From a884644c01f673a96f8716a0c1c3b5bdfa574694 Mon Sep 17 00:00:00 2001 From: Claude Date: Sat, 18 Jul 2026 09:07:10 +0000 Subject: [PATCH] Add Quant mixtures lesson + 5 practice questions Fills the empty word-problems/mixtures subtopic: a concept lesson (amount-conservation, dilution, replacement, alligation) plus five difficulty-progressing problem-solving questions (0017-0021) with worked explanations, distractor-trap analysis, and hints. Rebuilds the content index (32->37 questions, 8->9 lessons). Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_014KvdUA1eEHtttBQcWVCvLP --- .../quant/quant-word-problems-mixtures.md | 76 ++++++++ .../quant-word-problems-mixtures-0017.md | 49 +++++ .../quant-word-problems-mixtures-0018.md | 51 +++++ .../quant-word-problems-mixtures-0019.md | 50 +++++ .../quant-word-problems-mixtures-0020.md | 59 ++++++ .../quant-word-problems-mixtures-0021.md | 53 ++++++ docs/data/content.json | 175 +++++++++++++++++- 7 files changed, 509 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-word-problems-mixtures.md create mode 100644 content/questions/quant/quant-word-problems-mixtures-0017.md create mode 100644 content/questions/quant/quant-word-problems-mixtures-0018.md create mode 100644 content/questions/quant/quant-word-problems-mixtures-0019.md create mode 100644 content/questions/quant/quant-word-problems-mixtures-0020.md create mode 100644 content/questions/quant/quant-word-problems-mixtures-0021.md diff --git a/content/lessons/quant/quant-word-problems-mixtures.md b/content/lessons/quant/quant-word-problems-mixtures.md new file mode 100644 index 0000000..e02eccc --- /dev/null +++ b/content/lessons/quant/quant-word-problems-mixtures.md @@ -0,0 +1,76 @@ +--- +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. diff --git a/content/questions/quant/quant-word-problems-mixtures-0017.md b/content/questions/quant/quant-word-problems-mixtures-0017.md new file mode 100644 index 0000000..807c4d8 --- /dev/null +++ b/content/questions/quant/quant-word-problems-mixtures-0017.md @@ -0,0 +1,49 @@ +--- +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. diff --git a/content/questions/quant/quant-word-problems-mixtures-0018.md b/content/questions/quant/quant-word-problems-mixtures-0018.md new file mode 100644 index 0000000..b38d523 --- /dev/null +++ b/content/questions/quant/quant-word-problems-mixtures-0018.md @@ -0,0 +1,51 @@ +--- +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. diff --git a/content/questions/quant/quant-word-problems-mixtures-0019.md b/content/questions/quant/quant-word-problems-mixtures-0019.md new file mode 100644 index 0000000..35124fd --- /dev/null +++ b/content/questions/quant/quant-word-problems-mixtures-0019.md @@ -0,0 +1,50 @@ +--- +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 \(\frac{18}{60+w}=0.20\). diff --git a/content/questions/quant/quant-word-problems-mixtures-0020.md b/content/questions/quant/quant-word-problems-mixtures-0020.md new file mode 100644 index 0000000..3c7fe56 --- /dev/null +++ b/content/questions/quant/quant-word-problems-mixtures-0020.md @@ -0,0 +1,59 @@ +--- +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. diff --git a/content/questions/quant/quant-word-problems-mixtures-0021.md b/content/questions/quant/quant-word-problems-mixtures-0021.md new file mode 100644 index 0000000..3cdb4f1 --- /dev/null +++ b/content/questions/quant/quant-word-problems-mixtures-0021.md @@ -0,0 +1,53 @@ +--- +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. diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..d50b7a8 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 21, + "lessons": 6 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 37, + "lessons": 9 }, "questions": [ { @@ -596,6 +596,156 @@ "Adding a constant relocates the whole set; only stretching or shrinking it (multiplying) changes the spread." ] }, + { + "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", + "prompt": "A lab technician combines 30 liters of a 10% saline solution with 20 liters of a 35% saline\nsolution. 