Question

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)?

• A. 840
• B. 240
• C. 890
• D. 320

Answer 1

Answer:

A. 840

Let the initial total mixture be (x) litres.

The ratio (A:B:C = 7:8:5), so total parts (= 20).

Initial quantities

• $(A = \frac{7}{20}x)$

• $(B = \frac{8}{20}x = \frac{2}{5}x)$

• $(C = \frac{5}{20}x = \frac{1}{4}x)$

---

Step 1: After draining 40 litres

Since mixture is uniform, each component reduces in the same ratio:

Remaining fraction:
$1 - \frac{40}{x}$

So after draining:

• $(A = \frac{7}{20}x \left(1 - \frac{40}{x}\right))$

• $(B = \frac{2}{5}x \left(1 - \frac{40}{x}\right))$

---

Step 2: Additions

• Add 15 L of A

• Add 5 L of C

So final amounts:

• $A_f = \frac{7}{20}x\left(1 - \frac{40}{x}\right) + 15$
  $B_f = \frac{2}{5}x\left(1 - \frac{40}{x}\right)$

---

Step 3: Given condition

A is 25 litres less than B:

• $B_f - A_f = 25$
  

Substitute:

• $\frac{2}{5}x\left(1 - \frac{40}{x}\right)$

• $\left[\frac{7}{20}x\left(1 - \frac{40}{x}\right) + 15\right]$
  = 25
  

---

Step 4: Simplify

• $Factor (\left(1 - \frac{40}{x}\right)):$

$\left(\frac{2}{5} - \frac{7}{20}\right)x\left(1 - \frac{40}{x}\right) - 15 = 25$

Compute coefficient:

• $\frac{2}{5} = \frac{8}{20}$
  $\Rightarrow \frac{8}{20} - \frac{7}{20} = \frac{1}{20}$
  

So:

• $\frac{1}{20}x\left(1 - \frac{40}{x}\right) - 15 = 25$
  $\frac{1}{20}(x - 40) - 15 = 25$
  
  $\frac{x - 40}{20} = 40$
  
  $x - 40 = 800$
  
  x = 840
  

  Final Answer:

Initial quantity of mixture = 840 litres