Mixture Problem: Find the Value of P

Total mixture = 95 L

Initial ratio of milk : water = 15 : 4

Milk = \( \frac{15}{19} \times 95 = 75 \, \text{L} \)
Water = \( \frac{4}{19} \times 95 = 20 \, \text{L} \)

Step 1: Let P litres of mixture be removed

Milk removed = \( \frac{15}{19} \times P = \frac{15P}{19} \)
Water removed = \( \frac{4}{19} \times P = \frac{4P}{19} \)

Remaining milk = \( 75 - \frac{15P}{19} \)
Remaining water = \( 20 - \frac{4P}{19} \)

After adding 18 L of water:
New water = \( 20 - \frac{4P}{19} + 18 = 38 - \frac{4P}{19} \)

Step 2: According to the new ratio

\( \frac{75 - \frac{15P}{19}}{38 - \frac{4P}{19}} = \frac{3}{2} \)

Step 3: Cross multiply

\( 2 \left( 75 - \frac{15P}{19} \right) = 3 \left( 38 - \frac{4P}{19} \right) \)

\( 150 - \frac{30P}{19} = 114 - \frac{12P}{19} \)

\( 36 = \frac{18P}{19} \)

\( P = \frac{36 \times 19}{18} = \boxed{38} \)

✅ Final Answer: P = 38 litres

Milk-Water Mixture Problem

Given: 30 litres mixture contains 10% water ⇒ Water = 3 litres, Milk = 27 litres

Let x litres of milk be added. Total mixture becomes (30 + x) litres. Water remains 3 litres.

We want water to be 2% of the new mixture:

$\frac{3}{30+x}=\frac{2}{100}$

Solve:

3 / (30 + x) = 2 / 100

⇒ 3 × 100 = 2 × (30 + x)

⇒ 300 = 60 + 2x

⇒ 2x = 240 ⇒ x = 120

✅ Final Answer: 120 litres of milk should be added.