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:\n\n- \\(0.10 \\times 30 = 3\\) liters of salt\n- \\(0.35 \\times 20 = 7\\) liters of salt\n\nTotal salt is \\(3 + 7 = 10\\) liters in \\(30 + 20 = 50\\) liters of solution:\n\n\\[\\frac{10}{50} = 0.20 = 20\\%.\\]\n\nSo the mixture is **20%** saline.\n\n**The trap:** choice **(C) 22.5%** is the plain average of 10% and 35%. That average is only\ncorrect when the two volumes are equal. Here there is more of the weaker (10%) solution, so the\nresult is pulled below the midpoint, toward 20%.", + "hints": [ + "Don't average the two percentages — figure out how many liters of *salt* each solution", + "Add the salt amounts, then divide by the total volume of the combined solution." + ] + }, + { + "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", + "prompt": "A container holds 50 liters of a solution that is 12% acid. How many liters of **pure acid**\nmust 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\nof acid *and* \\(x\\) liters of volume.\n\n- Acid before: \\(0.12 \\times 50 = 6\\) liters. After adding: \\(6 + x\\).\n- Total volume after: \\(50 + x\\).\n\nSet the new concentration to 20%:\n\n\\[\\frac{6 + x}{50 + x} = 0.20.\\]\n\nMultiply out: \\(6 + x = 0.20(50 + x) = 10 + 0.2x\\), so \\(0.8x = 4\\) and \\(x = 5\\).\n\nAdding **5** liters of pure acid works. Check: acid \\(= 11\\) liters in \\(55\\) liters total,\nand \\(\\tfrac{11}{55} = 20\\%\\). ✓\n\n**The trap:** choice **(A) 4** comes from forgetting that the added acid also increases the total\nvolume — 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." + ] + }, + { + "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", + "prompt": "A tank contains 60 liters of a 30% alcohol solution. How many liters of **water** must be added\nto dilute it to a 20% alcohol solution?", + "explanation": "Adding water changes the total volume but adds **no** alcohol, so the amount of alcohol stays\nfixed while the denominator grows.\n\nAlcohol present: \\(0.30 \\times 60 = 18\\) liters, and this never changes.\n\nLet \\(w\\) be the liters of water added. The new total volume is \\(60 + w\\), and we need:\n\n\\[\\frac{18}{60 + w} = 0.20.\\]\n\nSo \\(60 + w = \\dfrac{18}{0.20} = 90\\), giving \\(w = 30\\).\n\nAdd **30** liters of water. Check: \\(18\\) liters of alcohol in \\(90\\) liters is \\(20\\%\\). ✓\n\n**The trap:** choice **(B) 18** is the (unchanging) amount of alcohol, not the water to add;\nchoice **(C) 20** tempts you to knock off \"10 percentage points\" of the 60 liters. Anchor on the\nfixed 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\\)." + ] + }, + { + "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", + "prompt": "A 40-liter tank is completely filled with a 25% antifreeze solution. Some of the solution is\ndrained off and replaced with an equal volume of **pure antifreeze**, leaving the tank full at\n40% 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\nexactly what you took out, the total volume stays **40 liters** throughout — that is what makes\nthis a \"replacement\" problem rather than a simple addition.\n\nDraining \\(x\\) liters removes solution at its *current* 25% concentration, so it removes\n\\(0.25x\\) liters of antifreeze. Antifreeze remaining:\n\n\\[0.25 \\times 40 - 0.25x = 10 - 0.25x.\\]\n\nNow add \\(x\\) liters of pure antifreeze (100%):\n\n\\[10 - 0.25x + x = 10 + 0.75x.\\]\n\nThis must equal \\(40\\%\\) of the still-40-liter tank, i.e. \\(0.40 \\times 40 = 16\\):\n\n\\[10 + 0.75x = 16 \\implies 0.75x = 6 \\implies x = 8.\\]\n\nSo **8** liters were drained and replaced. Check: after draining 8 L you have \\(10 - 2 = 8\\) L of\nantifreeze; adding 8 L pure gives \\(16\\) L in \\(40\\) L \\(= 40\\%\\). ✓\n\n**The trap:** it's tempting to treat draining \\(x\\) liters as removing \\(x\\) liters of pure\nantifreeze, or to let the volume change. Draining removes solution at 25%, and the replacement\nkeeps 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)," + ] + }, + { + "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", + "prompt": "A merchant blends tea worth $8 per kilogram with tea worth $14 per kilogram to produce a mixture\nworth $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\nblend's value:\n\n\\[8c + 14d = 10(c + d).