Three boys split a bag of nuts by age — but the age clue hides a sneaky catch
Three boys were given a bag of 770 nuts to split in proportion to their ages, which totalled 17½ years. As often as Herbert took 4 nuts, Robert took 3; as often as Herbert took 6, Christopher took 7. How many nuts did each boy get, and how old was each of them?
Reveal the answer
Herbert got 264 nuts, Robert 198, and Christopher 308 — ages 6, 4½, and 7. Scaling Herbert's two ratios to a common share of 12 gives Herbert 12, Robert 9 and Christopher 14 nuts per 'round' of 35; 770 ÷ 35 = 22 rounds, so multiply 12, 9 and 14 by 22. The boys' ages must sit in that same 12:9:14 ratio, and since the ages sum to 17½ — exactly half of 12+9+14 — the ages are simply half of 12, 9 and 14: 6, 4½, and 7.
— Henry Dudeney, Amusements in Mathematics — Thomas Nelson & Sons, 1917, Puzzle No. 50
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