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If x and y are positive integers and x/4 < y < x/2, what is the smalle
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15 Feb 2022, 06:31
1
Carcass wrote:
If x and y are positive integers and \(\frac{x}{4} < y < \frac{x}{2}\), what is the smallest possible value for x?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
\(1\) is the smallest positive integer value of \(y\), which means our inequality becomes: \(\frac{x}{4} < 1 < \frac{x}{2}\).
From here, if we scan the answer choices, we can see that \(x = 3\) (the smallest answer choice) satisfies the given inequality, since we get: \(\frac{3}{4} < 1 < \frac{3}{2}\).
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Re: If x and y are positive integers and x/4 < y < x/2, what is the smalle
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23 Feb 2022, 12:45
Silly me, don't know for what reason I keep multiplying 4 with x to get middle term and wondering why no equation is satisfying. I would have just selected y = 1 and tested different scenarios.