agni_h1 wrote:If x^2 + 5y = 49 . Is y an integer ?

(1) 1<x<4
(2) x^2 is an integer

I selected C. But the correct answer is E.

Please help.

One way to solve this question is by counter-example.

Statement 1: 1<x<4
Here, we can see that if x=2, then y is an integer. So, when x=2 the answer to the target question is YES.

At this point, we can see that most other values of x (other than x=3) will mean that y is not an integer. However, before we choose a value for x, let's take a quick look at statement 2: x^2 is an integer.

Notice that the first value for x we chose (x=2) is an integer when squared. So, this value of x satisfies statements 1 and 2.
If we can find another value of x that satisfies both statements and that value of x is such that y is not an integer (i.e., the answer to the target question is NO), then we can solve this question very quickly.

Notice that x=root3 satisfies both statements 1 and 2.

So, for statement 1, if x = root3, then y is not an integer.
So statement 1 is INSUFFICIENT

Here's the bonus of choosing numbers that satisfy both statements.

Statement 2: x^2 is an integer
x=2 means that y is an integer.
x=root3 means that y is not an integer.
So statement 2 is INSUFFICIENT

Since we used the same two x values for both statements, we know that the statements combined are
INSUFFICIENT, which means the answer is E

agni_h1 wrote:Thanks. I did think about picking numbers with root. Just another small query if you don't mind.

To prove that statement 1 alone is Insufficient I considered values like 1.5 and 2.5 because it is not mentioned anywhere that x is an integer. Is that the right approach ? These numbers do not satisfy the equation when we take statement 2 into consideration.So, I had assumed C to be the answer.I had not considered picking roots.

Our goal when picking numbers is to find numbers that satisfy a given condition but result in different answers to the target question.

For example, consider the relatively simple question::
If x is greater than 0, is x a prime number?
(1) x is odd

To show that statement 1 is not sufficient by counter-example, we need to find values of x that satisfy statement 1, but yield different answers to the target question.

So, given the condition in statement 1, x could equal 3 in which case the answer to the target question is YES (x is prime)

Now we need to find a value of x that yields an answer of NO.

Well, given the condition in statement 1, x could equal 15 in which case the answer to the target question is NO (x is not prime)

Since statement 1 yields two conflicting answers to the target question, it is not sufficient.

Now let's examine your approach.

While x=1.5 and x=2.5 do satisfy the condition in statement 1 (1<x<4 ), these two values of x don't yield different answers to the target question.

If x=1.5, then y is not an integer
If x=2.5, then y is not an integer
At this point, we cannot be certain that statement 1 is insufficient, since these two values of x do not yield conflicting answers to the target question.