6  Averages, Mixtures and Alligations

6.1 Introduction

The concept of averages and mixtures is fundamental in quantitative aptitude.
- Average represents the “equal distribution” of a total.
- Mixture problems combine items of different values (price, concentration, speed, marks, etc.) into one mean value.
- Alligation is a powerful shortcut for mixture questions, saving time in exams.

We will build from basics → advanced applications → exam-style practice.

6.2 Average Basics

6.2.1 Definition

If \(n\) numbers are \(x_1,x_2,\dots,x_n\), then
\[ \text{Average} = \frac{x_1+x_2+\cdots+x_n}{n}. \]

  • Total = Average × Number of items.
  • If one value changes, average changes proportionally.

6.2.2 Properties

  • If all numbers increase/decrease by \(k\), average increases/decreases by \(k\).
  • If all numbers multiplied/divided by \(k\), average is multiplied/divided by \(k\).
  • Average lies between the smallest and largest of the data.
  • If average = \(a\), and total numbers \(n\), then sum = \(an\).

6.3 Weighted Average

When items have different frequencies/weights,
\[ \text{Weighted Average} = \frac{w_1x_1+w_2x_2+\cdots+w_nx_n}{w_1+w_2+\cdots+w_n}. \]

Example: Average marks of two groups:
Group A = 30 students avg 60, Group B = 20 students avg 75.
Overall avg = \(\frac{30\cdot60+20\cdot75}{30+20}=\frac{1800+1500}{50}=66\).

6.4 Replacement Problems

If a number in the group is replaced,
Change in total = (New − Old).
Change in average = (Change in total)/n.

Example: Avg age of 10 students = 15. One student aged 20 replaced by new student. Avg becomes 14.5. New student’s age?
Old sum=150. New sum=145. Change=−5. New age=20+(−5)=15.

6.5 Mixtures and Alligation

6.5.1 Mixture Average

If two ingredients A and B are mixed in ratio \(m:n\) with values \(V_A\) and \(V_B\), then
\[ \text{Mean Value} = \frac{mV_A+nV_B}{m+n}. \]

6.5.2 Alligation Rule

Shortcut for mixtures:
If two varieties with values \(V_1\) and \(V_2\) form a mean \(M\),
then ratio = \((V_2-M):(M-V_1)\).

Diagram (alligation cross):

6.5.3 Examples

Ex 1 (Basic mixture)
Mix 2 kg rice @ 30/kg with 3 kg @ 40/kg. Avg price = (2×30+3×40)/5=36.

Ex 2 (Alligation)
Mix water (0%) with milk (100%) to get 80% milk. Ratio water:milk=(100−80):(80−0)=20:80=1:4.

Ex 3 (Profit mixture)
Two tea types at 30 and 40 per kg mixed to sell at 35. Ratio=(40−35):(35−30)=5:5=1:1.

Ex 4 (Alcohol solution)
5 L of 60% alcohol mixed with 15 L of 20% alcohol. New % = (5×60+15×20)/20=30%.

6.6 Successive Replacement (Container Problems)

When a solution is partially replaced multiple times:
If a container of volume \(V\) has solution, and each time \(R\) is replaced, after \(n\) operations,
\[ \text{Final quantity} = V\Big(1-\frac{R}{V}\Big)^n. \]

Example: Container 20 L milk. Each time 5 L replaced with water, repeated 2 times. Milk left=20×(1−5/20)2=20×(0.75)2=11.25 L.

6.7 Speed/Work/Marks via Average

  • Average Speed (two journeys equal distance):
    \[\text{Avg Speed}=\frac{2xy}{x+y}\]
    (harmonic mean).

  • Work rate average: If A completes in a days, B in b days, then daily work = (1/a+1/b).

  • Average marks: Weighted averages across groups.

6.8 Solved Examples

  1. Find average of 20, 30, 50, 80.
    = (20+30+50+80)/4=180/4=45.

  2. Average of 40 students=50. A new student joins, avg=51. New student’s marks?
    Old total=2000. New total=2091. New student=91.

  3. A mixture of milk 70% and water 30% is obtained by mixing milk 80% with water 20%. Find ratio.
    By alligation: (80−70):(70−20)=10:50=1:5.

  4. Container 40 L has 25% milk. How much milk added to make 50% milk?
    Milk=10 L. Want 20 L. Add 10 L milk.

  5. Average speed: 60 km/h one way, 40 km/h return. Avg=2×60×40/(60+40)=48 km/h.

  6. A mixture of 2 kinds of sugar worth 25/kg and 35/kg sold at 30/kg with 20% profit. Find ratio.
    Effective CP=30/1.2=25. Ratio=(35−25):(25−25)=10:0 → Only 25/kg sugar used.

6.9 Common Traps

  • Confusing arithmetic mean with harmonic mean.
  • Forgetting to multiply weights in weighted average.
  • Misusing alligation (always subtract diagonally).
  • In replacement, remember change applies to total, not just one value.
  • In container problems, ratio repeats exponentially.

6.10 Practice Set – Level 1

  1. Average of 6 numbers is 8. Total?
  2. Add 80 to 5 numbers avg 10. New avg?
  3. Marks of 3 subjects=50, 60, 70. Find avg.
  4. Average of 40 students=35. One absent with 20 marks. Correct avg?
  5. Divide 300 into 3 parts with avg 30,40,50.

6.11 Practice Set – Level 2

  1. Avg of 20 numbers=25. If 18 numbers total 450, find last 2.
  2. A container 100 L has 40% milk. 20 L removed, replaced with water. New %?
  3. Avg of 10 numbers=20. If each +5, new avg?
  4. Tea 30/kg, 40/kg mixed to get 35/kg. Ratio?
  5. A car goes 40 km/h, returns 60 km/h. Avg speed?

6.12 Practice Set – Level 3

  1. A:B:C share profit in ratio of investment 2:3:4. Profit=18,000. Find shares.
  2. 2 alloys with gold 80% and 60% mixed to get 70%. Ratio?
  3. A container has 50 L pure milk. Each time 10 L replaced by water, repeated thrice. Milk left?
  4. A person travels at 12 km/h for half distance, 6 km/h for half time. Find avg speed.
  5. Two classes: 30 students avg 60, 20 students avg 70. Combined avg?

6.13 Answer Key (Outline)

Level 1: 1) 48; 2) 22; 3) 60; 4) 35.38; 5) 90,120,90.
Level 2: 6) 25,25; 7) 32%; 8) 25; 9) 1:1; 10) 48.
Level 3: 11) 4000,6000,8000; 12) 1:1; 13) 27.34 L; 14) 8 km/h; 15) 64.

6.14 Summary

  • Average = Total ÷ Number of items.
  • Weighted average = sum of weighted values ÷ total weights.
  • Alligation rule: shortcut ratio from mean value.
  • Successive replacement → exponential formula.
  • Average speed = harmonic mean.
  • Applications: mixtures, marks, partnership, container problems.

Mastering averages, mixtures, and alligation equips you to crack a wide variety of aptitude questions with speed and accuracy.