MathRevolution wrote:
[GMAT math practice question]
(Number Properties) \(x\) and \(y\) are positive integers. \(\sqrt{\frac{y}{x}}\) is rounded to the closest integer to \(3\). What is the value of \(x + y\)?
1) \(x\) and \(y\) are relatively prime.
2) \(|x - y| = 50.\)
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Since we have \(2\) variables (\(p\) and \(q\)) 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) together give us that:
Since \(\sqrt{\frac{y}{x}}\) is rounded to the closest integer to \(3\), we have \(2.5 ≤ \sqrt{\frac{y}{x}} < 3.5 or 6.25 ≤ \frac{y}{x} < 12.25\) (by squaring all parts of the equation).
Since \(\frac{y}{x} > 1\), we have \(y > x\) and \(y = x + 50.\)
\(6.25 ≤ \frac{y}{x} < 12.25\)
⇔ \(6.25 ≤ \frac{[x + 50] }{ x} < 12.25\)
⇔ \(6 .25 ≤ 1 +\frac{ 50}{x} < 12.25\) (dividing the middle portion by \(x\))
⇔ \(5.25 ≤ \frac{50}{x} < 11.25\) (subtracting 1)
⇔ \(\frac{1}{11.25}< \frac{x}{50} ≤ \frac{1}{5.25}\) (inversing all portions)
⇔ \(\frac{50}{11.25}< x ≤ \frac{50}{5.25}\) (multiplying by \(50\))
⇔ \(4.444… < x ≤ 9.523\) (simplifying)
⇔ \(5 ≤ x ≤ 9.\) (rounding_
The possible pairs of \((x, y)\) are \((5, 55), (6, 56), (7, 57), (8, 58)\) and \((9, 59)\) since \(y = x + 50\) from condition 2).
When we consider condition 1), we have two pairs of \((x, y)\) which are \(x = 7, y = 57\) and \(x = 9, y = 59.\)
The answer is not unique, and the conditions are not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.
Both conditions 1) and 2) together are not sufficient.
Therefore, E is the answer.
Answer: E
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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