Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.8 AMC 10/12 Find the answer Italy

Problem:

Three quantities $X = a + 7b$ , $Y = 2a + 5b$ , $Z = 4a + 2b$ are given, where $a$ and $b$ are positive real numbers. Then:

Pick one

Solution

Solution:

The answer is (E). Setting $a = 3$ and $b = 1$ gives $X = 10$ , $Y = 11$ , $Z = 14$ (hence $X < Y < Z$ ), while setting $a = b = 1$ gives $X = 8$ , $Y = 7$ , $Z = 6$ (hence $X > Y > Z$ ). This allows us to conclude that the relation between any two of the three numbers $X, Y, Z$ depends on the particular values of $a$ and $b$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.