Total mixture = 95 L
Initial ratio of milk : water = 15 : 4
Milk = 1519×95=75L
Water = \frac{4}{19} \times 95 = 20 \, \text{L}
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}
\frac{75 - \frac{15P}{19}}{38 - \frac{4P}{19}} = \frac{3}{2}
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}
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:
Solve:
3 / (30 + x) = 2 / 100
⇒ 3 × 100 = 2 × (30 + x)
⇒ 300 = 60 + 2x
⇒ 2x = 240 ⇒ x = 120
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