Official Syllabus Item: 8
Commercial Mathematics
Averages, Arithmetic Mean & Weighted Aggregates
Master Arithmetic Mean, sum invariant laws, inclusion/exclusion/replacement formulas, misread data corrections, cricket averages, and average speed for RIMC Entrance Exam.
Overview & Fundamental Concept
The concept of Average (Arithmetic Mean) is a fundamental measure of central tendency and an essential problem-solving weapon in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 8 of Chapter 8 in the official RIMC brochure, questions on averages feature consistently in Part A (5 Marks) testing natural number series, data deviations, and missing observation deductions, as well as in Part B (10 Marks) testing multi-tiered group dynamics: inclusion of instructors, crew member replacements, examination data corrections, weighted class aggregates, cricket batting/bowling statistics, and harmonic average speeds. In the descriptive exam booklet, RIMC examiners demand formal declarations of total sums ($S = N \times \bar{X}$), explicit arithmetic step balancing, tabular organization of before-and-after scenarios, and boxed final answers with standard physical units.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 8 | Average: Arithmetic mean, sum invariant property, group changes, data corrections, and applied word problems | Core commercial arithmetic syllabus item. Assesses aggregate balancing, statistical corrections, dynamic population tracking, and weighted metrics. |
1. Definition, Conceptual Foundation & The Invariant Sum Principle
The Average (or Arithmetic Mean) of a set of numerical observations is the single representative value obtained by dividing the sum of all observations by the total count of observations:
$$\text{Average } (\bar{X}) = \frac{\text{Sum of all observations } (\sum x_i)}{\text{Total number of observations } (N)}$$
The Invariant Sum Principle: In descriptive problem solving, the total sum of the values is the primary bridge connecting initial and altered group states:
$$\text{Sum of Observations } (S) = \text{Average } (\bar{X}) \times \text{Number of Observations } (N)$$
Whenever members are added, removed, or replaced, calculate the Total Sum Before and the Total Sum After. The difference between these two sums reveals the unknown quantity directly.
The Four Fundamental Algebraic Invariance Theorems
- Addition Theorem: If each observation is increased by a constant $k$, the new average increases by $k$: $\bar{X}_{\text{new}} = \bar{X} + k$.
- Subtraction Theorem: If each observation is decreased by a constant $k$, the new average decreases by $k$: $\bar{X}_{\text{new}} = \bar{X} - k$.
- Multiplication Theorem: If each observation is multiplied by a non-zero constant $k$, the new average is multiplied by $k$: $\bar{X}_{\text{new}} = k \times \bar{X}$.
- Division Theorem: If each observation is divided by a non-zero constant $k$, the new average is divided by $k$: $\bar{X}_{\text{new}} = \frac{\bar{X}}{k}$.
Special Averages of Natural Number Series (Direct RIMC Shortcuts)
- First $n$ Natural Numbers: $\text{Average} = \frac{n + 1}{2}$. (e.g., average of first 99 natural numbers $= \frac{99 + 1}{2} = 50$).
- First $n$ Even Numbers: $\text{Average} = n + 1$. (e.g., average of first 50 even numbers $= 51$).
- First $n$ Odd Numbers: $\text{Average} = n$. (e.g., average of first 40 odd numbers $= 40$).
- Any Arithmetic Progression (AP): $\text{Average} = \frac{\text{First Term} + \text{Last Term}}{2}$.
2. Group Dynamics: Inclusion, Exclusion & Replacement
A large portion of RIMC Part B questions involve dynamic group adjustments:
| Dynamic Event | Mathematical Mechanics | Descriptive Exam Formula |
|---|---|---|
| Inclusion (New person joins) | Count increases from $N$ to $N + 1$. | $$\text{Value of New Member} = \text{New Average} + [N_{\text{old}} \times (\bar{X}_{\text{new}} - \bar{X}_{\text{old}})]$$ |
| Exclusion (Person leaves) | Count decreases from $N$ to $N - 1$. | $$\text{Value of Excluded Member} = \text{Old Sum} - \text{New Sum}$$ |
| Replacement (One swapped for another) | Total count $N$ remains strictly unchanged. | $$\text{Value of New} = \text{Value of Replaced} \pm [N \times \Delta \bar{X}]$$ ($+$ if average increases, $-$ if average decreases) |
3. Misread Data Corrections & Statistical Audits
In examination evaluation questions, certain marks or weights are inadvertently misread (e.g., reading $83$ instead of $38$). The protocol for rectifying such data is:
$$\text{True Sum} = \text{Incorrect Sum} - \text{Incorrect Reading} + \text{Correct Reading}$$
$$\text{Correct Average} = \frac{\text{True Sum}}{N}$$
Alternatively, using net deviation: $\text{Correct Average} = \text{Incorrect Average} + \frac{\text{Correct} - \text{Incorrect}}{N}$.
