Mixture or solution
A combined quantity containing two or more components. The total quantity is the sum of component quantities when no loss occurs.
\[V=V_1+V_2+\cdots\]
9 SSC CGL · Mathematics
Mixture and Alligation
Handwritten-style notes for SSC CGL Tier I and Tier II.
01MAP IT!
The Big Picture — Graphic Mind Map
Every mixture question tracks an additive quantity—solute, pure metal or total cost—inside a combined total.
MIXTURE
&
ALLIGATION
&
ALLIGATION
1. Compositionsolute, solvent, purity and concentration
2. Weighted Meantotal useful content divided by total quantity
3. Alligationcross-difference ratio for two components
4. Change the Mixadd, remove, evaporate or dilute
5. Replacementsurviving fraction after repeated cycles
6. Applicationsmilk-water, alloys, purity and price
Conservation is the hidden spine
Choose the quantity that is actually conserved or added—pure substance, metal, salt, milk, or total cost. Add its contribution from every component.
\[\boxed{\text{total content}=\sum(\text{quantity}\times\text{concentration})}\]
\[\text{mean concentration}=\frac{q_1c_1+q_2c_2+\cdots+q_nc_n}{q_1+q_2+\cdots+q_n}\]
02ZERO TO EXPERT
The Foundation (Basics) — Complete Concept Build
Define the useful component first; all formulas then become weighted-average or conservation equations.
AMixture vocabulary
Solute and solvent
The tracked substance is the solute; the carrier is the solvent. In milk-water questions, either may be tracked if used consistently.
Pure content
\[\text{pure content}=V\times\frac{C}{100}\]
Here \(C\%\) is the concentration by the stated measure.
Concentration
\[C\%=\frac{\text{pure component}}{\text{total mixture}}\times100\]
Impurity
\[\text{impurity}\%=100\%-\text{purity}\%\]
Ratio to percentage
If solute : solvent is \(a:b\), then
\[C_{\text{solute}}=\frac{a}{a+b}\times100\%\]
Concentration is a part-to-whole fraction
Never divide by solvent alone. The denominator is the entire mixture unless the question explicitly asks for a component ratio.
\[\text{fraction of solute}=\frac{\text{solute}}{\text{solute}+\text{solvent}}\]
BWeighted average — the universal method
Two components
\[M=\frac{q_1c_1+q_2c_2}{q_1+q_2}\]
Many components
\[M=\frac{\sum_{i=1}^{n}q_ic_i}{\sum_{i=1}^{n}q_i}\]
Total contribution
\[(q_1+q_2)M=q_1c_1+q_2c_2\]
This equation is safer than memorized shortcuts in irregular cases.
Mean lies between extremes
\[\min(c_i)\le M\le\max(c_i)\]
With positive amounts of different strengths, the inequalities are strict.
Closer to the larger quantity
The mean is pulled toward the component present in greater quantity. This is a fast option check.
Units must match
Concentration, price per unit and purity can all be weighted, but their quantities must use a compatible weight or volume unit.
The weighted mean is a balance point
Large quantity means a longer pull. The final mean settles between the component values, closer to the heavier side.
\[q_1(M-c_1)=q_2(c_2-M)\quad(c_1\lt M\lt c_2)\]
CRule of alligation — two-component shortcut
Set the values
Let lower value be \(L\), higher value be \(H\), and desired mean be \(M\), with
\[L\lt M\lt H\]
Cross differences
\[\boxed{q_L:q_H=(H-M):(M-L)}\]
Why it works
\[q_L(M-L)=q_H(H-M)\]
The deficits below the mean balance the excesses above it.
Opposite difference gives the amount ratio
The amount of the lower-value component is paired with the distance from the mean to the higher value, and vice versa.
\[\text{lower amount}:\text{higher amount}=\text{upper gap}:\text{lower gap}\]
Mean outside the interval?
No positive mixture of only \(L\) and \(H\) can have a mean outside \([L,H]\). An outside target signals impossible data or an additional process.
Mean equals an endpoint
If \(M=L\), the amount of \(H\) must be zero; if \(M=H\), the amount of \(L\) must be zero.
DFrom ratio to actual quantities
Total mixture known
If alligation gives \(a:b\) and total is \(V\):
\[q_1=V\frac{a}{a+b},\qquad q_2=V\frac{b}{a+b}\]
One component known
If \(q_1:q_2=a:b\):
\[q_2=q_1\frac ba\]
Difference known
If \(a\gt b\) and quantity difference is \(D\):
\[V=D\frac{a+b}{a-b}\]
EMixing three or more components
Weighted-mean equation
\[M=\frac{q_1c_1+q_2c_2+q_3c_3}{q_1+q_2+q_3}\]
This is always valid under additive quantities.
Group then combine
Find the mean of a subgroup and its total amount; treat that subgroup as one new component.
