How much water is to be added to 75 ml of alcohol so that the mixture contains 25% of alcohol?

Correct — B, 225 ml. Adding water changes the total but not the alcohol, so fix on the quantity that stays constant. The alcohol is 75 ml before and after, and it has to end up as 25 per cent of the mixture. If the final mixture is M, then 75 = 0.25M, so M = 300 ml. The water added is 300 − 75 = 225 ml. The same result follows from the ratio: at 25 per cent alcohol the mixture is one part alcohol to three parts water, and three times 75 is 225.

• (a)100 ml — That gives 175 ml of mixture holding 75 ml of alcohol, about 43 per cent — far above the target.
• (c)250 ml — 325 ml of mixture with 75 ml of alcohol works out at roughly 23 per cent, just under the mark.
• (d)125 ml — The answer for a 37.5 per cent mixture, not a 25 per cent one — 75 out of 200. It attracts anyone who reads the target as one-third rather than one-quarter.

Every dilution problem has an invariant. Adding water leaves the solute untouched, so write the solute as a fixed quantity and let the total float. Once the final concentration is fixed, the final total follows at once, and the quantity added is the difference between the two totals. Working with the fraction that does not change is what keeps these problems short.

The percentage refers to the mixture, not to the alcohol, and that is where the item does its damage. If alcohol is one-fourth of the whole, water must be three-fourths, which is three times the alcohol rather than the same as it. A candidate who reads 25 per cent as a ratio of water to alcohol will produce a much smaller answer and find it waiting among the options.

• In a dilution the solute stays constant; only the total volume changes.
• Final volume = solute ÷ final fraction, here 75 ÷ 0.25 = 300 ml.
• Water to be added = final volume − initial volume = 300 − 75 = 225 ml.
• A 25 per cent solution is one part solute to three parts solvent.
• The same relation solves the reverse problem — how much must be evaporated to raise a concentration.

• Reading 25 per cent as the ratio of alcohol to water instead of alcohol to mixture.
• Giving the final volume, 300 ml, as the answer when the question asks for the water added.

A one-step dilution item that punishes a misread of what the percentage is a percentage of.

No directly related past PYQ was found.

• practice — not a real PYQ

  How much water must be added to 40 ml of alcohol to make a mixture that is 20% alcohol?

  • (a)120 ml
  • (b)160 ml
  • (c)200 ml
  • (d)80 ml

  Answer(b) 160 ml — the mixture must be 40 ÷ 0.20 = 200 ml, so 160 ml of water goes in.

• practice — not a real PYQ

  A 300 ml mixture is 25% alcohol. How much alcohol must be added to raise it to 40% alcohol?

  • (a)60 ml
  • (b)75 ml
  • (c)80 ml
  • (d)100 ml

  Answer(b) 75 ml — the water is fixed at 225 ml and must become 60 per cent of the new total, so the total is 375 ml.