4. Weighted Aggregates, Cricket Statistics & Average Speed
Advanced applied models in RIMC require specific definitions:
Weighted / Combined Average
If group 1 has $n_1$ members with mean $\bar{x}_1$ and group 2 has $n_2$ members with mean $\bar{x}_2$:
- $$\bar{X}_{\text{combined}} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$
Cricket Averages
- Batting Average: $\text{Batting Average} = \frac{\text{Total Runs Scored}}{\text{Number of Times Dismissed (Innings - Not Outs)}}$.
- Bowling Average: $\text{Bowling Average} = \frac{\text{Total Runs Conceded}}{\text{Total Wickets Taken}}$. (A lower bowling average indicates superior performance).
Average Speed (Harmonic Mean Principle)
- $$\text{Average Speed} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$$
- If a journey covers two equal distances at speeds $u\text{ km/h}$ and $v\text{ km/h}$, the average speed is NOT $\frac{u+v}{2}$, but the Harmonic Mean:
$$\text{Average Speed} = \frac{2uv}{u + v}$$
Illustrated Conceptual Diagrams & Visual Schematics
The Balance Beam Principle of Average & Speed Mean Architecture
Average (Arithmetic Mean): $\text{Average} = \frac{\text{Sum of Observations}}{\text{Number of Observations}}$. Sum of deviations from mean is always ZERO ($\sum (x_i - \bar{x}) = 0$). For equal distance round-trips at speeds $u$ and $v$, the harmonic Average Speed is $\frac{2uv}{u+v}$ (never arithmetic mean).
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: Declare Group Parameters & Initial Sum
1 - 2 Marks
State the number of observations $N_1$ and initial average $\bar{X}_1$, and compute the initial total sum $S_1 = N_1 \times \bar{X}_1$.
Booklet Layout:
Given:\nNumber of students $N_1 = 30$\nInitial average age $\bar{X}_1 = 14\text{ years}$\n$$\text{Total Age of 30 Students } S_1 = 30 \times 14 = 420\text{ years}$$
Given:\nNumber of students $N_1 = 30$\nInitial average age $\bar{X}_1 = 14\text{ years}$\n$$\text{Total Age of 30 Students } S_1 = 30 \times 14 = 420\text{ years}$$
Step 2: Transition Mechanics & New Total Sum
2 - 3 Marks
State the new count $N_2$ and new average $\bar{X}_2$, and compute the new total sum $S_2 = N_2 \times \bar{X}_2$.
Booklet Layout:
When the teacher's age is included:\nNew number of persons $N_2 = 30 + 1 = 31$\nNew average age $\bar{X}_2 = 14 + 1 = 15\text{ years}$\n$$\text{New Total Age } S_2 = 31 \times 15 = 465\text{ years}$$
When the teacher's age is included:\nNew number of persons $N_2 = 30 + 1 = 31$\nNew average age $\bar{X}_2 = 14 + 1 = 15\text{ years}$\n$$\text{New Total Age } S_2 = 31 \times 15 = 465\text{ years}$$
Step 3: Difference Formulation & Unknown Isolation
2 - 3 Marks
Formulate the algebraic difference between sums to isolate the unknown member's value.
Booklet Layout:
$$\text{Age of Teacher} = S_2 - S_1 = 465 - 420 = 45\text{ years}$$
$$\text{Age of Teacher} = S_2 - S_1 = 465 - 420 = 45\text{ years}$$
Step 4: Boxed Final Value with Proper Units
1 Mark
State the concluding answer in a clear sentence with standard physical units enclosed in a box.
Booklet Layout:
$$\mathbf{\text{Answer: The teacher is } 45\text{ years old.}}$$
$$\mathbf{\text{Answer: The teacher is } 45\text{ years old.}}$$
Examiner Traps & Common Blunders (Marks Penalties)
Average Speed Simple Arithmetic Mean Trap -3 to -4 Marks
Common Mistake: Computing average speed for a round trip at $40\text{ km/h}$ going and $60\text{ km/h}$ returning as $\frac{40 + 60}{2} = 50\text{ km/h}$.
Cadet Antidote: Average speed is Total Distance / Total Time. For equal distances, use harmonic mean: $\frac{2(40)(60)}{40 + 60} = \frac{4800}{100} = 48\text{ km/h}$.
