Mixture Problems - College Algebra

1. A laboratory technician needs to create 400 mL of a 35% alcohol solution by mixing pure alcohol (100% alcohol) with a 20% alcohol solution. However, the technician accidentally uses a 15% alcohol solution instead of the 20% solution. To correct this and still achieve 400 mL of 35% alcohol solution, how much additional pure alcohol must be added to the incorrect mixture?
   Answer: 25 mL First, find what the correct mixture should have been: Let $x$ = mL of pure alcohol and $y$ = mL of 20% solution. We have $x + y = 400$ and $1.00x + 0.20y = 0.35(400) = 140$. From the first equation: $y = 400 - x$. Substituting: $x + 0.20(400 - x) = 140$, so $x + 80 - 0.20x = 140$, giving $0.80x = 60$, thus $x = 75$ mL pure alcohol and $y = 325$ mL of 20% solution. Instead, the technician used 75 mL pure alcohol and 325 mL of 15% solution, creating: $75(1.00) + 325(0.15) = 75 + 48.75 = 123.75$ mL of pure alcohol in 400 mL total, giving $\frac{123.75}{400} = 30.9375\%$ concentration. To get 35% in 400 mL, we need 140 mL of pure alcohol total. Currently have 123.75 mL, so need $140 - 123.75 = 16.25$ mL more. But this changes the total volume. Let $z$ = additional pure alcohol added. New total volume is $400 + z$, pure alcohol amount is $123.75 + z$. For 35% concentration: $\frac{123.75 + z}{400 + z} = 0.35$. Solving: $123.75 + z = 0.35(400 + z) = 140 + 0.35z$, so $123.75 + z = 140 + 0.35z$, giving $0.65z = 16.25$, thus $z = 25$ mL.
2. A paint store mixes two paints: Paint X contains 40% pigment and Paint Y contains 10% pigment. A customer orders 60 gallons of paint with 25% pigment content. After mixing, the store realizes they made an error and actually created a 28% pigment blend. If they used the correct total volume of 60 gallons, how many more gallons of Paint X did they use than intended?
   Answer: 6 gallons First find the intended mixture for 25% pigment: Let $x$ = gallons of Paint X and $y$ = gallons of Paint Y. We have $x + y = 60$ and $0.40x + 0.10y = 0.25(60) = 15$. From the first equation: $y = 60 - x$. Substituting: $0.40x + 0.10(60 - x) = 15$, so $0.40x + 6 - 0.10x = 15$, giving $0.30x = 9$, thus $x = 30$ gallons of Paint X and $y = 30$ gallons of Paint Y. Now find the actual mixture that produced 28% pigment: Let $a$ = gallons of Paint X actually used and $b$ = gallons of Paint Y actually used. We have $a + b = 60$ and $0.40a + 0.10b = 0.28(60) = 16.8$. From the first equation: $b = 60 - a$. Substituting: $0.40a + 0.10(60 - a) = 16.8$, so $0.40a + 6 - 0.10a = 16.8$, giving $0.30a = 10.8$, thus $a = 36$ gallons. The difference is $36 - 30 = 6$ gallons more Paint X than intended.
3. A chemistry student mixes Solution A (8% acid) with Solution B (20% acid) to create 150 mL of a 14% acid solution. The student then realizes that an additional 50 mL of 14% solution is needed for the experiment. If the student maintains the same ratio of Solution A to Solution B, what is the total amount of Solution A that will be used for both the original and additional mixtures combined?
   Answer: 100 mL First, find the amounts needed for the original 150 mL mixture: Let $x$ = mL of Solution A and $y$ = mL of Solution B. We have $x + y = 150$ and $0.08x + 0.20y = 0.14(150) = 21$. From the first equation: $y = 150 - x$. Substituting: $0.08x + 0.20(150 - x) = 21$, so $0.08x + 30 - 0.20x = 21$, giving $-0.12x = -9$, thus $x = 75$ mL of Solution A and $y = 75$ mL of Solution B. The ratio is $\frac{x}{y} = \frac{75}{75} = 1:1$. For the additional 50 mL of 14% solution maintaining the same 1:1 ratio: 25 mL of Solution A and 25 mL of Solution B. Verification: $0.08(25) + 0.20(25) = 2 + 5 = 7$ mL pure acid in 50 mL total = 14% ✓. Total Solution A used: $75 + 25 = 100$ mL.
4. A pharmacist needs to prepare 200 mL of a 12% saline solution by mixing a 20% solution with distilled water (0% saline). Due to measurement constraints, the pharmacist can only measure in increments of 5 mL. What is the minimum amount of the 20% solution needed to achieve a concentration as close as possible to 12%?
   Answer: 120 mL Let $x$ = mL of 20% solution. Then $(200-x)$ = mL of water. The equation is: $0.20x + 0(200-x) = 0.12(200)$, so $0.20x = 24$, giving $x = 120$ mL exactly. Since 120 is divisible by 5, this is achievable with the measurement constraint. Checking nearby values in 5 mL increments: 115 mL gives concentration $\frac{0.20(115)}{200} = \frac{23}{200} = 11.5\%$, and 125 mL gives concentration $\frac{0.20(125)}{200} = \frac{25}{200} = 12.5\%$. The differences from 12% are: 115 mL gives $|11.5 - 12| = 0.5\%$, 120 mL gives $|12 - 12| = 0\%$, 125 mL gives $|12.5 - 12| = 0.5\%$. Therefore 120 mL is optimal and achievable.

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