Add Quant mixtures lesson + 5 practice questions

content/lessons/quant/quant-word-problems-mixtures.md

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+---
+id: quant-word-problems-mixtures
+section: quant
+topic: word-problems
+subtopic: mixtures
+title: "Mixtures & Solutions"
+tags: [mixtures, concentration, alligation, weighted-average]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Mixture problems ask what happens to a concentration when you combine solutions, add a pure
+ingredient, dilute with water, or swap part of a batch. They look varied, but almost all of them
+yield to a single habit: **track the amount of the pure component, not the percentage.**
+
+## Core concepts
+
+**The one equation that matters.**
+
+\[\text{amount of component} = \text{concentration} \times \text{total volume}\]
+
+Percentages don't add; **amounts do.** When you mix solutions, the total amount of the component
+is conserved — sum the amounts, then divide by the new total volume to get the new concentration.
+
+**Special ingredients.**
+
+- **Pure component** (e.g. pure acid, pure antifreeze) is at concentration \(100\% = 1\).
+- **Water** contributes volume but \(0\) of the component — it dilutes.
+
+**Alligation (a shortcut for two-solution blends).** To hit a target concentration \(m\) by mixing
+a weaker solution \(a\) with a stronger one \(b\), the quantities are in the ratio
+
+\[\frac{\text{amount of } a}{\text{amount of } b} = \frac{b - m}{m - a}\]
+
+The mean sits *closer* to whichever solution you use more of — this is just the weighted-average
+idea read backwards.
+
+## Worked examples
+
+**Combining two solutions.** Mix 30 L of 10% saline with 20 L of 35% saline.
+Amounts of salt: \(0.10\times 30 = 3\) L and \(0.35\times 20 = 7\) L, so \(10\) L of salt in
+\(50\) L total:
+
+\[\frac{10}{50} = 20\%.\]
+
+Note it is **not** the plain average \(\tfrac{10+35}{2}=22.5\%\) — there's more of the weaker
+solution, so the result is pulled toward 10%.
+
+**Adding a pure ingredient.** You have 50 L of 12% acid and want 20% acid. Add \(x\) L of pure
+acid. The acid amount becomes \(0.12\times 50 + x = 6 + x\); the volume becomes \(50 + x\):
+
+\[\frac{6 + x}{50 + x} = 0.20 \implies 6 + x = 10 + 0.2x \implies 0.8x = 4 \implies x = 5.\]
+
+## Common traps
+
+- **Averaging the percentages.** \(\tfrac{a+b}{2}\) is right only when the two volumes are equal.
+  Otherwise weight by volume — or just add the amounts.
+- **Forgetting the volume also changes.** When you add acid or water, both the numerator *and*
+  the denominator move. Don't hold the total fixed unless the problem says you removed the same
+  amount you added.
+- **"Replace" problems.** Draining \(x\) liters of a solution removes \(x\) liters *of the
+  component's current concentration*, not \(x\) liters of pure component. Compute what's left,
+  then add the replacement.
+
+## Key takeaways
+
+- Convert every percentage to an **amount** (\(\text{concentration}\times\text{volume}\)) and add
+  those; amounts are conserved, percentages are not.
+- Water is 0%, pure ingredient is 100% — plug those in like any other solution.
+- For a two-way blend to a target, alligation gives the mixing ratio directly: distances from
+  the mean, cross-assigned.