Miscounting Total Members After Inclusion/Exclusion -2 Marks
Common Mistake: Multiplying new average by $30$ instead of $31$ when a teacher joins $30$ students.
Cadet Antidote: Always explicitly write $N_2 = N_1 + 1$ before multiplying.
Forgetting to Divide by Total Count in Weighted Averages -2 Marks
Common Mistake: Averaging the averages directly: $\frac{75 + 85}{2} = 80$ when section A has 40 students and section B has 60 students.
Cadet Antidote: Compute weighted sum $\frac{40(75) + 60(85)}{40 + 60} = \frac{3000 + 5100}{100} = 81$.
Cricket Batting Invariant with 'Not Out' Innings -2 to -3 Marks
Common Mistake: Dividing total runs by total innings played when the batsman remained 'not out' in certain innings.
Cadet Antidote: Batting average divisor is the number of COMPLETED DISMISSALS, not total matches played.
Cadet Topper Mental Verification Shortcuts
The 1-Line Replacement Formula
When 1 person replaces another in a group of $N$: $\text{New} = \text{Replaced} + (N \times \Delta \bar{X})$.
In a crew of 8, a man weighing 60 kg is replaced, raising the average by 1.5 kg: $\text{New} = 60 + (8 \times 1.5) = 60 + 12 = 72\text{ kg}$ in 2 seconds!
Net Deviation Shortcut for Misread Values
Instead of recalculating the entire sum: $\bar{X}_{\text{new}} = \bar{X}_{\text{old}} + \frac{\text{Correct} - \text{Incorrect}}{N}$.
Average of 50 marks was 40. Mark 83 was misread as 38: $\bar{X} = 40 + \frac{83 - 38}{50} = 40 + \frac{45}{50} = 40.9$ instantly.
Consecutive AP Middle Element Law
For any odd number of consecutive integers or AP terms, the average is EXACTLY the median (middle) term.
Average of 5 consecutive odd numbers is 27. The numbers are: $23, 25, \mathbf{27}, 29, 31$.
Key Formulas & Conversion Cheat Sheet
Basic Average Formula
$$\bar{X} = \frac{\sum x_i}{N}$$
Arithmetic mean equals sum of observations divided by count.
Sum of Observations
$$S = \bar{X} \times N$$
Total sum equals average times observation count.
Group Inclusion Formula
$$X_{\text{new}} = \bar{X}_{\text{new}} + [N_{\text{old}} \times (\bar{X}_{\text{new}} - \bar{X}_{\text{old}})]$$
Calculates incoming member's value directly.
Group Replacement Formula
$$X_{\text{incoming}} = X_{\text{outgoing}} \pm (N \times \Delta \bar{X})$$
Direct relationship between replaced and replacing member.
Corrected Average Formula
$$\bar{X}_{\text{correct}} = \frac{S_{\text{incorrect}} - \text{Wrong} + \text{Right}}{N}$$
Corrects arithmetic mean after misread data.
Combined Weighted Average
$$\bar{X}_{\text{combined}} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2}$$
Merges two separate groups with known sizes and means.
Harmonic Average Speed Formula
$$v_{\text{avg}} = \frac{2uv}{u + v}$$
Average speed for round trip of equal distance at speeds u and v.
Average of First n Natural Numbers
$$\bar{X}_{\mathbb{N}} = \frac{n + 1}{2}$$
Arithmetic mean of the first n positive integers.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Teacher Inclusion in Class Average Age
5 Marks
The average age of $30\text{ students}$ in a class is $14\text{ years}$. If the class teacher's age is included, the average age of the group increases by $1\text{ year}$. Find the age of the class teacher.
Step-by-Step Model Solution:
Step 1: Compute Total Age of Students
2 Marks
Number of students $N_1 = 30$.
Average age of students $\bar{X}_1 = 14\text{ years}$.
$$\text{Total Age of 30 Students } S_1 = 30 \times 14 = 420\text{ years}$$
Average age of students $\bar{X}_1 = 14\text{ years}$.
$$\text{Total Age of 30 Students } S_1 = 30 \times 14 = 420\text{ years}$$
Step 2: Compute New Total Age with Teacher
2 Marks
Total persons including teacher $N_2 = 30 + 1 = 31$.
New average age $\bar{X}_2 = 14 + 1 = 15\text{ years}$.
$$\text{New Total Age } S_2 = 31 \times 15 = 465\text{ years}$$
New average age $\bar{X}_2 = 14 + 1 = 15\text{ years}$.
$$\text{New Total Age } S_2 = 31 \times 15 = 465\text{ years}$$
Step 3: Calculate Teacher's Age & Conclude
1 Mark
$$\text{Teacher's Age} = S_2 - S_1 = 465 - 420 = \mathbf{45\text{ years}}$$
$$\mathbf{\text{Answer: The class teacher is } 45\text{ years old.}}$$
$$\mathbf{\text{Answer: The class teacher is } 45\text{ years old.}}$$
Examiner Scoring Key: Award 2 marks for S1 = 420, 2 marks for S2 = 465, and 1 mark for the boxed final answer of 45 years.
