If x and y are positive, what is the value of x?

If x and y are positive, what is the value of x?

Notice that we are not told that \(x\) and \(y\) are integers only, we are just told that they are both positive.

(1) 200% of x equals to 400% of y --> \(\frac{200}{100}*x=\frac{400}{100}*y\) --> \(x=2y\). Not sufficient to get the single numerical value of \(x\).

(2) xy is the square of a positive integer --> \(xy=n^2\), for some positive integer \(n\). Not sufficient to get the single numerical value of \(x\).

(1)+(2) Since from (1) \(x=2y\) then from (2) \(x*\frac{x}{2}=n^2\) --> \(x^2=2n^2\) --> the value of \(x\) (x^2) is determined by the value of integer \(n\), so we still cannot get the single numerical value of \(x\). For example: if \(n=1\) then \(x=\sqrt{2}\) but if \(n=2\) then \(x=2\sqrt{2}\). Not sufficient.

Answer: E.

Hope it's clear.




Since we are not told that \(x\) and \(y\) are integers only, then from \(x=2y\) we cannot say whether \(x\) even:

\(x\) will be even if \(y=integer\);
\(x\) will be odd if \(y=\frac{odd}{2}\), for example if \(y=\frac{3}{2}\) then \(x=3=odd\);
\(x\) may not be an integer at all, for example if \(y=\frac{1}{3}\) then \(x=\frac{2}{3}\neq{integer}\).

In fact if we knew that \(x=even\) then \(x^2=2n^2\), from (1)+(2), would have only one integer solution \(x=0\) and \(n=0\), but this would contradict the given fact that \(x\) is positive.

Hope it's clear.
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(1)
200/100 * x = 400/100 * y
2x = 4y
X = 2y
=> X is even

(2)
Xy = 1,4,9,16,…

Not Sufficient.

(1) + (2)

Suppose:

2y^2 = 16
Y = 2root(2)
X = 4root(2)
Or
2y^2 = 4
Y = root(2)
X = 2root(2)

Not Sufficient.

So Answer is E.
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Hello Bunuel,

Thanks for this wonderful explanation!

However I'm still not able to figure out the meaning of the following statement



Can you please provide more clarity and let know which particular number property you are using to deduce the above statement?

Which particular number property - I think Bunuel is saying that 0 is neither positive nor negative.
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Sure. We know that \(n\) is a positive integer. Now, if \(x\) is an even integer, then \(x^2=2n^2\) should be true for some integers \(n\) and \(x\), but it's true only for one integer solution \(x=n=0\), which cannot be valid, since we are given that \(x\) is a positive number (0 is neither positive nor negative).

Hope it's clear.
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Hi Bunuel,

Thanks for the nice explanation . I would like to modify the question a bit & would like to present a twist to the same question.

If the new statement reads "If x and y are positive integers, what is the value of x" rather than "If x and y are positive, what is the value of x"
Now what will be the answer.

As per me the answer should be C because i believe no such value will exist.

Kindly enlighten us all.

In this case, for (1)+(2) we would have the same equation: \(x=2n^2\), which has only one integer solution for \(x\) and \(n\): \(x=n=0\) but since we are told that \(x\) is a positive integer then this solution is no good.

But we won't see such question on the real test where no value can satisfy the statements. So, the question in this case would be flawed.
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If x and y are positive, what is the value of x?

(1) 200% of x equals to 400% of y.-NS as we dont know the values of x,y

(2) xy is the square of a positive integer.--NS as we dont know the values of x,y and the new variable used
1& 2 together also will not help

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Since we have 2 variables (x and y) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

We have \(2x = 4y\) or \(x = 2y\) from condition 1).
Then \(xy = 2y \cdot y = 2y^2\).


If \(x = 2\sqrt{2}, y = \sqrt{2}\), then \(xy = 4 = 2^2\), which is a square of a positive integer \(2\).
If \(x = 6\sqrt{2}, y = 3\sqrt{2}\), then \(xy = 36 = 6^2\), which is a square of a positive integer \(6\).

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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