AMC 10 Math Olympiad USA Math Olympiad

Algebraic Equation | AMC-10A, 2001 | Problem 10

Try this beautiful problem from algebra, based on algebraic equations from AMC-10A, 2001. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra based on Algebraic Equation.

Algebraic Equation – AMC-10A, 2001- Problem 10

If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is

  • \(5\)
  • \(20\)
  • \(22\)
  • \(25\)
  • \(36\)

Key Concepts




Check the Answer

Answer: \(22\)

AMC-10A (2001) Problem 10

Pre College Mathematics

Try with Hints

The given equations are $xy=24$ and $xz=48$.we have to find out \(x+y+z\)

Now using two relations we can write \(\frac{xy}{xz}=\frac{24}{48}\)\(\Rightarrow 2y=z\)

Can you now finish the problem ……….

Now if we substitute the value $z = 2y$ into the equation $yz = 72$ then we will get \(2y^2=72\) \( \Rightarrow y=6\) and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$.

Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)

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