4.2 Averages, Change in Average, Batsman/Bowling Averages & Replacement Problems

The Arithmetic Mean and the Assumed Mean Method

The Standard Formulation

The arithmetic mean (commonly referred to simply as the average) of a set of $n$ numerical observations $x_1, x_2, \dots, x_n$ is the ratio of their sum to the total count:

$$A = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{x_1 + x_2 + \dots + x_n}{n}$$ $$\mathbf{\sum_{i=1}^{n} x_i = n \times A \iff \text{Total Sum} = \text{Number of Observations} \times \text{Average}}$$

The Fundamental Zero-Deviation Principle

The algebraic sum of the deviations of all observations from their true arithmetic mean is always identically zero:

$$\mathbf{\sum_{i=1}^{n} (x_i - A) = 0}$$

The Assumed Mean (Deviation) Shortcut Method

When calculating the average of large, closely grouped numbers (e.g., train speeds between $78$ and $94\text{ km/h}$, or station delay minutes between $140$ and $175\text{ min}$), summing multi-digit figures introduces arithmetic error. Instead, choose a convenient central benchmark called the assumed mean ($A_0$):

1. Define deviation for each value: $d_i = x_i - A_0$.
2. Calculate the average of the small deviations: $\bar{d} = \frac{\sum d_i}{n}$.
3. Compute the true average:

$$\mathbf{A = A_0 + \frac{\sum_{i=1}^{n} d_i}{n}}$$

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Weighted Average and Sub-Group Aggregation

When combining distinct groups of observations with different sizes and averages, simple averaging of averages produces incorrect results. One must compute the weighted average ($A_w$):

$$A_w = \frac{n_1 A_1 + n_2 A_2 + \dots + n_k A_k}{n_1 + n_2 + \dots + n_k} = \frac{\sum n_i A_i}{\sum n_i}$$

Two-Group Algebraic Balance

For two sub-groups of sizes $n_1$ and $n_2$ with averages $A_1$ and $A_2$ (where $A_1 < A_w < A_2$):

$$A_w = \frac{n_1 A_1 + n_2 A_2}{n_1 + n_2}$$ $$n_1 (A_w - A_1) = n_2 (A_2 - A_w)$$ $$\mathbf{\frac{n_1}{n_2} = \frac{A_2 - A_w}{A_w - A_1}}$$

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Replacement Problems (Constant Group Size)

In railway operations, replacement scenarios occur frequently: a worn coach in a rake is replaced by a newly overhauled coach, or a retiring gangman is replaced by a young recruit. Because the total number of entities $n$ remains constant, evaluating total sums is unnecessary.

Replacement Dynamics (n is Constant):
Old Member (V_out) Leaves   ---> Group Sum Drops by V_out
New Member (V_in) Enters    ---> Group Sum Increases by V_in
Net Change in Sum = V_in - V_out = n * (Delta A)

The Replacement Theorem

Let a person of weight/value $V_{\text{out}}$ leave a group of $n$ members, and a new person of weight/value $V_{\text{in}}$ enter. If the average of the group shifts by $\Delta A = A_{\text{new}} - A_{\text{old}}$:

$$\text{Net Sum Change} = V_{\text{in}} - V_{\text{out}} = n \times \Delta A$$ $$\mathbf{V_{\text{in}} = V_{\text{out}} + (n \times \Delta A)}$$

Use the signed change $\Delta A=A_{\text{new}}-A_{\text{old}}$ directly. A decrease already makes it negative; do not subtract that negative value a second time.

Multi-Person Simultaneous Replacement

When two or more members are replaced simultaneously:

$$\mathbf{\sum V_{\text{in}} = \sum V_{\text{out}} + (n \times \Delta A)}$$ $$\text{Average of New Entrants} = \frac{\sum V_{\text{in}}}{k} \quad (k = \text{number of entrants})$$

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Addition (Inclusion) and Departure (Exclusion) Dynamics

Unlike replacement, inclusion and exclusion alter the total population count $n$.

