| Stars And Pounds |
Let the operations '#' and '*' be defined as follows:
a * b = 1/a + 1/b
a # b = (a + b)/(a - b)
Find the value X such that:
(22 * X) # (X * 33) = 27 |
Problem Moderated by: Douglas |
| Problem Solution |
Let Y = 22 * X = 1/22 + 1/X, and
Z = X * 33 = 1/X + 1/33
Then Y # Z = 27
(Y + Z)/(Y-Z) = 27
Y + Z = 27Y - 27Z
28Z = 26Y
Now substituting our original equations:
28/33 + 28/X = 26/22 + 26/X
28/33 - 26/22 = 26/X - 28/X
Multiplying through by -66 gives:
84 - 52 = 132/X
22 = 132/X
X = 132/22
X = 6
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