How many milliliters of the 40% acid solution must the chemist add?

Question

A chemist has 40 milliliters of a solution that is 15% acid by volume. She will add some amount of a second solution that is 40% acid by volume. The two solutions combine without any loss of volume, and the resulting mixture must be 30% acid by volume.

Accepted Answer

C. 60. Let x be the milliliters of 40% solution added. Acid before: 0.15(40) + 0.40x. Acid after: 0.30(40 + x). So 6 + 0.40x = 12 + 0.30x, giving 0.10x = 6 and x = 60. Check: 40 mL + 60 mL = 100 mL containing 6 + 24 = 30 mL of acid, which is 30%. (A) 15 comes from writing 0.15(40) + 0.40x = 0.30(40) — forgetting that the added liquid increases the total volume of the mixture. (B) 24 comes from correctly finding that 6 additional mL of acid are needed but then dividing by 0.25 (the difference 40% − 15%) instead of by the correct 0.10 (the difference 40% − 30%). (D) 120 comes from solving 0.40x = 0.30(40 + x) — ignoring the acid already present in the original 40 mL.

Answer

A. 15 — You set 0.15(40) + 0.40x = 0.30(40), which keeps the total volume at 40 mL; adding x milliliters makes the new total 40 + x, so the right side must be 0.30(40 + x).

Answer

B. 24 — You correctly found that 6 more milliliters of acid are needed, but then divided by 0.25 (40% − 15%) instead of 0.10 (40% − 30%) — the added solution has to make up the gap between 40% and the target 30%.

Answer

D. 120 — You solved 0.40x = 0.30(40 + x), which ignores the 6 mL of acid already in the original 40 mL; that acid belongs on the left side too.

Problem-Solving and Data Analysis practice question

Answer: C. 60

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