1. Inclusion (A New Member Joins)

An initial cohort of $n$ members has an average $A_1$. A new member of value $V_{\text{new}}$ joins, increasing the cohort to $(n + 1)$ and changing the average to $A_2$:

$$\text{New Sum} = (n + 1) A_2 = n A_1 + V_{\text{new}}$$ $$V_{\text{new}} = (n + 1) A_2 - n A_1 = A_2 + n(A_2 - A_1)$$ $$\mathbf{V_{\text{new}} = A_1 + (n + 1) \times (A_2 - A_1)}$$

Shortcut Interpretation: The newcomer must provide the old average $A_1$, plus sufficient additional value to raise all $(n + 1)$ members by the increment $(A_2 - A_1)$.

2. Exclusion (A Member Leaves)

An initial cohort of $n$ members has average $A_1$. One member of value $V_{\text{left}}$ leaves, reducing the cohort to $(n - 1)$ and changing the average to $A_2$:

$$\text{New Sum} = (n - 1) A_2 = n A_1 - V_{\text{left}}$$ $$V_{\text{left}} = n A_1 - (n - 1) A_2 = A_1 + (n - 1)(A_1 - A_2)$$ $$\mathbf{V_{\text{left}} = A_1 - (n - 1) \times (A_2 - A_1)}$$

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Cricket Performance Metrics: Batting and Bowling Averages

Sports performance indices appear regularly on RRB competitive exams as practical tests of weighted averages and division nuances.

Batting Average Mechanics

A batsman's average is defined strictly as total runs scored divided by the number of times dismissed (completed innings):

$$\mathbf{\text{Batting Average} = \frac{\text{Total Runs Scored}}{\text{Total Innings Dismissed (Out)}} = \frac{\text{Total Runs}}{\text{Total Innings} - \text{Not Out Innings}}}$$

The Milestone Increment Formula

If a batsman has been dismissed $n$ times, scores $R$ in a further innings in which he is dismissed, and raises his batting average from $A$ to $(A+k)$:

$$(n + 1)(A + k) = n A + R$$ $$n A + n k + A + k = n A + R$$ $$\mathbf{R = A + (n + 1)k \iff A = R - (n + 1)k}$$

Bowling Average Mechanics

A bowler's average measures runs conceded per wicket taken:

$$\mathbf{\text{Bowling Average} = \frac{\text{Total Runs Conceded}}{\text{Total Wickets Taken}}}$$ $$\text{Bowling Strike Rate} = \frac{\text{Total Balls Bowled}}{\text{Total Wickets Taken}}, \quad \text{Economy Rate} = \frac{\text{Total Runs Conceded}}{\text{Total Overs Bowled}}$$ $$A_{\text{new}} = A_{\text{old}} - 0.4$$

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Kinematic Speed Averages: Harmonic vs. Arithmetic Mean

When a train travels across multiple track segments at varying speeds, the overall average speed depends on whether the segments share equal distances or equal travel times.

Case 1: Equal Distances Traversed (Harmonic Mean)

When a train traverses two equal distances $d$ at speeds $v_1$ and $v_2$:

$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}} = \mathbf{\frac{2 v_1 v_2}{v_1 + v_2}}$$

For three equal distances $d$ at speeds $v_1, v_2, v_3$:

$$\mathbf{S_{\text{avg}} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}}$$

Case 2: Equal Time Durations Traversed (Arithmetic Mean)

When a train runs for equal time durations $t$ at speeds $v_1$ and $v_2$:

$$\text{Average Speed} = \frac{v_1 t + v_2 t}{2t} = \mathbf{\frac{v_1 + v_2}{2}}$$

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Comprehensive Worked Operational Problems

Worked Problem 1: Replacement in Sectional Track Maintenance Gang

Scenario: A permanent way maintenance unit comprises 24 track gangmen. When one senior gangman weighing 58 kg retires and is replaced by a newly recruited track maintainer, the average weight of the 24-member gang increases by 1.25 kg. Calculate the weight of the newly recruited track maintainer.