Model 2: Correcting Misread Data in Examination Marks
5 Marks
The average marks of $40\text{ cadets}$ in an assessment was found to be $68$. Later, it was discovered that a mark of $53$ was misread as $83$. Find the correct average marks.
Step-by-Step Model Solution:
Step 1: Compute Incorrect Total Marks
2 Marks
Total cadets $N = 40$.
Incorrect average marks $= 68$.
$$\text{Incorrect Total Marks } S_{\text{inc}} = 40 \times 68 = 2,720$$
Incorrect average marks $= 68$.
$$\text{Incorrect Total Marks } S_{\text{inc}} = 40 \times 68 = 2,720$$
Step 2: Correct the Sum
1.5 Marks
Incorrect entry $= 83$. Correct entry $= 53$.
$$\text{Correct Total Marks } S_{\text{corr}} = 2720 - 83 + 53 = 2720 - 30 = 2,690$$
$$\text{Correct Total Marks } S_{\text{corr}} = 2720 - 83 + 53 = 2720 - 30 = 2,690$$
Step 3: Compute Correct Average Marks
1.5 Marks
$$\text{Correct Average} = \frac{S_{\text{corr}}}{N} = \frac{2690}{40} = \frac{269}{4} = \mathbf{67.25}$$
$$\mathbf{\text{Answer: The correct average mark is } 67.25}$$
$$\mathbf{\text{Answer: The correct average mark is } 67.25}$$
Examiner Scoring Key: Subtracting 83 and adding 53 to get 2690 carries 1.5 marks. Division by 40 to get 67.25 carries 1.5 marks.
Model 3: Crew Member Replacement
7 Marks
The average weight of an $8\text{-cadet}$ military rowing crew increases by $1.5\text{ kg}$ when one of the crew members, who weighs $60\text{ kg}$, is replaced by a new recruit. Find the weight of the new recruit.
Step-by-Step Model Solution:
Step 1: Declare Group Parameters & Initial Total
2 Marks
Let original average weight of 8 cadets be $\bar{X}\text{ kg}$.
$$\text{Total Weight of 8 Cadets } S_1 = 8\bar{X}\text{ kg}$$
$$\text{Total Weight of 8 Cadets } S_1 = 8\bar{X}\text{ kg}$$
Step 2: Formulate New Total Weight
2.5 Marks
Let weight of new recruit be $W\text{ kg}$.
When the $60\text{ kg}$ member is replaced by the new recruit:
$$\text{New Total Weight } S_2 = 8\bar{X} - 60 + W$$
Also, given that the new average increases by $1.5\text{ kg}$:
$$\text{New Average} = \bar{X} + 1.5$$
$$\text{New Total Weight } S_2 = 8(\bar{X} + 1.5) = 8\bar{X} + 12$$
When the $60\text{ kg}$ member is replaced by the new recruit:
$$\text{New Total Weight } S_2 = 8\bar{X} - 60 + W$$
Also, given that the new average increases by $1.5\text{ kg}$:
$$\text{New Average} = \bar{X} + 1.5$$
$$\text{New Total Weight } S_2 = 8(\bar{X} + 1.5) = 8\bar{X} + 12$$
Step 3: Equate Expressions & Solve for W
2.5 Marks
$$8\bar{X} - 60 + W = 8\bar{X} + 12$$
Subtract $8\bar{X}$ from both sides:
$$-60 + W = 12 \implies W = 60 + 12 = \mathbf{72\text{ kg}}$$
$$\mathbf{\text{Answer: The weight of the new recruit is } 72\text{ kg.}}$$
Subtract $8\bar{X}$ from both sides:
$$-60 + W = 12 \implies W = 60 + 12 = \mathbf{72\text{ kg}}$$
$$\mathbf{\text{Answer: The weight of the new recruit is } 72\text{ kg.}}$$
Examiner Scoring Key: Setting up 8X - 60 + W = 8X + 12 cancels 8X and yields W = 72 kg directly. Award full 7 marks.
Model 4: Combined Weighted Average of Two Squadrons
7 Marks
In an academy, Section A has $36\text{ cadets}$ with an average score of $75\%$ in mathematics, while Section B has $44\text{ cadets}$ with an average score of $85\%$. Find the overall combined average score of both sections taken together.
