A technician needs to prepare a disinfectant by mixing a 70%70\% isopropyl alcohol solution with some 50%50\% solution to obtain 65%65\% alcohol. If the technician uses 500mL500\operatorname{\mathrm{mL}} of the 50%50\% solution, how many mLmL of the 70%70\% solution must be added?

1500 mL ⁡ 1500\operatorname{\mathrm{m}\mathrm{L}} 1500 mL

A technician needs to prepare a disinfectant by mixing a 70 % 70\% isopropyl alcohol solution with some 50 % 50\% solution to obtain 65 % 65\% alcohol. If the technician uses 500 m L 500\operatorname{\mathrm{mL}} of the 50 % 50\% solution, how many m L mL of the 70 % 70\% solution must be added?

In this problem, a technician needs to prepare a disinfectant by mixing 70% isopropyl alcohol solution with some 50% solution in order to obtain 65% alcohol. So if the technician uses 500 milliliters of the 50% solution, how many milliliters of the 70% solution must be added? So to better understand this problem, let's draw a picture. Let's say this rectangle represents the total amount of solution, and we know that we want the solution to be 65%. But we don't know how much of this solution that we need, So we'll go ahead and name this a variable. Let's call it x. Now we know that this total amount of solution is gonna be made up of two parts. We're going to have part that is 50%, and we already know how much of this we need. We need 500 milliliters. And the other part is going to be 70%, but we don't know how much of this one we need. So let's go ahead and name this a variable and we'll call it y. Now to set up this equation, we're going to take our total amount and set it equal to the sum of our smaller amounts. So to get our total amount here, we're going to multiply our percent times our amount x. So we have our percent written as a decimal is 0.65 times x, and this is going to be equal to our first part, which is 50% written as a decimal of 0.5 times its amount of 500, and we're going to add that to our second part of 70% as a decimal, 0.7 times our second amount of y. Now that we have our initial equation, we're ready to start solving. But first, we need to get it in terms of one variable. So what's another way that we could write a relationship between our x and y quantity? Both of these represent amounts in milliliters. And we know that to get this total amount x, we could add our 500 milliliters to our y amount of milliliters. Now that we have this relationship, we can go ahead and take this part of the equation and plug it in 4x into our main equation. So doing that, we get 0.65 times 500 plus y is equal to 0.5 times 500, reduces to two fifty, plus 0.7 y. Now that we have one variable, we're ready to solve. So we're going to simplify by distributing this in and we have 0.65 times 500 is three twenty five plus 0.65 y is equal to two fifty plus 0.7 y. Now we're going to collect all variable terms to one side and all other terms to the other side. So let's subtract 0.65y from both sides and let's also subtract two fifty from both sides. And that leaves us with three twenty five minus two fifty is 75, and 0.7 minus 0.65 is 0.05 y. Lastly, we isolate by dividing both sides by 0.05 and we get that y is equal to 1,500. Now, what did y represent here? It represented the amount of 70% solution that we needed, which is what we're looking for. So we can state our answer as needing 1,500 milliliters of 70 isopropyl alcohol.

A technician needs to prepare a disinfectant by mixing a 70 % 70\% isopropyl alcohol solution with some 50 % 50\% solution to obtain 65 % 65\% alcohol. If the technician uses 500 m L ⁡ 500\operatorname{\mathrm{mL}} of the 50 % 50\% solution, how many m L mL of the 70 % 70\% solution must be added?

1500 mL ⁡ 1500\operatorname{\mathrm{m}\mathrm{L}} 1500 mL

600 mL ⁡ 600\operatorname{\mathrm{m}\mathrm{L}} 600 mL

667 mL ⁡ 667\operatorname{\mathrm{m}\mathrm{L}} 667 mL

67 mL ⁡ 67\operatorname{\mathrm{m}\mathrm{L}} 67 mL

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Multiple Choice

A technician needs to prepare a disinfectant by mixing a $70 percent sign$ isopropyl alcohol solution with some $50 percent sign$ solution to obtain $65 percent sign$ alcohol. If the technician uses $500 m L$ of the $50 percent sign$ solution, how many $m L$ of the $70 percent sign$ solution must be added?

$1500\operatorname{\mathrm{m}\mathrm{L}}$

$600\operatorname{\mathrm{m}\mathrm{L}}$

$667\operatorname{\mathrm{m}\mathrm{L}}$

$67\operatorname{\mathrm{m}\mathrm{L}}$

Verified step by step guidance

Define the variable for the unknown volume of the 70% isopropyl alcohol solution to be added. Let this volume be $x$ mL.

Write an expression for the total amount of pure alcohol from each solution. From the 70% solution, the pure alcohol amount is \$0.70x$, and from the 50% solution (500 mL), it is $0.50 \times 500$ mL.

Write an expression for the total volume of the final mixture, which is $x + 500$ mL.

Set up an equation based on the desired concentration of 65% alcohol in the final mixture: the total pure alcohol divided by the total volume equals 0.65, so $\frac{0.70x + 0.50 \times 500}{x + 500} = 0.65$.

Solve the equation for $x$ by multiplying both sides by $(x + 500)$, expanding, combining like terms, and isolating $x$ to find the volume of the 70% solution needed.

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