content/questions/quant/quant-word-problems-mixtures-0017.md

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+---
+id: quant-word-problems-mixtures-0017
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: easy
+tags: [mixtures, concentration, weighted-average]
+choices:
+  A: "15%"
+  B: "20%"
+  C: "22.5%"
+  D: "25%"
+  E: "45%"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A lab technician combines 30 liters of a 10% saline solution with 20 liters of a 35% saline
+solution. What is the concentration of salt in the resulting mixture?
+
+## Explanation
+
+Percentages don't add — **amounts** do. Convert each solution to its actual amount of salt:
+
+- \(0.10 \times 30 = 3\) liters of salt
+- \(0.35 \times 20 = 7\) liters of salt
+
+Total salt is \(3 + 7 = 10\) liters in \(30 + 20 = 50\) liters of solution:
+
+\[\frac{10}{50} = 0.20 = 20\%.\]
+
+So the mixture is **20%** saline.
+
+**The trap:** choice **(C) 22.5%** is the plain average of 10% and 35%. That average is only
+correct when the two volumes are equal. Here there is more of the weaker (10%) solution, so the
+result is pulled below the midpoint, toward 20%.
+
+## Hints
+
+- Don't average the two percentages — figure out how many liters of *salt* each solution
+  contributes.
+- Add the salt amounts, then divide by the total volume of the combined solution.

content/questions/quant/quant-word-problems-mixtures-0018.md

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+---
+id: quant-word-problems-mixtures-0018
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: medium
+tags: [mixtures, pure-component, concentration]
+choices:
+  A: "4"
+  B: "5"
+  C: "6"
+  D: "8"
+  E: "10"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A container holds 50 liters of a solution that is 12% acid. How many liters of **pure acid**
+must be added so that the resulting solution is 20% acid?
+
+## Explanation
+
+Let \(x\) be the liters of pure acid added. Pure acid counts as 100% acid, so it adds \(x\) liters
+of acid *and* \(x\) liters of volume.
+
+- Acid before: \(0.12 \times 50 = 6\) liters. After adding: \(6 + x\).
+- Total volume after: \(50 + x\).
+
+Set the new concentration to 20%:
+
+\[\frac{6 + x}{50 + x} = 0.20.\]
+
+Multiply out: \(6 + x = 0.20(50 + x) = 10 + 0.2x\), so \(0.8x = 4\) and \(x = 5\).
+
+Adding **5** liters of pure acid works. Check: acid \(= 11\) liters in \(55\) liters total,
+and \(\tfrac{11}{55} = 20\%\). ✓
+
+**The trap:** choice **(A) 4** comes from forgetting that the added acid also increases the total
+volume — i.e. solving \(6 + x = 0.20\times 50\). You must let the denominator grow too.
+
+## Hints
+
+- Pure acid raises both the amount of acid *and* the total volume — track both.
+- Write \(\frac{6 + x}{50 + x} = 0.20\) and solve for \(x\); don't hold the total at 50.

content/questions/quant/quant-word-problems-mixtures-0019.md

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+---
+id: quant-word-problems-mixtures-0019
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: medium
+tags: [mixtures, dilution, water]
+choices:
+  A: "10"
+  B: "18"
+  C: "20"
+  D: "30"
+  E: "40"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A tank contains 60 liters of a 30% alcohol solution. How many liters of **water** must be added
+to dilute it to a 20% alcohol solution?
+
+## Explanation
+
+Adding water changes the total volume but adds **no** alcohol, so the amount of alcohol stays
+fixed while the denominator grows.
+
+Alcohol present: \(0.30 \times 60 = 18\) liters, and this never changes.
+
+Let \(w\) be the liters of water added. The new total volume is \(60 + w\), and we need:
+
+\[\frac{18}{60 + w} = 0.20.\]
+
+So \(60 + w = \dfrac{18}{0.20} = 90\), giving \(w = 30\).
+
+Add **30** liters of water. Check: \(18\) liters of alcohol in \(90\) liters is \(20\%\). ✓
+
+**The trap:** choice **(B) 18** is the (unchanging) amount of alcohol, not the water to add;
+choice **(C) 20** tempts you to knock off "10 percentage points" of the 60 liters. Anchor on the
+fixed alcohol amount instead.
+
+## Hints
+
+- Water is 0% alcohol — the number of liters of alcohol is the same before and after.
+- Keep the alcohol amount constant in the numerator and solve \(\frac{18}{60+w}=0.20\).

