Given that integer x>1 and x is not a perfect square, which of the following could be a perfect square?

Can someone clear this up please? Does it mean that it's not necessary in all cases that the number should form a perfect square, but if it does in one instance, it has to be considered?

Could anyone help with the quickest solution here?

First and foremost an important thing to notice here is that the stem says the word could be , which means that the possible answer can be a perfect square or NOT.

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

While plugging in value for A), why have you excluded 4*4 ?

Because 4 itself is precluded from being a possible value of x, since x cannot be a perfect square and of course 2 x 2 = 4, making 4 a perfect square.

Ahh. Damn. Missed that little information. Thanks for the response.

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