There is a 32 l mixture of milk and water. The ratio of milk and water in this mixture is 5 : 3. How many litres of water should be added to this mixture so that the resultant mixture has 50% water in it?

Correct — B. Split the 32 l in the ratio 5 : 3, which is 8 equal parts of 4 l.

Milk = 5 × 4 = 20 l
Water = 3 × 4 = 12 l

Only water is poured in, so the milk stays at 20 l.
Water is to be 50%, so milk is the other 50% — the two are equal.
New total = 2 × 20 = 40 l
Water added = 40 − 32 = 8 l → option (b)

Check: water becomes 12 + 8 = 20 l in a 40 l mixture, which is exactly half.

Why the others are wrong

• (a)
  4 — Adding 4 l gives 16 l of water in 36 l of mixture — 44.4%, still short of half.
• (c)
  6 — Adding 6 l gives 18 l of water in 38 l — 47.4%. Close, but the total grows with every litre poured in, so it does not reach 50%.
• (d)
  5 — Adding 5 l gives 17 l of water in 37 l — 45.9%, under half on the same count.

In an add-one-component sum, the untouched component is the anchor. Only water is poured in here, so the milk is fixed at 20 l from start to finish.

Once water is to be 50%, milk must be the other 50%, so the two are equal. A fixed 20 l of milk therefore forces a total of 40 l, and the rest is subtraction.

Working from the fixed quantity avoids the equation (12 + x)/(32 + x) = 1⁄2, which is where the growing total is usually forgotten.

Key facts

• 5 : 3 over 32 l is 8 parts of 4 l — milk 20 l, water 12 l.
• Pouring in water leaves the milk quantity unchanged and only cuts its share.
• 50% water means milk and water are equal, so the new total is twice the milk.
• 12 + 8 = 20 l of water in a 40 l mixture is the 50% the stem asks for.

Common traps

• Solving (12 + x)/32 = 1⁄2 and forgetting that the total rises by x as well.
• Halving the original 32 l and answering 16.
• Recomputing the milk after the addition, when only the water changed.

No directly related past PYQ was found.