How Quantities Combine: Understanding Data - Mixtures

In what ratio must rice at Rs 9.30 per kg be mixed with rice at Rs 10.80 per kg so that the mixture is worth Rs 10.00 per kg?

We use the Rule of Alligation. Given: Cheaper Price ($c$) = Rs 9.30 Dearer Price ($d$) = Rs 10.80 Mean Price ($m$) = Rs 10.00

Calculation of differences: $d - m = 10.80 - 10.00 = 0.80$ $m - c = 10.00 - 9.30 = 0.70$

Ratio: $\text{Ratio} = \frac{d - m}{m - c} = \frac{0.80}{0.70} = \frac{8}{7}$

The ratio is $8:7$.

The Rule of Alligation allows us to find the ratio by taking the difference between the component prices and the target mean price. The cheaper quantity corresponds to the difference $(d-m)$ and the dearer quantity to $(m-c)$.

A container contains 40 litres of milk. From this container, 4 litres of milk were taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

This is a replacement problem using the iterative formula: $\text{Final Quantity} = a \left(1 - \frac{x}{a}\right)^n$ Where: $a = 40$ litres (Initial quantity) $x = 4$ litres (Quantity replaced each time) $n = 3$ (Total number of operations: the first time + two further times)

Calculation: $\text{Final Quantity} = 40 \left(1 - \frac{4}{40}\right)^3$ $\text{Final Quantity} = 40 \left(1 - \frac{1}{10}\right)^3$ $\text{Final Quantity} = 40 \left(\frac{9}{10}\right)^3$ $\text{Final Quantity} = 40 \times \frac{729}{1000}$ $\text{Final Quantity} = \frac{29160}{1000} = 29.16 \text{ litres}$

Every time a part of the mixture is replaced with water, the concentration of milk reduces by a factor of $(1 - \frac{x}{a})$. After $n$ iterations, the initial volume is multiplied by this factor $n$ times.