ExamShortcut
🔍Log in

Mixtures & Alligation

🔒 Log in to track
high importance~1 Q in Tier 119 formulas⚡ 15 shortcuts5 subtopics
OverviewLearn (5)Formulas (19)Examples (5)Practice (66)
Subtopic 2 of 5·← Mixture Concentration & Amounts·Replacement & Repeated Operations →

Rule of Alligation

🔒 Log in to track

Alligation mixes two ingredients of known 'strength' (price, %, speed, average) to hit a target mean:

cheaper quantity:dearer quantity=(dearer−mean):(mean−cheaper)\text{cheaper quantity} : \text{dearer quantity} = (\text{dearer} - \text{mean}) : (\text{mean} - \text{cheaper})

It is the weighted average solved backwards: vˉ=n1v1+n2v2n1+n2\bar{v} = \frac{n_1 v_1 + n_2 v_2}{n_1 + n_2}, so n1n2=v2−vˉvˉ−v1\frac{n_1}{n_2} = \frac{v_2 - \bar{v}}{\bar{v} - v_1}.

Works whenever the quantity mixes linearly — prices, percentages (milk, acid, alcohol), average ages, average salaries, average speeds for equal times.

The mean must always lie BETWEEN the two strengths; the distances give the quantities, crossing over.

📖 Notes🧩 5 question types⚡ Formulas🎯 14 practice Q

Detailed notes

The rule of alligation

When two ingredients at 'prices' CC (cheaper) and DD (dearer) are mixed to give a mean MM, the mixing ratio is fixed by the balance: quantity of dearerquantity of cheaper=M−CD−M\frac{\text{quantity of dearer}}{\text{quantity of cheaper}} = \frac{M - C}{D - M} Or the tilted diagram: write C and D at the top, M in the middle; each lower arm subtracts diagonally — left arm =M−C= M - C, right arm =D−M= D - M — and the arms are the ratio of quantities of the side you came from. Tea at ₹320 and ₹250 to sell at ₹300: dearer =300−250320−300=5020=52= \frac{300-250}{320-300} = \frac{50}{20} = \frac{5}{2} of the cheaper → 2:5. The mean must always lie between C and D; if your ratio comes out negative, you have inverted an arm.

Why it works

Let x and y be the quantities: Cx+Dyx+y=M\frac{Cx + Dy}{x + y} = M. Cross-multiplying: x(D−M)=y(M−C)x(D - M) = y(M - C) — exactly the arms. Alligation is one linear equation wearing a picture.

The same rule everywhere

Any weighted mean splits into an alligation:

  • Speeds: equal distances at speeds u and v → average =2uvu+v= \frac{2uv}{u+v}; the time ratio is v−AA−u\frac{v - A}{A - u}.
  • Class averages: boys average 70, girls 75, class 72 → boys : girls =(75−72):(72−70)=3:2= (75-72):(72-70) = 3:2.
  • Interest rates: two investments at 5% and 8% averaging 6.25% → amounts in 7:5.
  • Profit per cents: items sold at 20% and 10% profit with a 14% overall → cost split 3:2. In each case the 'price' is the quantity being averaged and the arms give the weights.

Using quantities, not just ratios

When one actual quantity is known, scale the arms. 25 kg of sugar at ₹24 must be mixed with sugar at ₹42 to average ₹30: arms give cheaper : dearer =(42−30):(30−24)=2:1= (42-30):(30-24) = 2:1, so dearer =25= 25 kg — the cheaper arm is double the dearer one.

Strengths and fractions

Solution strengths work identically: 20% and 50% acid mixed to get 30% → 50−3030−20=2010=2:1\frac{50-30}{30-20} = \frac{20}{10} = 2:1 (dilute : strong). For two full mixtures, first reduce each to the fraction of the key component, then alligate on those fractions.

Common slips

  • Writing the arms upside down: cheaper : dearer =(D−M):(M−C)= (D-M):(M-C).
  • Alligating on the wrong quantity (alligate times for equal distances, distances for equal times).
  • Averaging three prices pairwise by gut — pair them so each pair brackets the mean, or use the ledger formula.

Quick revision

  • Dearer : cheaper =(M−C):(D−M)= (M-C) : (D-M).
  • The mean sits between the two prices; arms subtract diagonally.
  • Known quantity × arm ratio = actual weights.
  • Works on prices, speeds, marks, rates, profit %, strengths.
  • Verify: recompute the mean from the mixed ratio.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Two-ingredient alligation for the ratiovery common2 practice Q
How to spot it:

Two unit prices are given with the target mean price; the mixing ratio is asked.

D:C=(M−C):(D−M)D:C = (M - C) : (D - M)
  1. Identify cheaper, dearer and mean; check C < M < D.
  2. Subtract diagonally: (M−C)(M-C) and (D−M)(D-M).
  3. State the ratio of the dearer to cheaper — and scale by any known quantity.

Why: the weighted-mean equation rearranges into exactly these arms.

Example: In what ratio should tea at ₹320 per kg be mixed with tea at ₹250 per kg to get a mixture worth ₹300 per kg?

D:C =(300−250):(320−300)=50:20=5:2= (300-250):(320-300) = 50:20 = 5:2.

