A tank has a mixture of solutions A, B, and C in the respective ratio of 7:8:5. 40 litres of this mixture is drained out, and subsequently, 15 litres of solution A and 5 litres of solution C are added to the tank. If the resultant quantity of solution A is 25 litres less than the resultant quantity of solution B, what was the initial quantity of mixture in the tank (in litres)?

Correct — A.
Let the tank hold 7x, 8x and 5x litres of A, B and C, a total of 20x.

Drain 40 litres in the ratio 7 : 8 : 5:
A loses 40 × 7⁄20 = 14, B loses 40 × 8⁄20 = 16
Add 15 litres of A: A = 7x − 14 + 15 = 7x + 1, B = 8x − 16

Set B − A = 25: (8x − 16) − (7x + 1) = 25
Solve: x − 17 = 25, so x = 42
Initial mixture = 20 × 42 = 840 litres → option (a)

Why the others are wrong

• (b)
  240 — 240 gives x = 12: A ends at 84 − 14 + 15 = 85 and B at 96 − 16 = 80. A would be 5 litres more than B, the wrong way round.
• (c)
  890 — 890 gives x = 44.5, and the gap B − A = x − 17 = 27.5 litres, not the 25 the stem states.
• (d)
  320 — 320 gives x = 16: A = 112 − 14 + 15 = 113 and B = 128 − 16 = 112, so A exceeds B by 1 litre instead of trailing by 25.

Draining a well-mixed tank removes every component in its own ratio, so after the drain the tank is still 7 : 8 : 5, just smaller. Only the additions change the ratio.

Track the one quantity the stem asks about, the gap between B and A. It starts at 8x − 7x = x, the drain cuts it by 16 − 14 = 2, and the 15 litres of A cut it by 15. So x − 17 = 25 and x = 42.

Key facts

• Removing part of a well-mixed solution removes each component in the mixture's ratio.
• Here the drained 40 litres held 14 litres of A, 16 of B and 10 of C.
• Final quantities: A 295, B 320 and C 205 litres.

Common traps

• Taking the whole 40 litres out of one component instead of splitting it 14 : 16 : 10.
• Setting A − B = 25 when the stem says A is 25 litres less than B, which gives x = −8.

No directly related past PYQ was found.