is x > y?

[RPurewal]

courtesy of a student

Is x > y?

(1) x^2 > y
(2) √x < y

[RPurewal]

statement (1)
this doesn't tell you anything, because (a) x^2 could be much greater than x, and also because (b) x^2 is positive even if x is negative.
illustrations:
if x = 5 and y = 10, then x is not greater than y, even though x^2 is.
if x = -2 and y = 0, then x is not greater than y, even though x^2 is (and so is |x|, in this case).
...but of course x and x^2 could also be greater than y; consider something like x = 100 and y = 0.
insufficient

statement (2)
can't tell from this one, either, because √x is smaller than x. so you could have both √x and x smaller than y (as in x = 1, y = 10), or you could have √x < y but x > y (as in x = 5, y = 10).
insufficient

(together)
it's difficult to combine these equations in any meaningful way, but you now have the combination √x < y < x^2. so plug in numbers, and watch what happens. don't just choose numbers at random, though; choose one set of numbers with y close to the low end (= √x) and one set of numbers with y close to the high end (= x^2).
first set: x = 9, y = 4 (this works because 3 < 4 < 81). here x > y.
second set: x = 9, y = 80 (this works because 3 < 80 < 81). here x < y.
insufficient

answer = e

[netcaesar]

What is the answer, C or E?

In the Mgmat exam the official answer is the following:

1) AND (2) SUFFICIENT: Let’s start with statement 1 and add the constraints of statement 2. From statement 1, we see that x has to be positive since we are taking the square root of x. There is no point in testing negative values for y since a positive value for x against a negative y will always yield a yes to the question. Lastly, we should consider x values between 0 and 1 and greater than 1 because proper fractions behave different than integers with regard to exponents. When we try to come up with x and y values that fit both conditions, we must adjust the two variables so that x is always greater than y.

Logically it also makes sense that if the cube and the square root of a number are both greater than another number than the number itself must be greater than that other number.

The correct answer is C.

Then, what is the answer, C or E?

Regards.

[experts]

Is x > y?

(1) x^2 > y
(2) sqrt(x) < y

Explanation: I will try to solve using plugging
STEP 1):
take the statement (1) alone :
1) put x = 4 and y = 9 then x^2 > y as 16 > 9 and x < y : FALSE
2) put x = 9 and y = 4 then x^2 > y as 81 > 4 and x > y : TRUE

thus (1) alone is insufficient. so A and D are eliminated

STEP 2):
take the statement (2) alone :
1) put x = 4 and y = 9 then sqrt(x) < y as 4 < 9 and x < y : FALSE
2) put x = 9 and y = 4 then sqrt(x) < y as 3 < 4 and x > y : TRUE

thus (2) alone is insufficient. so eliminate B also.

STEP 3):
Now take (1) and (2) together then
=> sqrt(x) < y < x^2

1) put x = 4 and y = 9 then sqrt(x) < y < x^2 as 2 < 9 < 16 and x < y : FALSE
2) put x = 9 and y = 4 then sqrt(x) < y < x^2 as 3 < 4 < 81 and x > y : TRUE

thus (1) and (2) together are not sufficient to give the answer. So C is eliminated.

THUS ANSWER IS E

[JonathanSchneider]

netcaesar, I believe you have copied that explanation from a slightly different problem. Note that you mention a "cube," which is not an element of this problem.

It is correct to say that Statement 2 tells us that x is positive, as we cannot take the square root of a negative number. However, even knowing that x is positive, we cannot be sure of the relationship between x and y. All that we know with both statements combined is that (root x) < y < x^2. However, we do not know where y will fall between (root x) and x^2, whether it will be closer to (root x) or x^2, etc.