Bunuel has already mentioned some solid approaches above. Let me add a couple more.

Are X > 0 & Y > 0

1) 2X - 2Y = 1

2) (X/Y) > 1

(BTW, if the second statement were (X/Y) > 0, then the answer would be E)

Ques: Are X and Y positive?

Statement (I): X - Y = 1/2. But this could be true for X and Y both positive e.g. X = 2.5, Y = 2, for X positive (=0.5), and Y equal to 0, for X = 0 and Y = -0.5, for X and Y both negative e.g. X = -2, Y = -2.5. Hence not sufficient.

Statement (II): X/Y > 1 means X and Y are either both positive or both negative. It also means that absolute value of X is greater than absolute value of Y. e.g. X = 2.5, Y = 2; or X = -2.5, Y = -2. This alone is also not sufficient.

Taking both together, if X and Y are both positive and absolute value of X is greater than absolute value of Y, X - Y can be 1/2 e.g. X = 2.5 and Y = 2

But if X and Y are both negative and absolute value of X is greater than absolute value of Y, X - Y will be negative. e.g. -2.5 -(-2) = -0.5.

Hence, if we combine the two statements, we are left with only X and Y both positive solution. Answer (C).

2. Is a(K) factor of b(M). K & M are exponents of a, b

1) A is a factor of b

2) K < M (Here K = M or K < M doesn't really matter)

Ques: Is \(b^M\) divisible by \(a^K\)?

Let us just think for a moment here. When will \(b^M\) be divisible by \(a^K\). When a is a factor of b. In other words, when all the prime factors of a are in b as well so that they can get canceled e.g. 12 is divisible by which numbers? By 2, 3, 4, 6 etc i.e. numbers whose prime factors are in 12 as well and the power of these prime factors is less than or equal to that found in 12.

12 = 2^2 x 3. So a number having at the most two 2s and one 3 will divide 12. That is why 4 divides 12 but 9 doesn't.

Also K should be less than or equal to M to be certain that \(b^M\) is divisible by \(a^K\). Let me explain using an example: If a = 4 and b = 12, \(12^5\) is divisible by \(4^5\) but not by \(4^6\) because 12 has one 4 and \(12^5\) has five 4s so it is divisible by \(4^5\). When I try to divide it by \(4^6\), I need six 4s but I don't have them so it is not divisible. (Check theory of divisibility for further clarification.)

Hence both statements together are needed to solve the question. Answer (C).

In the third question, putting X = (A + B) and then checking if I can split this to be equal to F(A) + F(B) is pretty much what I would do. I would try the functions where terms are added or multiplied first because these are generally more symmetrical.

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Karishma

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