different results (Odd/Even)
Why doesn't denominator matter?

vatish wrote: 6/2 and 6/3 will have different results (Odd/Even)
Why doesn't denominator matter?

ab^2/c is even => ab^2 is product of c and even number
=> ab^2 is even irrespective of nature of 'c' as ab^2 already has got an even factor.
That was my intention and not that denominator won't effect odd/even nature in general. As you rightly said, it effects indeed. But in the given problem, c won't have any effect.Had it given that the result is odd, then 'c' would have got some effect.

vatish wrote:6/2 and 6/3 will have different results (Odd/Even)
Why doesn't denominator matter?

Hi, those will indeed have different results, but since 6/2 doesn't satisfy the condition that ab^2/c is even, we can ignore that possibility.

Remember, we take the equation in the question stem as a given; accepting that the equation is true, we now check to see which statement MUST also be true.

Let's go through the statements once more:

I) ab is even.

From odd/even rules, we know that an odd number divided by an integer NEVER produces an even integer. Accordingly, the numerator must be even. For ab^2 to be even, either a or b^2 must be even. Since b is an integer (given by the question stem), b^2 will have the same oddness/evenness as b; accordingly, either a or b MUST be even. Finally, as long as one of them is even, ab will always be even: MUST be true, eliminate all choices missing (I) - (B) is gone.

II) ab>0

We know that ab^2/c is positive. For that to be true, the numerator and the denominator must have the same sign. However, they could both be either positive OR negative. Accordingly, ab^2 could be either positive or negative, depending on c. ab>0 COULD be true, but it is NOT a must be true; eliminate all choices that include (II) - C and E are gone.

Sadly, we still have 2 choices, so we have to test (III).

III) c is even

As long as the numerator is even, c could be either even or odd and still give us a positive even result. Picking numbers shows this to be true:

ab^2 = 8, c = 2
8/2 is a positive even integer, so we've followed the rules.

ab^2 = 6, c = 3
6/3 is a positive even integer, so we've followed the rules.

(Note: we actually only needed to check to see if c could be odd, since our goal on roman numeral MUST BE TRUE questions is to show that a statement COULD BE FALSE, allowing us to eliminate it.)

Since c COULD be odd, (III) is NOT a must be true; eliminate (D) and choose (A)!