Given:
A 100 oz. of 60% iodine solution
We need to add the water to get a 55% solution
Let the number of ounces of water = x
There is an inverse relationship between the percentage of solution and the total amount
So, after adding x ounces, there is (x+100) oz. of the solution
We can write the following ratio:
![(x+100)/(100)=(60)/(55)](https://img.qammunity.org/2023/formulas/mathematics/college/t1bwn90tnaop2ps16f467hj4shiw7v3xax.png)
Solve for x, use the cross product:
![\begin{gathered} x+100=(60*100)/(55) \\ \\ x=(60*100)/(55)-100=9.0909 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/25vl7rlc85dxycyth95cpsx4s4cc84lm93.png)
Rounding to the nearest hundredth
So, the answer will be 9.1 oz.