Step-by-Step Model Solution:
Step 1: Compute Total Aggregate Scores for Each Section
3 Marks
For Section A:
Number of cadets $n_1 = 36$, Mean score $\bar{X}_1 = 75$.
$$\text{Total Score of Section A } = 36 \times 75 = 2,700$$
For Section B:
Number of cadets $n_2 = 44$, Mean score $\bar{X}_2 = 85$.
$$\text{Total Score of Section B } = 44 \times 85 = 3,740$$
Number of cadets $n_1 = 36$, Mean score $\bar{X}_1 = 75$.
$$\text{Total Score of Section A } = 36 \times 75 = 2,700$$
For Section B:
Number of cadets $n_2 = 44$, Mean score $\bar{X}_2 = 85$.
$$\text{Total Score of Section B } = 44 \times 85 = 3,740$$
Step 2: Compute Combined Total Score and Total Cadets
2 Marks
$$\text{Combined Total Score} = 2700 + 3740 = 6,440$$
$$\text{Combined Number of Cadets} = n_1 + n_2 = 36 + 44 = 80$$
$$\text{Combined Number of Cadets} = n_1 + n_2 = 36 + 44 = 80$$
Step 3: Calculate Combined Average & Conclude
2 Marks
$$\bar{X}_{\text{combined}} = \frac{6440}{80} = \frac{644}{8} = 80.5\%$$
$$\mathbf{\text{Answer: The combined average score of both sections is } 80.5\%.}$$
$$\mathbf{\text{Answer: The combined average score of both sections is } 80.5\%.}$$
Examiner Scoring Key: Calculating 36 * 75 = 2700 and 44 * 85 = 3740 carries 3 marks. Division by 80 to get 80.5% carries 2 marks.
Model 5: Cricket Batsman Innings Average Target
10 Marks
A cricket batsman has a certain average of runs for $16\text{ innings}$. In the $17\text{th inning}$, he scores $85\text{ runs}$, thereby increasing his batting average by $3\text{ runs}$. What is his new batting average after the $17\text{th inning}$?
Step-by-Step Model Solution:
Step 1: Declare Variable for Original Average
2.5 Marks
Let the batsman's batting average after $16\text{ innings}$ be $x$ runs.
$$\text{Total Runs in 16 Innings} = 16 \times x = 16x$$
$$\text{Total Runs in 16 Innings} = 16 \times x = 16x$$
Step 2: Formulate Total Runs after 17 Innings
2.5 Marks
Runs scored in $17\text{th inning} = 85$.
$$\text{Total Runs in 17 Innings} = 16x + 85$$
Also, his average after $17\text{ innings}$ increased by $3\text{ runs}$:
$$\text{New Average} = (x + 3)\text{ runs}$$
$$\text{Total Runs in 17 Innings} = 17(x + 3) = 17x + 51$$
$$\text{Total Runs in 17 Innings} = 16x + 85$$
Also, his average after $17\text{ innings}$ increased by $3\text{ runs}$:
$$\text{New Average} = (x + 3)\text{ runs}$$
$$\text{Total Runs in 17 Innings} = 17(x + 3) = 17x + 51$$
Step 3: Equate Expressions & Solve for Original Average x
3 Marks
$$17x + 51 = 16x + 85$$
Transpose variable and constants:
$$17x - 16x = 85 - 51 \implies x = 34\text{ runs}$$
Thus, his average after $16\text{ innings}$ was $34\text{ runs}$.
Transpose variable and constants:
$$17x - 16x = 85 - 51 \implies x = 34\text{ runs}$$
Thus, his average after $16\text{ innings}$ was $34\text{ runs}$.
Step 4: Compute New Average & Conclude
2 Marks
$$\text{New Average after 17 Innings} = x + 3 = 34 + 3 = \mathbf{37\text{ runs}}$$
Check: $16(34) = 544$. Total with 17th inning: $544 + 85 = 629$. New average: $629 / 17 = 37$ (Matches perfectly!).
$$\mathbf{\text{Answer: His new average after the } 17\text{th inning is } 37\text{ runs.}}$$
Check: $16(34) = 544$. Total with 17th inning: $544 + 85 = 629$. New average: $629 / 17 = 37$ (Matches perfectly!).
$$\mathbf{\text{Answer: His new average after the } 17\text{th inning is } 37\text{ runs.}}$$
Examiner Scoring Key: Do not stop at x = 34. The question specifically asked for his NEW average (x + 3 = 37). Stating 37 earns full 10 marks.
