10.1 Averages, Weighted Averages & Age Problems

10.1 Averages, Weighted Averages & Age Problems

Quantitative Aptitude in banking examinations tests numerical dexterity, logical deduction, and operational speed. In the SBI Clerk (Junior Associate) preliminary and main examinations, arithmetic problems centered on averages, weighted means, and chronological age relationships appear reliably as individual standalone word problems (1 to 3 questions) and serve as the mathematical foundation for Data Interpretation (DI) caselets and tabular charts. Mastering shortcut formulas and structural balancing techniques eliminates tedious long-division calculations and preserves critical exam time.

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1. Mathematical Foundations & The Net Deviation Principle

The arithmetic mean (or average) of a discrete dataset is defined as the total aggregate sum of all observations divided by the total number of observations:

$$\text{Average } (\bar{x}) = \frac{\text{Total Sum of Observations } (\sum x_i)}{\text{Total Number of Observations } (N)}$$ $$\text{Aggregate Sum } (\sum x_i) = N \times \bar{x}$$

While direct summation works for small numbers, commercial banking datasets often feature large, clustered values (e.g., daily branch transaction counts, customer deposit sizes, or employee productivity logs). In competitive examinations, calculating large sums induces arithmetic errors. The Deviation Method (Assumed Mean Method) bypasses large summations.

The Net Deviation Principle

Let $A$ represent an assumed arbitrary mean chosen close to the center of the observations. The deviation $d_i$ of any observation $x_i$ from $A$ is:

$$d_i = x_i - A$$

The true arithmetic mean is then obtained by adjusting the assumed mean by the average of the deviations:

$$\bar{x} = A + \frac{\sum d_i}{N}$$

Deviation Method Calculation Box

Consider five customer service counters at an SBI administrative hub handling the following daily voucher volumes: 42, 48, 55, 61, and 64.

$$\bar{x} = 50 + \frac{+20}{5} = 50 + 4 = \mathbf{54 \text{ vouchers}}$$

By balancing deviations, two-digit addition replaces multi-digit column summation, yielding the answer in under 15 seconds.

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2. Weighted Average Framework

When combining distinct groups with different average values and unequal group sizes, a simple average of averages yields incorrect results. The weighted average accounts for the relative weight or frequency of each constituent group.

Mathematical Formulation

If $k$ distinct groups have sizes $w_1, w_2, \dots, w_k$ and respective means $\bar{x}_1, \bar{x}_2, \dots, \bar{x}_k$, the combined weighted average $\bar{x}_w$ is:

$$\bar{x}_w = \frac{w_1 \bar{x}_1 + w_2 \bar{x}_2 + \dots + w_k \bar{x}_k}{w_1 + w_2 + \dots + w_k} = \frac{\sum_{i=1}^k w_i \bar{x}_i}{\sum_{i=1}^k w_i}$$

Key Ratio Simplification Shortcut

In the weighted average formula, absolute weights ($w_i$) can be replaced by their lowest integer ratio. If an SBI branch holds Rs. 30 crore in retail deposits at 6.5% interest, Rs. 20 crore in senior citizen deposits at 7.25% interest, and Rs. 50 crore in corporate term deposits at 5.75% interest, the weights ratio is:

$$w_1 : w_2 : w_3 = 30 : 20 : 50 = 3 : 2 : 5$$ $$\bar{x}_w = \frac{3(6.5) + 2(7.25) + 5(5.75)}{3 + 2 + 5} = \frac{19.5 + 14.5 + 28.75}{10} = \frac{62.75}{10} = \mathbf{6.275\%}$$

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3. Inclusion, Exclusion, and Replacement Dynamics

Questions involving an individual joining, leaving, or replacing an existing member of a group are standard in the SBI Clerk exam.

1. Inclusion Dynamics (New Member Joins)

When a new person joins a group of $N$ individuals with initial average $\bar{x}_{\text{old}}$, altering the group size to $N + 1$ and changing the average to $\bar{x}_{\text{new}}$:

$$\text{Value of Incoming Member} = \bar{x}_{\text{old}} + (N + 1) \times (\bar{x}_{\text{new}} - \bar{x}_{\text{old}})$$

Alternative perspective: The incoming entrant must bring their own share equal to the new average, plus or minus the cumulative adjustment contributed to the initial $N$ members: $\text{Value} = \bar{x}_{\text{new}} + N \times (\bar{x}_{\text{new}} - \bar{x}_{\text{old}})$. Both expressions are mathematically equivalent.

2. Exclusion Dynamics (Existing Member Leaves)

When one individual leaves a group of $N$ members, reducing the count to $N - 1$:

$$\text{Value of Departing Member} = \bar{x}_{\text{old}} - (N - 1) \times (\bar{x}_{\text{new}} - \bar{x}_{\text{old}})$$

3. Replacement Dynamics (Count Invariant)

When one member leaves and is immediately replaced by a new entrant, the total count $N$ remains strictly invariant:

$$\text{Value of New Entrant} - \text{Value of Replaced Member} = N \times (\bar{x}_{\text{new}} - \bar{x}_{\text{old}})$$ $$\text{Value of New Entrant} = \text{Replaced Member's Value} + N \times \Delta\bar{x}$$

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4. Chronological Age Problems & The Invariant Difference Principle

Age arithmetic problems require translating past, present, and future chronological relationships into linear equations.

The Core Axiom of Age Arithmetic

$$\text{The difference in age between two individuals remains constant at all points in time.}$$

If Person A is 6 years older than Person B today, Person A was 6 years older 15 years ago and will remain 6 years older 25 years in the future ($A_t - B_t = \text{constant}$). Total elapsed chronological time shifts every individual's age by the exact same scalar: after $t$ years, both individuals have aged by $+t$ years.

Chronological Transition Matrix

Ratio Equating Technique

When past and future age ratios are specified, align the unit differences:

1. Suppose the ratio of ages of A to B was $a_1 : b_1$ at time $t_1$, and becomes $a_2 : b_2$ at time $t_2$.
2. Compute difference in parts: $\Delta_1 = |a_1 - b_1|$ and $\Delta_2 = |a_2 - b_2|$.
3. Normalize the ratios by multiplying ratio 1 by $\Delta_2$ and ratio 2 by $\Delta_1$, ensuring that the invariant age difference corresponds to the exact same number of ratio units.
4. Equate the change in individual ratio parts directly to the elapsed calendar time $(t_2 - t_1)$.