Underdetermined warning
One mean equation with three unknown quantities does not determine a unique ratio. Additional information is essential.
Three streams require enough constraints
Total quantity and total content supply two equations. If three individual quantities are unknown, another independent relation is required.
\[q_1+q_2+q_3=V,\qquad q_1c_1+q_2c_2+q_3c_3=VM\]
FDilution, strengthening and evaporation
Add pure solvent
Solute remains constant. If initial volume is \(V\), concentration \(C\%\), and solvent added is \(x\):
\[C'=\frac{VC}{V+x}\]
Add pure solute
Pure content and total both rise by \(x\):
\[C'=100\frac{VC/100+x}{V+x}\]
Evaporate pure solvent
If \(x\) units of solvent evaporate and solute does not:
\[C'=\frac{VC}{V-x}\]
Solvent needed for target \(M\)
\[\frac{VC}{V+x}=M\Rightarrow x=V\left(\frac CM-1\right)\]
Use consistent percentage numbers for \(C\) and \(M\).
Solvent to remove for target \(M\)
\[\frac{VC}{V-x}=M\Rightarrow x=V\left(1-\frac CM\right)\]
Applicable when only solvent leaves.
Track the unchanged component
When solvent alone is added or removed, solute quantity is the invariant. Equate solute before and after.
\[V\frac C{100}=V'\frac{C'}{100}\]
GSingle replacement
Withdraw a well-mixed sample
Removing \(x\) from total \(V\) removes the same fraction \(x/V\) of every component.
Fraction left
\[f=1-\frac{x}{V}\]
Each original component is multiplied by \(f\).
Refill restores volume
After withdrawing \(x\), add \(x\) of the replacement liquid so total volume returns to \(V\).
Withdraw → survive → refill
Replacement has three distinct stages. The withdrawn sample has the current composition, not the original composition after the first cycle.
\[\text{old content after one cycle}=\text{old content}\left(1-\frac{x}{V}\right)\]
HRepeated replacement — the survival formula
Original liquid left
\[\boxed{Q_n=Q_0\left(1-\frac{x}{V}\right)^n}\]
For constant vessel volume \(V\), equal replacement \(x\), and thorough mixing before every withdrawal.
Original liquid replaced
\[Q_{\text{replaced}}=Q_0\left[1-\left(1-\frac{x}{V}\right)^n\right]\]
Unequal withdrawals
\[Q_n=Q_0\prod_{i=1}^{n}\left(1-\frac{x_i}{V}\right)\]
Assumes volume is restored to \(V\) after each cycle.
Replacement behaves like compound decay
Every cycle multiplies what remains by the same survival factor. This is why the exponent is the number of cycles.
\[f^n=\underbrace{f\times f\times\cdots\times f}_{n\text{ cycles}}\]
IGeneral replacement with a non-pure refill
Concentration recurrence
If refill concentration is \(C_r\), current concentration is \(C_k\), and \(f=1-x/V\):
\[C_{k+1}=fC_k+(1-f)C_r\]
Closed form
\[C_n=C_r+(C_0-C_r)f^n\]
The mixture moves toward the refill concentration.
Pure solvent refill
Set \(C_r=0\):
\[C_n=C_0f^n\]
JPrice mixtures and profit links
Mean cost price
\[M=\frac{q_1p_1+q_2p_2}{q_1+q_2}\]
Alligation by price
\[q_{\text{cheap}}:q_{\text{dear}}=(p_{\text{dear}}-M):(M-p_{\text{cheap}})\]
Target cost from selling price
If mixture sells at \(S\) with profit \(g\%\):
\[M=\frac{100S}{100+g}\]
Use this target mean cost in alligation.
Free water as zero-price component
\[p_{\text{water}}=0\]
Alligation can mix milk priced at \(p\) with water at zero to reach a desired mixture cost.
Free water sold at milk’s cost price
If milk : water \(=m:w\), and the mixture is sold at the cost price of pure milk:
\[\text{profit}\%=\frac wm\times100\%\]
Price is also a concentration
Replace “percentage strength” with “cost per unit.” Total cost is additive, so the same weighted-mean and alligation structure applies.
\[\text{total cost}=\sum(q_i\times p_i)\]
KPurity and alloy applications
Pure metal content
\[\text{pure metal}=W\frac{P}{100}\]
Use weight \(W\) and purity \(P\%\).
Alloy with two metals
If metal \(A:B=a:b\), percentage of \(A\) is
\[\frac{a}{a+b}\times100\%\]
Mixing alloys
Track one metal at a time. Its percentage in each alloy becomes the concentration used in weighted average.
Hallmark/purity language
Translate any purity scale supplied in the question into a fraction of pure substance before mixing.
Impurity removal
If a sample of the whole mixture is removed, pure and impure parts leave in the current proportion.
Component check
\[\sum\text{component quantities}=\text{total mixture}\]
Calculate the complementary component as a final verification.
