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

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==Comment==
 
==Comment==
 
It would be nicer if the organizers chose a more difficult substitution, say <math>x=4.7</math>, to penalize those solutions that do not take advantage of the distributive property.
 
It would be nicer if the organizers chose a more difficult substitution, say <math>x=4.7</math>, to penalize those solutions that do not take advantage of the distributive property.
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==Comment==
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They should also have made the numbers fit in, for the distributive property.
  
 
== See also ==
 
== See also ==

Revision as of 21:31, 17 April 2017

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,

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

Comment

It would be nicer if the organizers chose a more difficult substitution, say $x=4.7$, to penalize those solutions that do not take advantage of the distributive property.

Comment

They should also have made the numbers fit in, for the distributive property.

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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