Brain-Teaser 473: Gouttes d’Or
From The Sunday Times, 21st June 1970 [link]
The conical glasses at the Hôtel d’Or hold one eighth of a litre and are embellished from base to brim with a picture of Bacchus. In making the hotel’s famous Gouttes d’Or, sold at NF 5.20, it was customary to fill the glass up to Bacchus’ neck (4/5th of the way up the glass) and then add an olive; the profit on raw materials was a modest 100 per cent.
Gaston, the barman, has decided that he must make the more realistic profit of 400 per cent. So now the olive is put in first and then liquor is added up to the level of Bacchus’ chest (3/5ths of the way up the glass). The price has been reduced to NF 4.80.
Gaston explained: “Ze olives are standard sized and cost me one centime a cc. Yes, I could put in ze olive after measuring out the liquid — but zen it would be necessary to charge …”
How much?
This is a corrected version of the originally published puzzle. In the original version the price of the first glass was incorrectly given as NF 5.30 (instead of NF 5.20). The puzzle can still be solved in the same way, but the numbers don’t work out as nicely.
This puzzle is included in the book Sunday Times Brain Teasers (1974).
[teaser473]
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Jim Randell 8:03 am on 16 April 2019 Permalink |
If the conical glass has a height of h and a radius (at the top) of r, then a completely full glass will have a volume of:
So, if it was filled to a height of (k/5) h (and hence a radius of (k/5) r, then the volume of liquid is:
So a 4/5 full glass contains 64 cc, and a 3/5 full glass 27 cc.
If olives have a volume of x cc (and cost 1 centime per cc) and the liquor has a cost of y centimes per cc, and the profit is 100% (i.e. the cost of the drink is twice the cost of the raw materials), then for the first glass we have:
And for the second glass we have a 400% profit (i.e. the cost of the drink is 5 times the cost of the raw materials), on a drink with the olive added before the liquor we have:
(assuming the olive is submerged, and so displaces its own volume of liquid).
These two equations have a reasonable solution at x = 4, y = 4. (And an unreasonable solution where x = 219, which is a very large olive, and wouldn’t fit in the glass).
So the cost of a 3/5 glass, if the olive was added last, and the profit multiple is 5, would be:
Solution: The drink would cost NF 5.60.
Here is a Python program that solves the puzzle numerically. It runs in 81ms.
from enigma import (Polynomial, fdiv, find_zero, printf) # for a 4/5 full glass, with olive added last, the cost is k1, profit multiple is m1 # for a 3/5 full glass, with olive added first, the cost is k2, profile multiple is m2 # return (<olive volume>, <liquor cost>, <answer>) def solve(k1, m1, k2, m2): # volumes for a 4/5 and 3/5 glass (v4, v3) = (4**3, 3**3) # create a polynomial that gives the cost of liquor in terms of the volume of an olive # using the first equation: # k1 = m1 * (x + v4 * y) q = Polynomial([fdiv(k1, m1 * v4), fdiv(-1, v4)]) # create the polynomial whose roots are the volume of an olive # using the second equation: # k2 = m2 * (x + (v3 - x) * y) p = q * Polynomial([v3, -1]) - Polynomial([fdiv(k2, m2), -1]) # find a solution with a reasonably sized olive r = find_zero(p, 0, 10) x = r.v y = q(x) # cost of a 3/5 glass, with an olive last, profit multiple m2 k = m2 * (x + v3 * y) return (k, x, y) # solve for the required amounts (k, x, y) = solve(520, 2, 480, 5) printf("cost = NF {f:.02f} [x = {x:.02f}, y = {y:.02f}]", f=0.01 * k)If we change the first argument to [[
solve()]] on line 30 to 530 we get the solution to the puzzle as originally set: approximately NF 5.72.LikeLike