Model 6: Harmonic Average Speed Round Trip
10 Marks
A military transport convoy travels from Depot A to Base B at an average speed of $40\text{ km/h}$, and immediately returns from Base B to Depot A along the exact same route at an average speed of $60\text{ km/h}$.
(a) Calculate the average speed of the convoy for the entire round trip.
(b) Explain why this average speed is strictly less than the simple arithmetic mean of the two speeds.
(a) Calculate the average speed of the convoy for the entire round trip.
(b) Explain why this average speed is strictly less than the simple arithmetic mean of the two speeds.
Step-by-Step Model Solution:
Step 1: Declare Variable for One-Way Distance
2.5 Marks
Let the one-way distance between Depot A and Base B be $D\text{ km}$.
Total distance for the round trip $= D + D = 2D\text{ km}$.
Total distance for the round trip $= D + D = 2D\text{ km}$.
Step 2: Compute Times Taken for Outward and Return Journeys
2.5 Marks
Time taken from A to B: $t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{D}{40}\text{ hours}$.
Time taken from B to A: $t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{D}{60}\text{ hours}$.
Total time for the round trip $T = t_1 + t_2$:
$$\text{LCM}(40, 60) = 120 \implies T = \frac{3D + 2D}{120} = \frac{5D}{120} = \frac{D}{24}\text{ hours}$$
Time taken from B to A: $t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{D}{60}\text{ hours}$.
Total time for the round trip $T = t_1 + t_2$:
$$\text{LCM}(40, 60) = 120 \implies T = \frac{3D + 2D}{120} = \frac{5D}{120} = \frac{D}{24}\text{ hours}$$
Step 3: Calculate Average Speed (Part a)
2.5 Marks
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2D}{\frac{D}{24}} = 2D \times \frac{24}{D} = 2 \times 24 = \mathbf{48\text{ km/h}}$$
Step 4: Pedagogical Explanation (Part b)
2.5 Marks
The simple arithmetic mean of the two speeds is $\frac{40 + 60}{2} = 50\text{ km/h}$.
However, the convoy spent more time travelling at the slower speed of $40\text{ km/h}$ ($t_1 = \frac{D}{40}$) than at the faster speed of $60\text{ km/h}$ ($t_2 = \frac{D}{60}$). Because speed is distance divided by time, the slower speed is weighted more heavily in the time domain, pulling the true average speed down to the harmonic mean of $\mathbf{48\text{ km/h}}$.
However, the convoy spent more time travelling at the slower speed of $40\text{ km/h}$ ($t_1 = \frac{D}{40}$) than at the faster speed of $60\text{ km/h}$ ($t_2 = \frac{D}{60}$). Because speed is distance divided by time, the slower speed is weighted more heavily in the time domain, pulling the true average speed down to the harmonic mean of $\mathbf{48\text{ km/h}}$.
Examiner Scoring Key: Showing Total Time = D/24 carries 2.5 marks. Deriving Average Speed = 48 km/h carries 2.5 marks. The time-weighting explanation carries 2.5 marks.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
Find the average of the first $10\text{ prime numbers}$.
Model Descriptive Solution:
Step 1: List the Primes
2 Marks
The first 10 prime numbers are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$.
Step 2: Sum & Divide by 10
3 Marks
$$\text{Sum} = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129$$
$$\text{Average} = \frac{129}{10} = 12.9$$
$$\text{Average} = \frac{129}{10} = 12.9$$
Final Answer: $12.9$
Question 2
5 Marks
The average of $5\text{ numbers}$ is $27$. If one number is excluded, the average becomes $25$. Find the excluded number.
Model Descriptive Solution:
Step 1: Compute Initial and Altered Sums
2.5 Marks
Initial Sum of 5 numbers $= 5 \times 27 = 135$.
Sum of 4 remaining numbers $= 4 \times 25 = 100$.
Sum of 4 remaining numbers $= 4 \times 25 = 100$.
Step 2: Calculate Excluded Number
2.5 Marks
$$\text{Excluded Number} = 135 - 100 = 35$$
Final Answer: $35$
Question 3
5 Marks
Find the average of all odd numbers between $20$ and $40$.
Model Descriptive Solution:
Step 1: Identify First and Last Odd Numbers
2 Marks
Odd numbers: $21, 23, 25, \dots, 39$. This is an AP with first term $a = 21$ and last term $l = 39$.
Step 2: Compute Average
3 Marks
$$\text{Average} = \frac{a + l}{2} = \frac{21 + 39}{2} = \frac{60}{2} = 30$$
Final Answer: $30$
Question 4
7 Marks
The average age of a family of $6\text{ members}$ is $25\text{ years}$. If the age of the youngest member is $5\text{ years}$, what was the average age of the family at the time of the birth of the youngest member?
