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Quiz of the Month (June 2005)

Hector C. Parr

***

SOLUTION TO LAST MONTH'S QUIZ

  1.  12.65 ft
  2.  11.28 ft
  3.  1.09 ft
Notes. 
  1.  By similar triangles, OA/OB = CX/CB = 3/9
        So OA = 4, and AB = 12.6491 by Pythagoras.

  2.  Let OA = a, OB = b
   By similar triangles, (b-3)/3  = b/a  -  (i)
   By Pythagoras,         a2 + b2 = 144  -  (ii)
        From (i), 3a + 3b = ab
        Squaring, 9a2 + 18ab + 9b2 = a2b2
       From (ii), 9a2        + 9b2 = 1296   
    Subtracting, (ab)2 - 18(ab) - 1296 = 0
   Solving for (ab) and rejecting negative root,
        ab = 46.10795
   From (i), a = 3b/(b-3)
   So 3b2/(b-3) = ab = 46.10795
    Solving and rejecting smaller root,
         b = 11.2827

  3.  a = 46.10795 / 11.2827 = 4.0866
       So DA = 1.0866

THIS MONTH'S QUIZ

(As you will recall, Ruritanian coins are minted only in the following denominations:
1c, 2c, 5c, 10c, 20c, 50c.)

When shopping in Ruritania I always tender a whole number of Ruros. The change is equally likely to be any number of Rurocents from 0c to 99c, and the checkout staff always give the change using the least possible number of coins.

1. What is the probability that the change includes a 1c coin?

2. What is the probability that the change includes at least one 20c coin?

3. What is the probability that the change consists of five or more coins?

***

© Hector C. Parr (2005)

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