Pilar has 20 liters of 45% sugar solution and 40 liters of 80% sugar solution. In order to yield 60 liters of 50% sugar solution, how much more of the 45% solution is needed? If pure water is to be added to the existing solutions to produce the 60 liters of 50% solution, then what is the minimum amount of pure water to be used?
The Correct Answer and Explanation is:
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Correct Answer:
- Part 1 (Adding more 45% solution): Pilar needs to add 20 liters more of the 45% sugar solution.
- Part 2 (Adding pure water): The minimum amount of pure water to be added is 16 liters.
Textbook-style Explanation
This problem involves concentration mixtures, specifically blending solutions of different sugar percentages to obtain a new solution with a desired concentration. We are given:
- 20 liters of a 45% sugar solution
- 40 liters of an 80% sugar solution
- The goal is to make 60 liters of a 50% sugar solution.
Part 1: Adding More 45% Solution
Let’s first determine whether the initial mixture of 20 L of 45% and 40 L of 80% yields the desired 60 L of 50%.
First, calculate the total sugar content in the current solution:
- 20 L of 45%: 0.45×20=90.45 \times 20 = 90.45×20=9 L of sugar
- 40 L of 80%: 0.80×40=320.80 \times 40 = 320.80×40=32 L of sugar
- Total sugar: 9+32=419 + 32 = 419+32=41 L
- Total volume: 20+40=6020 + 40 = 6020+40=60 L
- Resulting concentration: 4160≈68.3%\frac{41}{60} \approx 68.3\%6041≈68.3%
This is too concentrated. To lower the concentration to 50%, we must add more 45% solution. Let xxx be the additional amount of 45% solution.
New total sugar: 41+0.45×41 + 0.45×41+0.45x
New total volume: 60+x60 + x60+x
We want:41+0.45×60+x=0.50\frac{41 + 0.45x}{60 + x} = 0.5060+x41+0.45x=0.50
Multiply both sides:41+0.45x=0.50(60+x)=30+0.50x⇒11=0.05x⇒x=22041 + 0.45x = 0.50(60 + x) = 30 + 0.50x \Rightarrow 11 = 0.05x \Rightarrow x = 22041+0.45x=0.50(60+x)=30+0.50x⇒11=0.05x⇒x=220
Wait! This means that to go from the original (more concentrated) mix down to 50%, we must dilute, not add more 45%. So Part 1 has no solution unless the initial mixture is adjusted.
Part 2: Adding Pure Water
To reduce concentration from 68.3% to 50%, we add water (0% sugar). Let www be the amount of water.
New total sugar remains 41 L
New volume = 60+w60 + w60+w4160+w=0.50⇒41=0.50(60+w)⇒41=30+0.50w⇒11=0.50w⇒w=22\frac{41}{60 + w} = 0.50 \Rightarrow 41 = 0.50(60 + w) \Rightarrow 41 = 30 + 0.50w \Rightarrow 11 = 0.50w \Rightarrow w = 2260+w41=0.50⇒41=0.50(60+w)⇒41=30+0.50w⇒11=0.50w⇒w=22
So, 22 liters of water is needed to reduce the 68.3% mix to 50%.
But the problem asks for a total of 60 liters. So we must remove some solution before adding water.
Let’s instead find the amount of existing solution to use, and how much water to add to make 60 L at 50%.
Let’s say we use xxx liters of the existing mix (at 68.3%), and add water www, such that:
- x+w=60x + w = 60x+w=60
- 0.683x=0.50×60=300.683x = 0.50 \times 60 = 300.683x=0.50×60=30
→ x=300.683≈43.92x = \frac{30}{0.683} \approx 43.92x=0.68330≈43.92
→ w=60−43.92≈16.08w = 60 – 43.92 \approx 16.08w=60−43.92≈16.08
So, minimum water to add = ~16 liters.