\\]\n\nExpand: \\(8c + 14d = 10c + 10d\\), so \\(4d = 2c\\), giving \\(\\dfrac{c}{d} = \\dfrac{4}{2} = 2\\).\n\nThe ratio cheaper : dearer is **2 : 1**.\n\n**Alligation shortcut:** the mean price $10 is $2 above the cheaper tea and $4 below the dearer\ntea. Cross-assign those distances: cheaper : dearer \\(= 4 : 2 = 2 : 1\\). The mixture sits closer\nto $8, so you need *more* of the cheaper tea — which matches a 2:1 ratio.\n\nCheck: 2 kg at $8 plus 1 kg at $14 is \\($16 + $14 = $30\\) for 3 kg, or $10 per kg. ✓\n\n**The trap:** choice **(A) 1:2** reverses the ratio — a classic alligation error where the\ndistances get assigned to the wrong solution. Sanity-check by asking which tea you need *more* of:\nsince $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 —" + ] + }, { "id": "quant-word-problems-overlapping-sets-0006", "section": "quant", @@ -1004,6 +1154,23 @@ "status": "in-review", "body": "## Overview\n\nDescriptive statistics on GMAT Focus Quant is almost entirely about two ideas: where a data set is\n**centered** (mean, median, mode) and how **spread out** it is (range, standard deviation). The\narithmetic is light — the exam tests whether you understand what each measure does, especially how\noutliers and transformations affect them. Master a handful of moves and this topic becomes free points.\n\n## Core concepts\n\n**Mean (arithmetic average).**\n\n\\[\\text{mean} = \\frac{\\text{sum of values}}{\\text{count}}\\]\n\nThe single most useful rearrangement is **sum = mean × count**. Most \"average\" problems are really\nabout the *sum*: find it, adjust it, and divide again.\n\n**Weighted average.** When groups have different sizes, you cannot just average the group averages —\nweight each by its count:\n\n\\[\\text{overall mean} = \\frac{n_1\\bar{x}_1 + n_2\\bar{x}_2}{n_1 + n_2}\\]\n\nThe result always lands *between* the two group averages, pulled toward the larger group.\n\n**Median.** The middle value once the data is sorted.\n\n- **Odd count:** the single middle value.\n- **Even count:** the average of the two middle values.\n\nAlways sort first. The median ignores *how far away* extreme values are — only their position matters.\n\n**Mode.** The value that appears most often. A set can have no mode, one mode, or several.\n\n**Range.** \\(\\text{range} = \\text{max} - \\text{min}\\). A crude measure of spread that depends on only\ntwo numbers.\n\n**Standard deviation (SD).** Measures the *typical distance of values from the mean*. You will almost\nnever compute it by hand on the GMAT; you reason about it:\n\n- SD is **0** only when every value is identical.\n- SD depends on **spread from the mean, not on the size of the numbers**. \\(\\{1,2,3\\}\\) and\n \\(\\{101,102,103\\}\\) have the *same* SD.\n- **Adding a constant** to every value shifts the set but leaves the spread — and therefore the SD —\n **unchanged**.\n- **Multiplying** every value by \\(k\\) multiplies the SD by \\(|k|\\).\n\n## Worked examples\n\n**Sum trick.** The average of 6 numbers is 22. A seventh number is added and the average becomes 25.\nThe seventh number? Old sum \\(= 22 \\times 6 = 132\\); new sum \\(= 25 \\times 7 = 175\\); the added number\nis \\(175 - 132 = 43\\). (Notice it is far above 25 — it had to *pull the average up* across all seven\nvalues.)\n\n**Outlier: mean vs. median.** For \\(\\{4, 7, 9, 10, 90\\}\\), the mean is \\(120/5 = 24\\) but the median is\n\\(9\\). The lone large value \\(90\\) drags the mean far above the median while leaving the median\nuntouched. When a distribution has a long tail, mean \\(\\ne\\) median.\n\n**Spread, not magnitude.** Which has the larger SD, \\(\\{2,4,6,8,10\\}\\) or \\(\\{102,104,106,108,110\\}\\)?\nNeither — they are identical shapes shifted by 100, so the SD is the same. Bigger numbers do not mean\nbigger spread.\n\n## Common traps\n\n- **Averaging averages.