Step-by-Step Mathematical Resolution:

1. Identify the Given Parameters:

   • Cohort size ($n$) $= 24$
   • Weight of retiring member ($V_{\text{out}}$) $= 58\text{ kg}$
   • Change in average ($\Delta A$) $= +1.25\text{ kg}$

2. Apply the Replacement Theorem:

$$V_{\text{in}} = V_{\text{out}} + (n \times \Delta A)$$ $$V_{\text{in}} = 58 + (24 \times 1.25)$$

1. Compute the Product:

$$24 \times 1.25 = 24 \times \frac{5}{4} = 6 \times 5 = 30\text{ kg}$$

1. Calculate Final Weight:

$$V_{\text{in}} = 58 + 30 = 88\text{ kg}$$

1. Verification via Total Sum: Let old sum be $S$. Old average $A = \frac{S}{24}$. New sum $S' = S - 58 + 88 = S + 30$. New average $A' = \frac{S + 30}{24} = A + \frac{30}{24} = A + 1.25\text{ kg (verified)}$.

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Worked Problem 2: Batsman Milestone Century Inning

Scenario: In a railway zonal inter-divisional cricket tournament, a locomotive pilot plays his $17^{\text{th}}$ tournament inning and scores 85 runs. This knock increases his career batting average by 3 runs per inning. Assuming he was dismissed in all 17 innings, calculate:

1. His batting average prior to the $17^{\text{th}}$ inning.
2. His batting average after the $17^{\text{th}}$ inning.

Step-by-Step Mathematical Resolution:

1. Define the Variables: Let the average after 16 completed innings be $A$.

   • Total runs after 16 innings $= 16 A$
   • Runs in $17^{\text{th}}$ inning $= 85$
   • New average after 17 innings $= A + 3$
   • Total runs after 17 innings $= 17(A + 3)$

2. Formulate the Algebraic Equality:

$$17(A + 3) = 16 A + 85$$ $$17 A + 51 = 16 A + 85$$

1. Isolate $A$:

$$17 A - 16 A = 85 - 51$$ $$A = 34\text{ runs per inning (prior average)}$$

1. Compute New Batting Average:

$$A_{\text{new}} = A + 3 = 34 + 3 = 37\text{ runs per inning}$$

1. Rapid Exam Shortcut Verification:

$$A_{\text{new}} = \text{Latest Score} - (n_{\text{prior}} \times \Delta A)$$ $$A_{\text{new}} = 85 - (16 \times 3) = 85 - 48 = 37$$ $$A_{\text{old}} = 37 - 3 = 34$$

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Worked Problem 3: Locomotive Speed Harmonic Mean Over Sectional Loop Runs

Scenario: A diesel-electric locomotive runs an inspection circuit divided into three equal consecutive track sections of 120 km each. The locomotive traverses the first section at $60\text{ km/h}$, the second section at $40\text{ km/h}$, and the third section at $30\text{ km/h}$. Calculate the average speed of the locomotive across the entire 360 km journey.

Step-by-Step Mathematical Resolution:

1. Identify the Equal Distance Condition: Because all three sections are of equal length ($d = 120\text{ km}$), the average speed is the harmonic mean of the three speeds:

$$v_1 = 60\text{ km/h}, \quad v_2 = 40\text{ km/h}, \quad v_3 = 30\text{ km/h}$$

1. Apply the 3-Way Harmonic Mean Formula:

$$S_{\text{avg}} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} = \frac{3}{\frac{1}{60} + \frac{1}{40} + \frac{1}{30}}$$

1. Determine the Common Denominator:

$$\text{LCM}(60, 40, 30) = 120$$ $$\frac{1}{60} + \frac{1}{40} + \frac{1}{30} = \frac{2 + 3 + 4}{120} = \frac{9}{120} = \frac{3}{40}$$

1. Compute Final Harmonic Speed:

$$S_{\text{avg}} = \frac{3}{\frac{3}{40}} = 3 \times \frac{40}{3} = 40\text{ km/h}$$

1. Verification via Time Summation:
   • Time for Section 1: $t_1 = \frac{120}{60} = 2\text{ hours}$
   • Time for Section 2: $t_2 = \frac{120}{40} = 3\text{ hours}$
   • Time for Section 3: $t_3 = \frac{120}{30} = 4\text{ hours}$
   • Total Time: $2 + 3 + 4 = 9\text{ hours}$
   • Total Distance: $120 + 120 + 120 = 360\text{ km}$
   • Average Speed: $\frac{360}{9} = 40\text{ km/h (exact match)}$.

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Common Exam Traps & Operational Pitfalls