A vessel initially contains 400 liters of acid A and 80 liters of acid B. A certain quantity of mixture is removed from the vessel, and then some liters of acid B are added to the remaining mixture. After this process, the quantity of acid A in the vessel exceeds the quantity of acid B by 250 liters.

Which of the following pairs of integral values (liters of mixture removed, liters of acid B added) satisfy this condition?

Question

A vessel initially contains 400 liters of acid A and 80 liters of acid B. A certain quantity of mixture is removed from the vessel, and then some liters of acid B are added to the remaining mixture. After this process, the quantity of acid A in the vessel exceeds the quantity of acid B by 250 liters.

Which of the following pairs of integral values (liters of mixture removed, liters of acid B added) satisfy this condition?

Accepted Answer

**Solution:**

1. Initially, the vessel contains 400 liters of acid A and 80 liters of acid B.

2. Total mixture = 400 + 80 = 480 liters.

3. The ratio of acid A to acid B initially is:
   $$\frac{400}{80} = 5 : 1$$

4. Let *X* liters of mixture be removed. Since the mixture is uniform, the amount of acid A removed is $$\frac{5}{6}X$$ and acid B removed is $$\frac{1}{6}X$$.

5. After removal, the remaining acid A is $$400 - \frac{5}{6}X$$ and acid B is $$80 - \frac{1}{6}X$$.

6. Then, *Y* liters of acid B is added. So, new amount of acid B becomes:
   $$80 - \frac{1}{6}X + Y$$

7. According to the problem, acid A exceeds acid B by 250 liters:
   $$400 - \frac{5}{6}X = 80 - \frac{1}{6}X + Y + 250$$

8. Simplify the equation:
   $$400 - \frac{5}{6}X = 330 - \frac{1}{6}X + Y$$

9. Rearranging terms:
   $$400 - 330 = \frac{5}{6}X - \frac{1}{6}X + Y$$
   $$70 = \frac{4}{6}X + Y$$
   $$70 = \frac{2}{3}X + Y$$

10. Multiply both sides by 3:
    $$210 = 2X + 3Y$$

11. Now, check which options satisfy this equation:

- Option A: (72, 22)
  $$2 	imes 72 + 3 	imes 22 = 144 + 66 = 210$$ ✔

- Option B: (60, 30)
  $$2 	imes 60 + 3 	imes 30 = 120 + 90 = 210$$ ✔

- Option C: (84, 24)
  $$2 	imes 84 + 3 	imes 24 = 168 + 72 = 240 
eq 210$$ ✘

- Option D: (48, 38)
  $$2 	imes 48 + 3 	imes 38 = 96 + 114 = 210$$ ✔

- Option E: (36, 48)
  $$2 	imes 36 + 3 	imes 48 = 72 + 144 = 216 
eq 210$$ ✘

12. Therefore, only options A, B, and D satisfy the condition.

**Answer:** Only options A, B, and D are possible.

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