A vessel contains a certain quantity of a solution of acid and water. When 3 litres of water is added to it, the new solution has 50% acid concentration. When 12 litres of acid is further added to this new solution, the final solution has 75% acid concentration. The ratio of water and acid in the original solution was:
Answer and solution
Correct answer: A
Solution
Answer: A — 1 : 2
1. Let the original amount of acid be A litres and water be W litres.
2. After adding 3 litres of water, the solution is 50% acid. This means the amount of acid and water are equal.
3. So,
4. A = W + 3. ...(1)
5. Next, 12 litres of acid is added. The final solution is 75% acid.
6. Therefore,
7. (A + 12)/(A + W + 15) = 3/4
8. Cross-multiplying:
9. 4(A + 12) = 3(A + W + 15)
4A + 48 = 3A + 3W + 45
A = 3W − 3. ...(2)
10. From equations (1) and (2):
11. W + 3 = 3W − 3
2W = 6
W = 3
12. So, A = 3 + 3 = 6.
13. The original ratio of water to acid is 3 : 6, or 1 : 2.
14. Answer: 1 : 2.
Why this method works:
A 50% acid mixture contains equal volumes of acid and water. After only acid is added, the water amount stays fixed, while 75% concentration requires acid to be three times that fixed water amount.
Independent check and common trap:
At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.
Apply: After water addition acid=water. Final acid: water=3: 1, so adding 12 acid equals 2 parts: each part 6. Originally water=6−3=3, acid 6 →1: 2.
Check: At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.
1. Let the original amount of acid be A litres and water be W litres.
2. After adding 3 litres of water, the solution is 50% acid. This means the amount of acid and water are equal.
3. So,
4. A = W + 3. ...(1)
5. Next, 12 litres of acid is added. The final solution is 75% acid.
6. Therefore,
7. (A + 12)/(A + W + 15) = 3/4
8. Cross-multiplying:
9. 4(A + 12) = 3(A + W + 15)
4A + 48 = 3A + 3W + 45
A = 3W − 3. ...(2)
10. From equations (1) and (2):
11. W + 3 = 3W − 3
2W = 6
W = 3
12. So, A = 3 + 3 = 6.
13. The original ratio of water to acid is 3 : 6, or 1 : 2.
14. Answer: 1 : 2.
Why this method works:
A 50% acid mixture contains equal volumes of acid and water. After only acid is added, the water amount stays fixed, while 75% concentration requires acid to be three times that fixed water amount.
Independent check and common trap:
At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.
G Strategy shortcut
Spot: Arithmetic – Mixtures and Alligation.Apply: After water addition acid=water. Final acid: water=3: 1, so adding 12 acid equals 2 parts: each part 6. Originally water=6−3=3, acid 6 →1: 2.
Check: At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.
G Strategy: Spot: Arithmetic – Mixtures and Alligation. Apply: After water addition acid=water. Final acid: water=3: 1, so adding 12 acid equals 2 parts: each part 6. Originally water=6−3=3, acid 6 →1: 2. Check: At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.