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**

algebra

Equation

sum

## 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 \((x+y+z)\)=\(4+6+12\)=\(22\)

## Other useful links

- https://www.cheenta.com/integers-and-divisors-isi-b-stat-entrance-tomato-98/
- https://www.youtube.com/watch?v=9VauLfKar70