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 = 11\ \mathrm{(D)}</cmath>
+
<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 16:25, 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 $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$?

$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$

Solution

By the distributive property,

\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}\]

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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
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