perplexus.info :: Tricks : Proving a witch weighs the same as a duck

Proving a witch weighs the same as a duck (Posted on 2005-02-21)

Suppose the witch weighs W pounds, and the duck weighs D pounds. Also, suppose their average weight is A pounds. (Their combined weight is 2A)

W + D = 2A. (Given)
W = 2A - D. (Subtract D)
W - 2A = -D (Subtract 2A)
W(W - 2A) = W(-D) (Multiply by W)
W(W - 2A) = (2A - D)(-D) (Substitute 2A-D for W using the given equation)
W² - 2AW = -2AD + D² (Distribute)
W² - 2AW + A² = D² - 2AD + A² (Add A²)
(W-A)(W-A)=(D-A)(D-A). (Factor)
(W-A)² = (D-A)². (An expression times itself equals the expression squared)
(W-A) = (D-A) (Square root)
W = D (Add A)

In other words, a witch weighs the same as a duck.

Where did I go wrong?

See The Solution Submitted by Dustin

Rating: 3.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Puzzle Solution

Comment 13 of 13 |

The anomaly stems from the step:

(W-A)^2 = (D-A)^2

=> W-A = D-A

In reality, however, we must have:

(W-A)^2 = (D-A)^2

=> W-A = D-A or, W-A = - (D-A)

Now, we know that :

A= Average (W, D) implying inter-alia that:

W >A >D

Then, considering the former CASE, we observe that:

W- A is a positive quantity, but D-A is a negative quantity. This is a contradiction.

Accordingly, the equality W-A = D-A is untenable.

Considering the latter case, we have:

W-A = -(D-A)

=> W+D = 2A

=> A =(W+ D)/2 = Average (W, D)

This equality is inclusive of the given conditions and therefore true.

Consequently, both the cases should have been considered and succinct analysis made of the foregoing before drawing the erroneous conclusion based on a single case.

Edited on July 16, 2022, 10:22 pm

Posted by K Sengupta on 2022-07-16 22:13:28

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