Difference between revisions of "2002 AMC 12B Problems/Problem 2"

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\qquad\mathrm{(D)}\ 11
 
\qquad\mathrm{(D)}\ 11
 
\qquad\mathrm{(E)}\ 12</math>
 
\qquad\mathrm{(E)}\ 12</math>
== Solution ==
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== 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>
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<cmath>(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}</cmath>.
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== Solution 2 ==
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Inputting 4 into <math>x</math> in the original equation,
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<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 ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:46, 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.

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 1

By the distributive property,

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

Solution 2

Inputting 4 into $x$ in the original equation,

\[[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \boxed{\mathrm{(D)}\ 11}\]

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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