Mixture & Alligation — PYPs

The Setup: Mixture problems just require tracking the absolute volumes of the components like inventory. We find the exact litres of water and milk we currently hold, merge them, then solve for the extra water needed to hit the target 25%. The key discipline is that adding pure water changes the water column *and* the total, but leaves the milk column frozen.
Step 1: Audit Mixture 1. Total $=50$ L, water is 30%.

$$\text{Water}=0.30\times 50=15\text{ L}, \qquad \text{Milk}=50-15=35\text{ L}$$

Step 2: Audit Mixture 2. Total $=80$ L, water is 20%.

$$\text{Water}=0.20\times 80=16\text{ L}, \qquad \text{Milk}=80-16=64\text{ L}$$

Step 3: Merge the inventory. Tip both into the same barrel:

$$\text{Total volume}=50+80=130\text{ L}, \quad \text{Water}=15+16=31\text{ L}, \quad \text{Milk}=35+64=99\text{ L}$$

Step 4: Calculate the required additive. Let $w$ litres of pure water go in. Both the water and the total grow by $w$, and a 25% water mix means water is $\frac{1}{4}$ of the total:

$$\frac{31+w}{130+w}=\frac{1}{4}$$

Cross-multiply and solve:

$$4(31+w)=130+w \implies 124+4w=130+w \implies 3w=6 \implies w=2$$

Step 5: Verify from the milk side. The milk column never moved, so it is an independent check. Final total $=130+2=132$ L, of which milk is still 99 L:

$$\frac{99}{132}=\frac{3}{4}=75\% \text{ milk}, \text{ hence } 25\% \text{ water}$$

The two routes agree, so $w=2$ is right.

Final Answer: 2

The Setup: This is a weighted average problem mapping two groups of distinct heights into a combined new average height. We can isolate the ratio of group sizes using the Alligation method.

Step 1: Define the parameters for Alligation.
Group 1 ($n$ persons): Average height = $160 \text{ cm}$.
Group 2 ($m$ persons): Average height = $172 \text{ cm}$.
Mixture (Combined group): Average height = $164 \text{ cm}$.

Step 2: Apply the Alligation cross-difference formula.
The ratio of the quantities ($n : m$) is inversely proportional to the distances their specific averages sit from the combined mixture mean.

$$n : m = (172 - 164) : (164 - 160)$$

$$n : m = 8 : 4$$

$$n : m = 2 : 1$$

Step 3: Evaluate the requested ratio.
We must explicitly provide the ratio $m : n$.
Since $n : m = 2 : 1$, the ratio $m : n$ is exactly $1 : 2$.

Final Answer: $1 : 2$

The Setup: A mixed volume fruit punch starts out with unequal ratios. After some is consumed, pure orange juice is added back to strictly raise its proportion to $50\%$ of the newly defined mixture.

Step 1: Establish the baseline volume ratios of the mixture.
Initial ingredients: $60\text{ml}$ orange juice (OJ), $40\text{ml}$ water (W). Subtotal = $100\text{ml}$.
Add $100\text{ml}$ apple juice (AJ). Grand total = $200\text{ml}$.
The proportional composition of the punch is:
$\text{OJ} = \frac{60}{200} = 30\%$
$\text{W} = \frac{40}{200} = 20\%$
$\text{AJ} = \frac{100}{200} = 50\%$

Step 2: Calculate volumes remaining after consumption.
Madhu drinks $50\text{ml}$ of the punch.
Remaining total volume = $200\text{ml} - 50\text{ml} = 150\text{ml}$.
Because mixtures are perfectly homogenous, the $30\%$ ratio is maintained in the leftover fluid.
Volume of OJ remaining = $30\%$ of $150 = 45\text{ml}$.

Step 3: Calculate the required OJ addition.
Let $x$ be the volume in $\text{ml}$ of pure OJ poured back into the punch.
The new total volume of OJ is $45 + x$. The new total volume of the entire punch is $150 + x$.
We require the new OJ ratio to be exactly $50\%$:

$$\frac{45 + x}{150 + x} = 0.5$$

$$45 + x = 0.5(150 + x)$$

$$45 + x = 75 + 0.5x$$

$$0.5x = 30 \implies x = 60\text{ml}$$

Final Answer: 60 ml

The Setup: We've got an alchemy minigame. First, we need to unlock the exact ratio ($a:b$) that balances the metals 1:1. Once we have those initial mix weights, we hit them with an Uno reverse card, swap the weights ($b:a$) for a new batch, and calculate the final drop rates for copper and zinc.
**Step 1: Find the Initial Mix Ratio ($a:b$)**
In alloy P, Copper is $5/7$ and Zinc is $2/7$.
In alloy Q, Copper is $3/7$ and Zinc is $4/7$.
Mix them in parts $a$ and $b$ to get equal Copper and Zinc:

$$\frac{5a}{7} + \frac{3b}{7} = \frac{2a}{7} + \frac{4b}{7}$$

Multiply by 7 to drop the denominators and balance the equation:

$$5a + 3b = 2a + 4b$$

$$3a = b \implies \frac{a}{b} = \frac{1}{3}$$

So the ratio $a:b$ is strictly $1:3$.
**Step 2: Form Alloy S (Ratio $b:a$)**
Now we mix them using the reverse ratio: 3 parts of P and 1 part of Q.
Calculate the total Copper:

$$\text{Cu} = 3\left(\frac{5}{7}\right) + 1\left(\frac{3}{7}\right) = \frac{15}{7} + \frac{3}{7} = \frac{18}{7}$$

Calculate the total Zinc:

$$\text{Zn} = 3\left(\frac{2}{7}\right) + 1\left(\frac{4}{7}\right) = \frac{6}{7} + \frac{4}{7} = \frac{10}{7}$$

Step 3: Final Proportion
The ratio of Cu to Zn in alloy S is:

$$\frac{18}{7} : \frac{10}{7}$$

Multiply by 7 and simplify by dividing by 2:

$$18 : 10 = 9 : 5$$

Final Answer: $9:5$