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2002 AMC 10B Problems

Revision as of 17:37, 26 December 2008 by James4l (talk | contribs) (New page: ==Problem 1== The ratio <math>2^{2001}\cdot{3^{2003}}\over6^{2002}</math> is: (A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/2 Solution == Problem 2 ==For the...)
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Problem 1

The ratio $2^{2001}\cdot{3^{2003}}\over6^{2002}$ is:

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 3/2

Solution

== Problem 2 ==For the nonzero numbers a, b, and c, define

(a,b,c)=$abc\over{a+b+c}$

Find (2,4,6).

(A) 1 (B) 2 (C) 4 (D) 6 (E) 24

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also

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