NIMCET Previous Year Questions (PYQs) – Mixture And Alligation

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

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

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

$36=\frac{18P}{19}$

$P=\frac{36\times 19}{18}=\boxed{{\displaystyle 38}}$

✅ 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.