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

(1) x = y-3. Clearly insufficient. Just notice from this statement that x<y.
(2) x and y are prime numbers. Also insufficient.

(1)+(2) As x<y and both are primes, y is some prime more than 2, so odd. Now, x={odd}-3={even} and as the only even prime is 2 then x=2 and y=5. Sufficient.

Answer: C.

Hope it's clear.
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Target question: What is the value of (x+y)²?

Given: x and y are positive integers

Statement 1: x = y - 3
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 1 and y = 4, in which case (x+y)² = (1+4)² = 5² = 25
Case b: x = 2 and y = 5, in which case (x+y)² = (2+5)² = 7² = 49
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.gmatprepnow.com/articles/dat ... lug-values

Statement 2: x and y are prime numbers
This statement doesn't FEEL sufficient either, so I'll TEST some values.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 2 and y = 2, in which case (x+y)² = (2+2)² = 4² = 16
Case b: x = 2 and y = 3, in which case (x+y)² = (2+3)² = 5² = 25
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x = y - 3
In other words, x + 3 = y
Or we can say that x + ODD = y
This means that one of the two values (x or y) is EVEN and one is ODD

Statement 2 tells us that x and y are prime numbers
If one of the values is EVEN and also PRIME, then one value must equal 2.
Since x is smaller than y, we can conclude that x = 2, which means y = 5
Now that we know the values of x and y, we can see that (x+y)² = (2+5)² = 7² = 49

Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
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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.
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I did understand the above but any other methods to solve this?

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