** With unequal group sizes, the simple average of two means is wrong — you must\n weight by count.\n- **Forgetting to sort before taking the median.** The median is positional; unsorted data gives a\n meaningless \"middle.\"\n- **Assuming mean = median.** True only for symmetric data. Any outlier or skew separates them, and the\n mean moves toward the tail.\n- **Thinking \"big numbers ⇒ big standard deviation.\"** SD measures distance from the mean, not the\n numbers' magnitude. Adding a constant changes the mean but never the SD.\n\n## Key takeaways\n\n- Turn average problems into **sum** problems: sum = mean × count.\n- Weighted averages weight by group size and always fall between the group means.\n- Sort first, then take the median; the median ignores how extreme the outliers are, but the mean does not.\n- Standard deviation is about spread from the mean: it is 0 only for identical values, unchanged by\n adding a constant, and independent of how large the numbers themselves are." }, + { + "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", + "body": "## Overview\n\nMixture problems ask what happens to a concentration when you combine solutions, add a pure\ningredient, dilute with water, or swap part of a batch. They look varied, but almost all of them\nyield to a single habit: **track the amount of the pure component, not the percentage.**\n\n## Core concepts\n\n**The one equation that matters.**\n\n\\[\\text{amount of component} = \\text{concentration} \\times \\text{total volume}\\]\n\nPercentages don't add; **amounts do.** When you mix solutions, the total amount of the component\nis conserved — sum the amounts, then divide by the new total volume to get the new concentration.\n\n**Special ingredients.**\n\n- **Pure component** (e.g. pure acid, pure antifreeze) is at concentration \\(100\\% = 1\\).\n- **Water** contributes volume but \\(0\\) of the component — it dilutes.\n\n**Alligation (a shortcut for two-solution blends).** To hit a target concentration \\(m\\) by mixing\na weaker solution \\(a\\) with a stronger one \\(b\\), the quantities are in the ratio\n\n\\[\\frac{\\text{amount of } a}{\\text{amount of } b} = \\frac{b - m}{m - a}\\]\n\nThe mean sits *closer* to whichever solution you use more of — this is just the weighted-average\nidea read backwards.\n\n## Worked examples\n\n**Combining two solutions.** Mix 30 L of 10% saline with 20 L of 35% saline.\nAmounts of salt: \\(0.10\\times 30 = 3\\) L and \\(0.35\\times 20 = 7\\) L, so \\(10\\) L of salt in\n\\(50\\) L total:\n\n\\[\\frac{10}{50} = 20\\%.\\]\n\nNote it is **not** the plain average \\(\\tfrac{10+35}{2}=22.5\\%\\) — there's more of the weaker\nsolution, so the result is pulled toward 10%.\n\n**Adding a pure ingredient.** You have 50 L of 12% acid and want 20% acid. Add \\(x\\) L of pure\nacid. The acid amount becomes \\(0.12\\times 50 + x = 6 + x\\); the volume becomes \\(50 + x\\):\n\n\\[\\frac{6 + x}{50 + x} = 0.20 \\implies 6 + x = 10 + 0.2x \\implies 0.8x = 4 \\implies x = 5.\\]\n\n## Common traps\n\n- **Averaging the percentages.** \\(\\tfrac{a+b}{2}\\) is right only when the two volumes are equal.\n Otherwise weight by volume — or just add the amounts.\n- **Forgetting the volume also changes.** When you add acid or water, both the numerator *and*\n the denominator move. Don't hold the total fixed unless the problem says you removed the same\n amount you added.\n- **\"Replace\" problems.** Draining \\(x\\) liters of a solution removes \\(x\\) liters *of the\n component's current concentration*, not \\(x\\) liters of pure component. Compute what's left,\n then add the replacement.\n\n## Key takeaways\n\n- Convert every percentage to an **amount** (\\(\\text{concentration}\\times\\text{volume}\\)) and add\n those; amounts are conserved, percentages are not.\n- Water is 0%, pure ingredient is 100% — plug those in like any other solution.\n- For a two-way blend to a target, alligation gives the mixing ratio directly: distances from\n the mean, cross-assigned." + }, { "id": "quant-word-problems-rates-speed-distance", "section": "quant",