If x and y are positive integers, what is the value of (x+y)

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are positive integers, what is the value of (x+y)^2

(1) x = y -3
(2) x and y are prime numbers.

There are 2 variables (x,y) and 2 equations are given by the conditions, so there is high chance (C) will become the answer.
Looking at the conditions together, only x=2, y=5 works, as only 2 works for even prime numbers, and even+odd=odd.
Hence, the answer becomes (C).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

I did understand the above but any other methods to solve this?

If x and y are positive integers, what is the value of (x+y)^2?
What is the value of \(x^2 + 2xy + y^2\)?

(1) x = y -3

Clearly we can get different values. INSUFFICIENT.

(2) x and y are prime numbers.

Again, clearly INSUFFICIENT.

(1&2) If x = y - 3 and both x and y are prime, then we can conclude x = 2 and y = 5. SUFFICIENT.

(1) \(x-y=-3;\) Lots values will give the result, such as; 2-5=-3, 4-7=-3, So, x (first value) and y (the second) value are different in the two cases. Insufficient.

(2) There are lots of primes numbers; Insufficient.

Considering both:
We know that \(Even -Odd=Odd\), There is only one even prime number and that is 2. So, x=2 and y=5, Sufficient.

The answer is \(C.\)

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