Difference between revisions of "2002 AMC 12B Problems/Problem 2"
(Added solution 2) |
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\qquad\mathrm{(D)}\ 11 | \qquad\mathrm{(D)}\ 11 | ||
\qquad\mathrm{(E)}\ 12</math> | \qquad\mathrm{(E)}\ 12</math> | ||
| − | == Solution == | + | == Solution 1 == |
By the distributive property, | By the distributive property, | ||
| − | <cmath>(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}</cmath> | + | <cmath>(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}</cmath>. |
| + | |||
| + | == Solution 2 == | ||
| + | Inputting 4 into <math>x</math> in the original equation, | ||
| + | |||
| + | <cmath>[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \boxed{\mathrm{(D)}\ 11}</cmath> | ||
== See also == | == See also == | ||
| Line 19: | Line 24: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:46, 9 August 2022
- The following problem is from both the 2002 AMC 12B #2 and 2002 AMC 10B #4, so both problems redirect to this page.
Contents
Problem
What is the value of 
when 
?
Solution 1
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)
.
Solution 2
Inputting 4 into 
in the original equation,
![\[[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \boxed{\mathrm{(D)}\ 11}\]](http://latex.artofproblemsolving.com/3/b/9/3b94bdd1f28f6ba5874344e48addcc518070cf1f.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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