content/questions/quant/quant-word-problems-mixtures-0020.md

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+---
+id: quant-word-problems-mixtures-0020
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: hard
+tags: [mixtures, replacement, concentration]
+choices:
+  A: "6"
+  B: "8"
+  C: "10"
+  D: "12"
+  E: "15"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A 40-liter tank is completely filled with a 25% antifreeze solution. Some of the solution is
+drained off and replaced with an equal volume of **pure antifreeze**, leaving the tank full at
+40% antifreeze. How many liters were drained and replaced?
+
+## Explanation
+
+Let \(x\) be the liters drained (and then replaced with pure antifreeze). Because you put back
+exactly what you took out, the total volume stays **40 liters** throughout — that is what makes
+this a "replacement" problem rather than a simple addition.
+
+Draining \(x\) liters removes solution at its *current* 25% concentration, so it removes
+\(0.25x\) liters of antifreeze. Antifreeze remaining:
+
+\[0.25 \times 40 - 0.25x = 10 - 0.25x.\]
+
+Now add \(x\) liters of pure antifreeze (100%):
+
+\[10 - 0.25x + x = 10 + 0.75x.\]
+
+This must equal \(40\%\) of the still-40-liter tank, i.e. \(0.40 \times 40 = 16\):
+
+\[10 + 0.75x = 16 \implies 0.75x = 6 \implies x = 8.\]
+
+So **8** liters were drained and replaced. Check: after draining 8 L you have \(10 - 2 = 8\) L of
+antifreeze; adding 8 L pure gives \(16\) L in \(40\) L \(= 40\%\). ✓
+
+**The trap:** it's tempting to treat draining \(x\) liters as removing \(x\) liters of pure
+antifreeze, or to let the volume change. Draining removes solution at 25%, and the replacement
+keeps the total fixed at 40 liters.
+
+## Hints
+
+- The volume stays 40 liters the whole time — you replace exactly what you remove.
+- Draining \(x\) liters removes \(0.25x\) liters of antifreeze (the *current* concentration),
+  not \(x\) liters of it.

content/questions/quant/quant-word-problems-mixtures-0021.md

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+---
+id: quant-word-problems-mixtures-0021
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: hard
+tags: [mixtures, alligation, ratio]
+choices:
+  A: "1:2"
+  B: "2:3"
+  C: "1:1"
+  D: "3:2"
+  E: "2:1"
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A merchant blends tea worth $8 per kilogram with tea worth $14 per kilogram to produce a mixture
+worth $10 per kilogram. In what ratio, cheaper tea to more expensive tea, should the two be mixed?
+
+## Explanation
+
+Let the blend use \(c\) kg of the $8 tea and \(d\) kg of the $14 tea. The total cost equals the
+blend's value:
+
+\[8c + 14d = 10(c + d).\]
+
+Expand: \(8c + 14d = 10c + 10d\), so \(4d = 2c\), giving \(\dfrac{c}{d} = \dfrac{4}{2} = 2\).
+
+The ratio cheaper : dearer is **2 : 1**.
+
+**Alligation shortcut:** the mean price $10 is $2 above the cheaper tea and $4 below the dearer
+tea. Cross-assign those distances: cheaper : dearer \(= 4 : 2 = 2 : 1\). The mixture sits closer
+to $8, so you need *more* of the cheaper tea — which matches a 2:1 ratio.
+
+Check: 2 kg at $8 plus 1 kg at $14 is \($16 + $14 = $30\) for 3 kg, or $10 per kg. ✓
+
+**The trap:** choice **(A) 1:2** reverses the ratio — a classic alligation error where the
+distances get assigned to the wrong solution. Sanity-check by asking which tea you need *more* of:
+since $10 is nearer $8, the cheaper tea dominates.
+
+## Hints
+
+- Set up \(8c + 14d = 10(c+d)\) and solve for the ratio \(c:d\).
+- Or use alligation: the distances from the mean ($2 and $4) give the ratio, but cross-assigned —
+  then check which side the mean is closer to.