Try this beautiful problem from Algebra based on Algebraic Equation.
If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is
algebra
Equation
sum
But try the problem first...
Answer: \(22\)
AMC-10A (2001) Problem 10
Pre College Mathematics
First hint
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 ..........
Second Hint
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 $x=4$.
Final Step
Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
Try this beautiful problem from Algebra based on Algebraic Equation.
If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is
algebra
Equation
sum
But try the problem first...
Answer: \(22\)
AMC-10A (2001) Problem 10
Pre College Mathematics
First hint
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 ..........
Second Hint
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 $x=4$.
Final Step
Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
