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 A LPP Puzzle (Posted on 2006-12-11)
Long ago, in Medieval Europe, there lived an unscrupulous trader named Escrocfleuve. One day he decided to utilise one of his false balances both in buying and selling of Item A, thereby gaining 11.44% more than the LPP on the item. It was noted that if the scale pans on which Item A was weighed were interchanged, Escrocfleuve's actual percentage profit would have been 10.4% less than the LPP on the said item. Determine the LPP on Item A in that transaction.

On another occasion, Escrocfleuve used another of his defective balances on the buying and selling of Item B. This time he gained 9.9% more than the LPP on the item. It was noted that if the scale pans on which Item B was weighed were interchanged, Escrocfleuve would still achieve an overall profit of 2.01% by the transaction. Determine the LPP on Item B in the other transaction.

Note: Legitimate Percentage Profit (LPP) is the profit achieved by a person in the buying and selling of an article with a true balance; assume, for the current problem, that LPP is always positive.

 See The Solution Submitted by K Sengupta Rating: 1.0000 (1 votes)

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 solution | Comment 1 of 2

Part I:

Let f be the factor by which the scale misrepresents.
Let L be the LPP expressed as a fraction rather than a percentage.

(1+L)f = 1 + L + .1144
(1+L)/f = 1 + L - .104

Multiplying the equations:

(1+L)^2 = 1 + L + .1144 + L^2 + 1.1144 L - .1158976 = 2.1144 L + .9985024 + L^2

1+L^2+2L = 2.1144 L + .9985024 + L^2

.1144 L = .0014976

L = .01309090... , or 1 17/55 %.

... and the scale is off by a factor of 1.112921751615219

Part II:

(1+L)f = 1 + L + .099
(1+L)/f = 1.0201

(1+L)^2 = 1.1210899 + 1.0201 L

1+2L+L^2 = 1.1210899 + 1.0201 L

L^2 + .9799L -.1210899 = 0

By the quadratic formula, the positive value for L is .111, or an 11.1% profit, and the scale is off by a factor of 1.089108910891089.

 Posted by Charlie on 2006-12-11 11:01:04

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