AMC 8 Math Olympiad USA Math Olympiad

Largest possible value | AMC-10A, 2004 | Problem 15

Try this beautiful problem from Number Theory based on largest possible value from AMC-10A, 2004. You may use sequential hints to solve the problem.

Try this beautiful problem from Number system: largest possible value

Largest Possible Value – AMC-10A, 2004- Problem 15

Given that \( -4 \leq x \leq -2\) and \(2 \leq y \leq 4\), what is the largest possible value of \(\frac{x+y}{2}\)

  • \(\frac {-1}{2}\)
  • \(\frac{1}{6}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)
  • \(\frac{1}{9}\)

Key Concepts

Number system



Check the Answer

Answer: \(\frac{1}{2}\)

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints

The given expression is \(\frac{x+y}{x}=1+\frac{y}{x}\)

Now \(-4 \leq x \leq -2\) and \(2 \leq y \leq 4\) so we can say that \(\frac{y}{x} \leq 0\)

can you finish the problem……..

Therefore, the expression \(1+\frac{y}x\) will be maximized when \(\frac{y}{x}\) is minimized, which occurs when \(|x|\) is the largest and \(|y|\) is the smallest.

can you finish the problem……..

Therefore in the region \((-4,2)\) , \(\frac{x+y}{x}=1-\frac{1}{2}=\frac{1}{2}\)

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