4.2 Averages, Simple/Compound Interest, Alligations & Mixtures
Key Takeaways
- The Assumed Mean method ($\bar{X} = A + \frac{\sum d_i}{N}$) and deviation balancing allow rapid calculation of composite averages without large multi-digit summations.
- Simple interest scales strictly linearly with elapsed time ($SI = \frac{PRT}{100}$), whereas Compound Interest compounds geometrically ($A = P(1 + R/100)^T$).
- The difference between $CI$ and $SI$ for 2 years is $D_2 = P\left(\frac{R}{100}\right)^2$, and for 3 years is $D_3 = P\left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right) = D_2\left(3 + \frac{R}{100}\right)$.
- The Alligation rule resolves binary mixture and weighted-average problems via the cross ratio $\frac{Q_{\text{cheaper}}}{Q_{\text{dearer}}} = \frac{d - m}{m - c}$.
- Successive dilution of a pure ingredient follows the power-decay law $Q_{\text{final}} = Q_{\text{initial}}\left(1 - \frac{y}{x}\right)^n$, where $y$ is the quantity withdrawn and replaced in each of $n$ cycles from a total volume $x$.
Averages, Simple/Compound Interest, Alligations & Mixtures
Industrial operations at Coal India Limited routinely involve statistical averaging of blast-hole depths, equipment utilization rates, financial depreciation, capital project discounting, and alligation blending of varied coal grades. This section focuses on rapid calculation techniques that replace tedious long-form arithmetic with deviation analysis, compound difference identities, and alligation cross matrices.
1. Statistical Averages & The Assumed Mean Method
Weighted Average
When different groups with sample sizes $n_1, n_2, \dots, n_k$ have respective arithmetic means $\bar{x}_1, \bar{x}_2, \dots, \bar{x}k$, the combined weighted average $\bar{X}{w}$ is:
Assumed Mean (Deviation Method)
For large data points, avoid brute-force addition by selecting an arbitrary assumed mean $A$ close to the center of the dataset. Calculate the deviation $d_i = x_i - A$ for each point:
Because the algebraic sum of deviations from the true arithmetic mean is always zero ($\sum (x_i - \bar{X}) = 0$), any net deviation $\sum d_i$ simply distributes equally across the $N$ observations.
Dynamic Group Changes (Insertion, Deletion, Replacement)
- Inclusion of a New Member:
- Exclusion of a Member:
- Replacement of an Existing Member:
2. Simple and Compound Interest Mechanics
Simple Interest ($SI$)
Simple interest accrues uniformly solely on the original principal $P$:
Where:
- $P = \text{Principal sum}$
- $R = \text{Annual interest rate in } %$
- $T = \text{Time duration in years}$
Compound Interest ($CI$)
Compound interest reinvests accrued interest into the principal base at each compounding cycle:
Where $k$ is the compounding frequency per year:
- Annually: $k = 1$
- Semi-annually (Half-yearly): $k = 2$, rate $= R/2$, periods $= 2T$
- Quarterly: $k = 4$, rate $= R/4$, periods $= 4T$
Difference Between $CI$ and $SI$
Questions regarding the difference $D = CI - SI$ for 2 and 3 years appear frequently in PSU examinations:
-
Two-Year Difference ($D_2$):
-
Three-Year Difference ($D_3$):
Quick Growth Rules of Thumb
- Rule of 72 (Doubling Time): Time required for capital to double under compound interest is approximately $T \approx \frac{72}{R%}$.
- Rule of 114 (Tripling Time): Time required to triple capital is approximately $T \approx \frac{114}{R%}$.
- Rule of 144 (Quadrupling Time): Time required to quadruple capital is approximately $T \approx \frac{144}{R%}$.
3. Alligations & Industrial Mixture Blending
The Alligation Rule
The rule of alligation is a modified cross-computation derived from weighted averages. When two ingredients of price/concentration $c$ (cheaper/lower) and $d$ (dearer/higher) are combined to produce a mixture of mean value $m$:
Applications in Coal Quality Blending
In coal preparation and power dispatch, two grades of coal with different ash percentages ($A_1%$ and $A_2%$) or calorific values ($CV_1$ and $CV_2$) are blended to meet strict thermal plant specifications ($A_{\text{target}}$):
Three-Component Alligation Framework
When three components with values $c_1 < c_2 < c_3$ are mixed to obtain a target mean $m$ (where $c_1 < m < c_3$), pair the components lying below the mean with those above the mean:
- Pair $(c_1, c_3)$ with difference weights: $Q_1 = |c_3 - m|$, $Q_{3,a} = |m - c_1|$.
- If $c_2 < m$, pair $(c_2, c_3)$ giving: $Q_2 = |c_3 - m|$, $Q_{3,b} = |m - c_2|$.
- Sum the allocated parts for component 3: $Q_3 = Q_{3,a} + Q_{3,b}$.
4. Repeated Replacement Dilution Formula
When a container initially contains $x$ units of a pure liquid (or material), and $y$ units are drawn off and replaced with an adulterant/water, and this process is repeated $n$ times in succession, the remaining quantity of pure liquid is given by:
The fraction of pure liquid remaining relative to the total container capacity is:
The ratio of pure liquid to replacement liquid (water) after $n$ operations is:
5. Equal Annual Installments (EMIs) Under SI and CI
Financial appraisal of heavy earthmoving machinery leases and capital loans requires calculating equal annual installment payments:
Installments Under Simple Interest
If a total debt of amount $A$ due after $n$ years is discharged in $n$ equal annual installments of $x$ each at simple interest rate $R%$:
Solving for the annual installment $x$:
Installments Under Compound Interest
If a principal loan $P$ borrowed today is repaid in $n$ equal annual installments of $x$ at annual compound interest rate $R%$:
For a 2-year repayment cycle, let $v = \frac{1}{1 + R/100}$:
The difference between the compound interest (compounded annually) and the simple interest on a principal sum of money invested for 3 years at an annual interest rate of 10% is Rs. 1,550. What is the initial principal sum (P)?
A Coal India washery blends Grade-F coal having 42% ash content with high-grade washed Grade-B coal having 18% ash content to fulfill a thermal power plant contract requiring a blend with exactly 26% ash content. If the total dispatch batch is 60,000 metric tonnes, how many metric tonnes of Grade-B coal must be blended?
A reservoir containing 80 liters of pure chemical additive is subjected to repeated dilution. Exactly 8 liters of the chemical is drawn off and replaced with pure water. This replacement operation is performed a total of 3 times. What is the remaining quantity of pure chemical additive in the reservoir after the third operation?