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Jim Randell 8:57 am on 29 August 2021 Permalink Reply
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  Jim Randell 8:58 am on 29 August 2021 Permalink | Reply

  Working in prices of pennies and weights of 1lb, we can try to determine the following values for each item:

  (number of items per lb)
  (price of 1lb of items)
  (weight per item) = 1 / (items per lb)
  (price per item) = (price of 1lb items) / (items per lb)

  Untangling the facts we are given:

  “2 lb. of grapes = 64 grapes; cost = 4s. = 48d”

  (grapes per lb) = 32
  (weight per grape) = 1/32
  (price of 1lb grapes) = 24
  (price per grape) = 3/4

  “An onion is half the price of a carrot, 24 times the weight of a pea, and a quarter the price of a pound of beans.”

  (price per carrot) = k
  (price per onion) = k/2

  (peas per lb) = p
  (weight per pea) = 1/p
  (onions per lb) = p/24
  (weight per onion) = 24/p
  (price of 1lb onions) = (p/24) × (k/2) = kp/48

  (price per onion) = (price of 1lb beans) / 4
  (price of 1lb beans) = 4 × (k/2) = 2k

  “The cost of 3 carrots is to that of a grape as a bean is to a pea in weight”

  (price of 3 carrots) / (price of 1 grape) = (weight of 1 bean) / (weight of 1 pea)
  (price of 3 carrots) × (weight of 1 pea) = (weight of 1 bean) × (price of 1 grape)
  (3k) × (1/p) = (weight of 1 bean) × (3/4)
  (weight of 1 bean) = 4k/p
  (beans per lb) = p/4k

  (price per bean) = (price of 1lb beans) / (beans per lb)
  (price per bean) = (2k) / (p/4k)
  (price per bean) = 8k²/p

  So the answer to the question: “how many beans in 5 lb?”, is: 5 × (p/4k).

  All we need to do now is determine k and p.

  “You get 6 times as many beans for the price of 6 carrots as you do grapes for the price of 2 lb. of onions.”

  (number of beans for (price of 6 carrots)) = 6× (number of grapes for (price of 2lb onions))
  (number of beans for 6k pence) = 6× (number of grapes for kp/24 pence)
  (6k) / (8k²/p) = 6× (kp/24) / (3/4)
  (kp/24) (8k/p) = (3/4)
  8k² / 24 = 3 / 4
  k² = 9/4
  k = 3/2

  “two dozen carrots make three [lbs]”

  (carrots per lb) = 24/3 = 8
  (weight per carrot) = 1/8
  (price of 1lb carrots) = 8 × (3/2) = 12

  This leaves us with:

  “Seven times the number of peas that weigh the same as the number of onions that cost as much as 2 lb. of grapes, is less by the number of beans that weigh 12 lb. and the number of pence in the price of 16 lb. of carrots, than the number of beans that weigh as many pounds as the difference between the number of grapes that weigh the same as half the number of onions as there are carrots in a pound and a half, and the number of beans that cost as many pennies as there are grapes in the weight of a gross of peas.”

  We can break this down into:

  X = 7A + B + C

  where:

  A = “the number of peas that weigh the same as the number of onions that cost as much as 2 lb. of grapes”
  B = “the number of beans that weigh 12 lb.”
  C = “the number of pence in the price of 16 lb. of carrots”
  X = “the number of beans that weigh as many pounds as the difference between the number of grapes that weigh the same as half the number of onions as there are carrots in a pound and a half, and the number of beans that cost as many pennies as there are grapes in the weight of a gross of peas.”

  A = (the number of peas that weigh (the weight of (the number of onions that cost (the price of 2lb of grapes))))
  A = (the number of peas that weigh (the weight of (the number of onions that cost 48 pence)))
  A = (the number of peas that weigh (the weight of 64 onions))
  A = (the number of peas that weigh 1536/p lbs)
  A = 1536

  B = (the number of beans that weigh 12lb)
  B = 12 × (p/6)
  B = 2p

  C = (the number of pence in the price of 16lb of carrots)
  C = 16 × 12
  C = 192

  X = (the number of beans that weigh as many pounds as (the difference between (the number of grapes that weigh the same as (half the number of onions as there are (carrots in a pound and a half))) and (the number of beans that cost as many pennies as there are (grapes in the weight of (a gross of peas))))

  X = (the number of beans that weigh as many pounds as (the difference between Y and Z))
  Y = (the number of grapes that weigh the same as (half the number of onions as there are (carrots in a pound and a half)))
  Z = (the number of beans that cost as many pennies as there are (grapes in the weight of (a gross of peas)))

  Y = (the number of grapes that weigh the same as (half the number of onions as (the number of carrots in a pound and a half)))
  Y = (the number of grapes that weigh the same as (half the number of onions as (12 carrots)))
  Y = (the number of grapes that weigh the same as (6 onions))
  Y = (the number of grapes that weigh 144/p lbs)
  Y = 4608/p

  Z = (the number of beans that cost as many pennies as there are (grapes in the weight of (a gross of peas)))
  Z = (the number of beans that cost as many pennies as there are (grapes in (the weight of 144 peas)))
  Z = (the number of beans that cost as many pennies as there are (grapes in 144/p lbs))
  Z = (the number of beans that cost 4608/p pennies)
  Z = 256

  There are probably more than 18 peas per lb, so:

  X = (the number of beans that weigh as many pounds as (256 − 4608/p))
  X = (256 − 4608/p) × (p/6)
  X = 128p/3 − 768

  Hence:

  X = 7A + B + C
  128p/3 − 768 = 7×1536 + 2p + 192
  128p/3 = 11712 + 2p
  p = 35136 / 122
  p = 288

  Solution: There are 240 beans in 5 lb.

  We can summarise the calculated information as:

  grapes: 32 grapes per lb; price per lb = 24d
  carrots: 8 carrots per lb; price per lb = 12d
  onions: 12 onions per lb; price per lb = 9d
  peas: 288 peas per lb; price per lb = ?
  beans: 48 beans per lb; price per lb = 3d

  Which means grapes and onions have the same unit price (3/4 d each).

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