Model Descriptive Solution:
Step 1: Present Total Age
2.5 Marks
Total present age of 6 members $= 6 \times 25 = 150\text{ years}$.
Step 2: Total Age 5 Years Ago
2.5 Marks
5 years ago, each of the other 5 members was 5 years younger, and the baby was just born (age 0).
$$\text{Total Age at birth} = 150 - (6 \times 5) = 150 - 30 = 120\text{ years}$$
$$\text{Total Age at birth} = 150 - (6 \times 5) = 150 - 30 = 120\text{ years}$$
Step 3: Average Age of the 5 Members
2 Marks
$$\text{Average Age} = \frac{120}{5} = 24\text{ years}$$
Final Answer: $24\text{ years}$
Question 5
7 Marks
The mean of $50\text{ observations}$ was $36$. It was later found that an observation $48$ was wrongly copied as $23$. Find the correct mean.
Model Descriptive Solution:
Step 1: Calculate Incorrect Sum
2.5 Marks
Incorrect Total Sum $= 50 \times 36 = 1800$.
Step 2: Correct the Sum & Recompute Mean
4.5 Marks
$$\text{Correct Sum} = 1800 - 23 + 48 = 1800 + 25 = 1825$$
$$\text{Correct Mean} = \frac{1825}{50} = \frac{365}{10} = 36.5$$
$$\text{Correct Mean} = \frac{1825}{50} = \frac{365}{10} = 36.5$$
Final Answer: $36.5$
Question 6
7 Marks
The average weight of $24\text{ students}$ in a military dormitory is $35\text{ kg}$. If the warden's weight is added, the average increases by $400\text{ grams}$. Find the weight of the warden.
Model Descriptive Solution:
Step 1: Compute Initial and New Totals
3.5 Marks
Initial Total Weight $= 24 \times 35 = 840\text{ kg}$.
New average $= 35 + 0.400 = 35.4\text{ kg}$.
Total persons $= 24 + 1 = 25$.
New average $= 35 + 0.400 = 35.4\text{ kg}$.
Total persons $= 24 + 1 = 25$.
Step 2: Solve for Warden's Weight
3.5 Marks
$$\text{New Total Weight} = 25 \times 35.4 = 885\text{ kg}$$
$$\text{Warden's Weight} = 885 - 840 = 45\text{ kg}$$
$$\text{Warden's Weight} = 885 - 840 = 45\text{ kg}$$
Final Answer: $45\text{ kg}$
Question 7
7 Marks
A car covers the first $30\text{ km}$ of its journey at a speed of $60\text{ km/h}$ and the remaining $20\text{ km}$ at a speed of $40\text{ km/h}$. Find the average speed for the entire $50\text{ km}$ journey.
Model Descriptive Solution:
Step 1: Compute Individual Time Intervals
3 Marks
$$t_1 = \frac{30}{60} = 0.5\text{ h} = 30\text{ mins}$$
$$t_2 = \frac{20}{40} = 0.5\text{ h} = 30\text{ mins}$$
$$t_2 = \frac{20}{40} = 0.5\text{ h} = 30\text{ mins}$$
Step 2: Total Distance & Total Time
4 Marks
Total Distance $= 30 + 20 = 50\text{ km}$.
Total Time $= 0.5 + 0.5 = 1.0\text{ hour}$.
$$\text{Average Speed} = \frac{50}{1} = 50\text{ km/h}$$
Total Time $= 0.5 + 0.5 = 1.0\text{ hour}$.
$$\text{Average Speed} = \frac{50}{1} = 50\text{ km/h}$$
Final Answer: $50\text{ km/h}$
Question 8
10 Marks
The average score of a batsman after $25\text{ innings}$ is $56\text{ runs}$. In the next inning, he scores $108\text{ runs}$. What is his new batting average?
Model Descriptive Solution:
Step 1: Calculate Total Runs after 25 Innings
3 Marks
$$\text{Total Runs} = 25 \times 56 = 1,400\text{ runs}$$
Step 2: Add 26th Inning Runs
3 Marks
$$\text{New Total Runs} = 1400 + 108 = 1,508\text{ runs}$$
Step 3: Compute New Average
4 Marks
$$\text{New Average} = \frac{1508}{26} = 58\text{ runs}$$
Final Answer: $58\text{ runs}$
Question 9
10 Marks
The average monthly expenditure of a family was ₹ $2,200$ during the first $3\text{ months}$, ₹ $2,550$ during the next $4\text{ months}$, and ₹ $3,120$ during the last $5\text{ months}$ of a year. If the total savings during the year was ₹ $1,260$, find the average monthly income of the family.