Type 2: Mixing solutions by strengthvery common2 practice Q
How to spot it:

Two solutions of given concentration are mixed to reach a target strength.

dilutestrong=S−MM−W  (W%,S% strengths)\frac{\text{dilute}}{\text{strong}} = \frac{S - M}{M - W}\ \ (W\%, S\%\ \text{strengths})
  1. Alligate on the strength fractions exactly as on prices.
  2. If a final volume is fixed, split it by the ratio.
  3. Watch which side is dilute — the arm (S−M)(S - M) belongs to it.

Why: strength is just the 'price' per litre of solution.

Example: Solutions of 20% and 50% acid are mixed to make a 30% solution. The ratio of the 20% solution to the 50% solution is:

(50−30):(30−20)=20:10=2:1(50-30):(30-20) = 20:10 = 2:1.

Type 3: Alligation on averages (speeds, marks, ages)common2 practice Q
How to spot it:

Two groups with different averages combine to a known overall average — find the group sizes' ratio.

n1n2=A2−AA−A1\frac{n_1}{n_2} = \frac{A_2 - A}{A - A_1}
  1. Treat the averages as the two prices and the overall as M.
  2. Subtract to the arms; the arms are the sizes' ratio.
  3. For speeds over equal distances, alligate the times, not the distances.

Why: the overall average is the size-weighted mean of the group averages.

Example: The average marks of boys in a class is 70 and of girls is 75. If the class average is 72, the ratio of boys to girls is:

(75−72):(72−70)=3:2(75-72):(72-70) = 3:2.

Type 4: Find the quantity to add (alligation form)common2 practice Q
How to spot it:

'How much of X must be added to Y of the mixture to reach a given strength or price?'

qaddedqexisting=Mold−MM−Madded\frac{q_{\text{added}}}{q_{\text{existing}}} = \frac{M_{\text{old}} - M}{M - M_{\text{added}}}
  1. Give the added ingredient its own 'price' (0 for water, 100 for pure acid, its cost price).
  2. Alligate between the old mixture and the addition.
  3. Solve for the unknown quantity.

Why: adding a zero-price ingredient is a two-price mixture.

Example: How much pure acid must be added to 100 litres of a 30% acid solution to make it 50% acid?

Acid =30= 30; 30+x100+x=12\frac{30+x}{100+x} = \frac{1}{2} → x=40x = 40 L. Alligation view: free water in the mix vs pure acid at 'price' 100%.

Type 5: Alligation on rates and profitscommon2 practice Q
How to spot it:

Two investments at different interest rates, or items sold at different profit per cents, with a known combined result.

amount1amount2=R2−RR−R1\frac{\text{amount}_1}{\text{amount}_2} = \frac{R_2 - R}{R - R_1}
  1. Alligate on the percentages (rate or profit) to split the total amount.
  2. Overall rate =total interesttotal sum×100= \frac{\text{total interest}}{\text{total sum}} \times 100 if not given.
  3. Convert back to actual parts by scaling the total.

Why: the combined percentage is the amount-weighted mean.

Example: ₹12,000 is invested partly at 5% and the rest at 8% simple interest. The total yearly interest is ₹750. Find each part.

Mean rate =75012000×100=6.25%= \frac{750}{12000} \times 100 = 6.25\% → 5% : 8% =(8−6.25):(6.25−5)=1.75:1.25=7:5= (8-6.25):(6.25-5) = 1.75:1.25 = 7:5 → ₹7,000 and ₹5,000.

Formulas

Alligation ratio
n1n2=v2−vˉvˉ−v1\frac{n_1}{n_2} = \frac{v_2 - \bar{v}}{\bar{v} - v_1}
Weighted average
vˉ=n1v1+n2v2n1+n2\bar{v} = \frac{n_1 v_1 + n_2 v_2}{n_1 + n_2}
Cheaper : dearer
(d−m):(m−c)(d - m) : (m - c)
Water as free ingredient
milk:water=(m−0):(c−m), c>m\text{milk} : \text{water} = (m - 0) : (c - m), \ c > m

Shortcut tricks

⚡ Cross over the distances

Dearer-minus-mean and mean-minus-cheaper swap sides — that's the whole rule.

Example: In what ratio must rice at ₹36/kg be mixed with rice at ₹24/kg so that the mixture costs ₹30/kg?

Cheaper : dearer =(36−30):(30−24)=1:1= (36-30) : (30-24) = 1 : 1.

⚡ Water costs zero

Dilution questions are alligation with one strength = 0.

Example: Water must be added to 60 litres of milk at ₹24/litre so that the mixture is worth ₹20/litre. Water to add:

Milk : water =(20−0):(24−20)=5:1= (20 - 0) : (24 - 20) = 5 : 1 ⇒ 60/5=1260/5 = 12 litres of water.

⚡ Alligation on averages

Any average — salary, age, marks — alligates the same way.

Example: A class of 50 has an average weight of 42 kg; boys average 45 kg, girls 40 kg. The number of girls is:

Boys : girls =(42−40):(45−42)=2:3= (42-40) : (45-42) = 2 : 3 ⇒ girls =30= 30.

Where students lose marks

  • Not crossing over — writing cheaper : dearer = (mean − cheaper) : (dearer − mean), which inverts the ratio.

  • Alligating a mean that lies outside the two values (impossible — check bounds first).

  • Forgetting water/pure ingredient has strength 0% or 100%, not 'no value'.

  • Using alligation where averaging is not linear (e.g. average speeds for unequal distances).

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.