Question

A drum contains a blend of three chemicals: P, Q, and R, in the ratio of 3:5:2. 30 litres of this blend are removed, and then 12 litres of chemical P and 8 litres of chemical Q are poured into the drum. If the resultant quantity of chemical Q is 20 litres more than the resultant quantity of chemical P, what was the initial total quantity of the mixture in the drum (in litres)?

• A. 100
• B. 120
• C. 150
• D. 180

Answer 1

Answer:

C. 150

Let the initial total mixture be (x) litres.

Step 1: Initial quantities

Ratio (P:Q:R = 3:5:2)

Total parts = 10

• $(P = \frac{3x}{10})$

• $(Q = \frac{5x}{10} = \frac{x}{2})$

• $(R = \frac{2x}{10} = \frac{x}{5})$

Step 2: 30 litres removed

Since mixture is uniform:

Removed amounts:

• $(P = 30 \times \frac{3}{10} = 9)$

• $(Q = 30 \times \frac{5}{10} = 15)$

• $(R = 30 \times \frac{2}{10} = 6)$

Remaining:

• $(P = \frac{3x}{10} - 9)$

• $(Q = \frac{x}{2} - 15)$

• $(R = \frac{x}{5} - 6)$

Step 3: Add chemicals

• Add 12 L of P

• Add 8 L of Q

New quantities:

• $(P = \frac{3x}{10} - 9 + 12 = \frac{3x}{10} + 3)$

• $(Q = \frac{x}{2} - 15 + 8 = \frac{x}{2} - 7)$

Step 4: Given condition

Q is 20 litres more than P:

$\frac{x}{2} - 7 = \left(\frac{3x}{10} + 3\right) + 20$

$\frac{x}{2} - 7 = \frac{3x}{10} + 23$

Step 5: Solve

Multiply by 10:

5x - 70 = 3x + 230
2x = 300
x = 150

Final Answer:

Initial total quantity = $(\boxed{150 \text{ litres}})$