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

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

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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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

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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