Difference between revisions of "2002 AMC 12B Problems/Problem 2"
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| + | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #2]] and [[2002 AMC 10B Problems|2002 AMC 10B #4]]}} | ||
== Problem == | == Problem == | ||
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\qquad\mathrm{(E)}\ 12</math> | \qquad\mathrm{(E)}\ 12</math> | ||
== Solution == | == Solution == | ||
| + | By the distributive property, | ||
| − | <cmath>(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = | + | <cmath>(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}</cmath> |
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=3|num-a=5}} | ||
{{AMC12 box|year=2002|ab=B|num-b=1|num-a=3}} | {{AMC12 box|year=2002|ab=B|num-b=1|num-a=3}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 17:25, 28 July 2011
- The following problem is from both the 2002 AMC 12B #2 and 2002 AMC 10B #4, so both problems redirect to this page.
Problem
What is the value of 
when 
?
Solution
By the distributive property,
![\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}\]](http://latex.artofproblemsolving.com/e/a/5/ea57cc064cbe645f5ccdaf5e4735e62e5cae147f.png)
See also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
