8.2 Percentages, Base Conversions & Successive Changes

8.2 Percentages, Base Conversions & Successive Changes

Percentage arithmetic forms the computational core of the State Bank of India (SBI) Junior Associate Quantitative Aptitude curriculum. It is not merely an isolated arithmetic chapter; percentage relationships serve as the mathematical engine for Data Interpretation (DI tables, charts, and caselets), Simple and Compound Interest, Profit and Loss, and Partnership profit distributions. In an SBI Clerk examination, proficiency in percentage calculations directly determines whether a candidate finishes the Numerical Ability section with minutes to spare or runs out of time on multi-step word problems.

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The Percentage Multiplier & Fraction Equivalents

A percentage is fundamentally a standardized fraction whose denominator is fixed at 100:

$$P\% = \frac{P}{100}$$

In high-speed banking calculations, calculating $16.67\%$ of Rs. 4,800 by multiplying $4800 \times 0.1667$ is computationally disastrous. Recognizing that $16.67\% = \frac{1}{6}$ reduces the calculation to a trivial mental division: $4800 / 6 = \text{Rs. } 800$.

Master Fraction-to-Percentage Conversion Table (1/2 to 1/20)

Candidates must commit these unit fractions and their decimal multipliers to memory:

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The Asymmetric Base Change Theorem

One of the most frequent traps in banking examinations is confusing the direction of percentage comparison. When evaluating two quantities $A$ and $B$, the denominator (base) is always the quantity following the preposition "than" or "as compared to".

Mathematical Formulation

If $A$ is $x\%$ more than $B$:

$$A = B \left(1 + \frac{x}{100}\right) = B \left(\frac{100 + x}{100}\right)$$

To find by what percentage $B$ is less than $A$, the reference base shifts from $B$ to $A$:

$$\text{Percentage Less} = \frac{A - B}{A} \times 100\% = \frac{B\left(\frac{x}{100}\right)}{B\left(\frac{100 + x}{100}\right)} \times 100\% = \left(\frac{x}{100 + x}\right) \times 100\%$$

The Reciprocal Fraction Shift Ladder

Expressing the upward change as a simple fraction $\frac{1}{n}$ unlocks an instantaneous reciprocal shortcut:

$$\text{Increase of } +\frac{1}{n} \iff \text{Decrease of } -\frac{1}{n + 1}$$ $$\text{Decrease of } -\frac{1}{n} \iff \text{Increase of } +\frac{1}{n - 1}$$

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Product Constancy Architecture ($A \times B = K$)

In commercial banking and macroeconomic problems, many governing formulas follow a bivariate multiplicative structure where the product remains invariant:

$$\text{Price} \times \text{Consumption} = \text{Total Expenditure}$$ $$\text{Transaction Volume} \times \text{Processing Fee Rate} = \text{Fee Revenue}$$ $$\text{Speed} \times \text{Time} = \text{Distance}$$

The Product Constancy Rule

If factor $A$ increases by a fraction $+\frac{a}{b}$, then factor $B$ must decrease by an exact fraction $-\frac{a}{a + b}$ to ensure that the composite product $K$ remains strictly constant:

$$A \left(1 + \frac{a}{b}\right) \times B \left(1 - \frac{a}{a + b}\right) = A \left(\frac{a + b}{b}\right) \times B \left(\frac{b}{a + b}\right) = A \times B = K$$

Application: If the procurement cost of cash-handling machinery at a bank's regional hub increases by $25\%$ ($+1/4$), by what percentage must the quantity purchased be reduced to keep total outlay unchanged?

• Upward shift $= +1/4$ (where $a = 1, b = 4$).
• Required downward shift $= -\frac{a}{a + b} = -\frac{1}{1 + 4} = -\frac{1}{5} = -20\%$.
• The hub must reduce purchases by exactly $20\%$.

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Successive Percentage Changes & Compounding

When a financial baseline changes sequentially by $a\%$ followed by $b\%$, the net compound percentage change cannot be determined by simple addition ($a + b$). Because the second percentage acts upon the inflated or deflated intermediate base, compounding occurs.

The Universal Net Formula

$$\text{Net Percentage Change} = \left(a + b + \frac{a \times b}{100}\right)\%$$

Sign Convention:

• If a change is an increase / appreciation, assign a positive sign ($+a$).
• If a change is a decrease / depreciation, assign a negative sign ($-a$).

Decimal Multiplier Method for Multi-Stage Changes

When more than two successive shifts occur (e.g., three consecutive quarters of portfolio expansion: $+10\%$, $-5\%$, and $+20\%$), the net factor is computed by multiplying successive decimal factors:

$$\text{Final Value} = \text{Initial Value} \times (1 + 0.10) \times (1 - 0.05) \times (1 + 0.20)$$ $$\text{Final Value} = \text{Initial Value} \times 1.10 \times 0.95 \times 1.20 = \text{Initial Value} \times 1.254$$

The composite growth over the three periods is $(1.254 - 1) \times 100\% = +25.4\%$.

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Population Dynamics & Asset Depreciation Compounding

Problems modeling municipal population growth or computer hardware depreciation in bank branches utilize discrete compound interest formulas:

$$V_n = V_0 \left(1 \pm \frac{r}{100}\right)^n$$

where $V_0$ is the initial base, $r$ is the annual rate of change, and $n$ is the number of elapsed years.

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Step-by-Step Worked Banking Examination Problems

Problem 1: Branch Target Base Change

Problem: In a fiscal performance review, SBI Branch X mobilized Rs. 72 Crore in Current Account Savings Account (CASA) deposits, which is $20\%$ higher than the CASA deposits mobilized by Branch Y. By what percentage is Branch Y's mobilization lower than Branch X's?

Solution:

1. Let Branch Y's mobilization be $B$. Then Branch X's mobilization is $A = B \times (1 + 0.20) = 1.20 B$.
2. We are given $A = 72$ Crore. Thus, $B = 72 / 1.20 = 60$ Crore.
3. We must find by what percentage $B$ is less than $A$. The base is $A$: $$\text{Percentage Less} = \frac{A - B}{A} \times 100\% = \frac{72 - 60}{72} \times 100\% = \frac{12}{72} \times 100\% = \frac{1}{6} \times 100\% = 16.67\%$$
4. Fraction Shortcut: Branch X is $+1/5$ ($+20\%$) higher than Y. By the reciprocal ladder, Branch Y is $-1/(5+1) = -1/6 = -16.67\%$ lower than X. Solved in 5 seconds.

Problem 2: Portfolio Multi-Stage Expansion

Problem: An SBI regional loan hub manages an agricultural credit portfolio of Rs. 250 Crore. The portfolio grows by $12\%$ in Year 1. In Year 2, due to severe regional drought, the portfolio contracts by $10\%$. What is the net value of the loan portfolio at the conclusion of Year 2?

Solution:

1. Compute the net successive percentage change: $$\text{Net} = a + b + \frac{ab}{100} = +12 - 10 + \frac{12 \times (-10)}{100} = 2 - 1.20 = +0.80\%$$
2. The net change is a modest expansion of $+0.80\%$.
3. Calculate the monetary value on the base of Rs. 250 Crore: $$\text{Final Value} = 250 \times \left(1 + \frac{0.80}{100}\right) = 250 \times 1.008 = \text{Rs. } 252 \text{ Crore}$$
4. The portfolio closes at exactly Rs. 252 Crore.