An alloy is just a solid mixture
The arithmetic is unchanged. Replace volume with weight and track the selected pure metal through every addition or removal.
LDecision table and conditions
| Question pattern | Best model | Invariant / condition |
|---|---|---|
| Two known strengths, target mean | Alligation | Mean must lie between strengths |
| Three or more components | Weighted-average equation | Need enough independent relations |
| Add pure solvent | Conservation of solute | Solute amount unchanged |
| Evaporate solvent | Conservation of solute | Only solvent leaves |
| Remove well-mixed sample | Proportional removal | Every component loses the same fraction |
| Repeated equal replacement | Survival factor power | Mix thoroughly and restore volume each cycle |
| Price mixture | Total-cost weighted mean | Use compatible quantity units |
| Alloys and purity | Track one pure component | Purity is a part-to-whole fraction |
03SAVE TIME
Short Tricks & Magic Formulas
Use cross differences for two components; return to conservation for every irregular case.
1Alligation in one line
Amount ratio
\[L:H=(H-M):(M-L)\]
Quantity fractions
\[\frac{q_L}{V}=\frac{H-M}{H-L},\qquad\frac{q_H}{V}=\frac{M-L}{H-L}\]
Weighted mean from ratio
If quantity ratio is \(a:b\):
\[M=\frac{aL+bH}{a+b}\]
2Difference method
Total from component difference
\[V=D\frac{a+b}{|a-b|}\]
One part from total
\[q_i=V\frac{a_i}{\sum a}\]
Check without recomputing
\[q_L(M-L)=q_H(H-M)\]
If the two sides match, the alligation ratio is correct.
3Replacement speed rules
Survival factor
\[f=1-\frac{x}{V}\]
After \(n\) cycles
\[Q_n=Q_0f^n\]
Number of cycles
If \(Q_n/Q_0=k\):
\[f^n=k\]
Use recognizable powers or logarithms only if the question permits their evaluation.
4Dilution and concentration shortcuts
Same solute, changed total
\[V_1C_1=V_2C_2\]
Valid when only solvent is added or removed.
Solvent added
\[x=V\left(\frac{C}{M}-1\right)\]
Solvent evaporated
\[x=V\left(1-\frac{C}{M}\right)\]
5Option elimination
Interval test
A positive weighted mean cannot be below every component or above every component.
Direction test
Adding pure solvent must lower solute concentration; evaporating solvent must raise it.
Cycle test
Repeated replacement removes less original liquid each time in absolute amount, though the same fraction of what remains is removed.
SSC mixture speed dashboard
Mark the tracked component, convert all strengths to fractions, select the model, cancel ratios, then check the mean interval.
\[\text{track}\to\text{convert}\to\text{model}\to\text{cancel}\to\text{check}\]
04DON'T RUSH
The SSC / TCS Traps — Red Flags 🚩
Distractors reverse the alligation ratio, remove pure liquid instead of mixture, or apply the survival formula without restoring volume.
Trap 1: cross-difference ratio reversed
- Lower amount receives the upper gap \(H-M\).
- Higher amount receives the lower gap \(M-L\).
- The larger amount pulls the mean closer to its own value.
Trap 2: withdrawn sample treated as pure liquid
- A well-mixed withdrawal contains every component in the current ratio.
- After the first cycle, use the current composition.
- Removing only solvent is a different operation and must be stated.
Trap 3: exponent formula used with changing volume
- \(Q_0(1-x/V)^n\) assumes volume returns to \(V\) after each cycle.
- The mixture must be uniform before each withdrawal.
- For unequal withdrawals, multiply the different survival factors.
Average of percentages taken directly
\((c_1+c_2)/2\) works only for equal quantities. Unequal amounts require a weighted mean.
Mean outside the range accepted
With positive amounts of two components, a target outside their values is impossible without another operation or component.
Ratio denominator mistaken
In solute : solvent \(=a:b\), solute concentration is \(a/(a+b)\), not \(a/b\).
05LOCK IT IN
Memory Hooks & Mnemonics
Attach each formula to a vessel, balance or survival picture.
“Concentration is part over whole”
Pure component belongs in the numerator; the entire mixture belongs in the denominator.
“Cross the gaps, not the amounts”
Lower quantity pairs with the upper gap; higher quantity pairs with the lower gap.
“More quantity pulls the mean”
The mean stays between component values and moves toward the larger amount.
“Each cycle keeps the same fraction”
Original content after \(n\) cycles is the initial content multiplied by the survival factor \(n\) times.
“Solvent changes; solute stays anchored”
When only solvent is added or evaporated, equate pure solute before and after.
“Price mixes like strength”
Total cost is additive, so price per unit follows the same weighted-average and alligation rules.
Pure content\(VC/100\)
Weighted meantotal content ÷ total quantity
Alligationopposite differences
Dilutionsolute stays constant
Replacementsurviving fraction power
Price mixturetrack total cost