Model Descriptive Solution:
Step 1: Compute Total Annual Expenditure
4 Marks
First 3 months $= 3 \times 2200 = ₹ 6,600$.
Next 4 months $= 4 \times 2550 = ₹ 10,200$.
Last 5 months $= 5 \times 3120 = ₹ 15,600$.
$$\text{Total Annual Expenditure} = 6600 + 10200 + 15600 = ₹ 32,400$$
Next 4 months $= 4 \times 2550 = ₹ 10,200$.
Last 5 months $= 5 \times 3120 = ₹ 15,600$.
$$\text{Total Annual Expenditure} = 6600 + 10200 + 15600 = ₹ 32,400$$
Step 2: Add Savings to Obtain Annual Income
3 Marks
$$\text{Total Annual Income} = \text{Expenditure} + \text{Savings} = 32400 + 1260 = ₹ 33,660$$
Step 3: Calculate Monthly Average Income
3 Marks
$$\text{Average Monthly Income} = \frac{33660}{12} = ₹ 2,805$$
Final Answer: Average Monthly Income $= ₹ 2,805$
Question 10
10 Marks
The average of $11\text{ numbers}$ is $60$. If the average of the first six numbers is $58$ and that of the last six numbers is $63$, find the sixth number.
Model Descriptive Solution:
Step 1: Total Sum of 11 Numbers
3 Marks
$$S_{11} = 11 \times 60 = 660$$
Step 2: Sum of First Six and Last Six Numbers
3 Marks
Sum of first 6 numbers $= 6 \times 58 = 348$.
Sum of last 6 numbers $= 6 \times 63 = 378$.
$$\text{Combined Sum of both groups} = 348 + 378 = 726$$
Sum of last 6 numbers $= 6 \times 63 = 378$.
$$\text{Combined Sum of both groups} = 348 + 378 = 726$$
Step 3: Isolate Sixth Number
4 Marks
Notice that the sixth number is included in both groups.
$$\text{Sixth Number} = (\text{Sum of first 6} + \text{Sum of last 6}) - S_{11} = 726 - 660 = 66$$
$$\text{Sixth Number} = (\text{Sum of first 6} + \text{Sum of last 6}) - S_{11} = 726 - 660 = 66$$
Final Answer: The sixth number is $66$.
High-Yield FAQs & Revision Notes
What is an arithmetic mean?
The quotient obtained by dividing the sum of all numerical values in a data set by the total number of values.
What happens to the average if every number is increased by 5?
The average also increases by exactly 5, as proven by the addition invariance theorem.
Why can't round-trip average speed be computed by (u + v)/2?
Because the vehicle spends more time travelling at the slower speed than at the faster speed. True average speed must be total distance divided by total time.
What is the formula for the average of first n natural numbers?
$\frac{n + 1}{2}$. For example, the average of the first 50 natural numbers is $\frac{50 + 1}{2} = 25.5$.
How does replacement affect group average?
$\text{New Member} = \text{Replaced Member} + (N \times \text{Increase in Average})$. If the average decreases, subtract instead.
How do you correct an average when a value was misread?
Find the net difference between the correct value and the wrong value, divide this difference by $N$, and add it to the original average.
What is a weighted average?
An average where each component value is multiplied by a weighting factor (like group size) before summing: $\frac{\sum w_i x_i}{\sum w_i}$.
In cricket, how is a batsman's average calculated?
Total runs scored divided by the number of completed dismissals (times out), not total innings.
What is the average of an Arithmetic Progression?
The average of any AP is simply $\frac{\text{First Term} + \text{Last Term}}{2}$.
How does this module map to the official RIMC syllabus?
Directly covers Serial No. 8 ('Average') of Chapter 8 broader syllabus in the official RIMC admission brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 8 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Average Formula | $\bar{X} = \frac{\sum x_i}{N}$. Invariant Sum: $\text{Sum} = N \times \bar{X}$. |
| Linear Changes | Adding, subtracting, multiplying, or dividing every value by $k$ alters the average by $k$. |
| Inclusion/Exclusion | $\text{Value} = |\text{New Sum} - \text{Old Sum}|$. Track total count carefully. |
| Replacement Rule | $\text{New} = \text{Old} \pm (N \times \Delta \bar{X})$. |
| Average Speed | $v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$. For equal segments: $\frac{2uv}{u+v}$. |
| Overlapping Terms | Sixth number $= (\text{Sum of first 6} + \text{Sum of last 6}) - \text{Sum of 11}$. |