Difference between revisions of "2003 AMC 12B Problems/Problem 9"

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== Problem 9 ==
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== Problem ==
 
Let <math>f</math> be a linear function for which <math>f(6) - f(2) = 12.</math> What is <math>f(12) - f(2)?</math>
 
Let <math>f</math> be a linear function for which <math>f(6) - f(2) = 12.</math> What is <math>f(12) - f(2)?</math>
  
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</math>
 
</math>
  
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==Solution 1==
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Since <math>f</math> is a linear function with slope <math>m</math>,
  
[[2003 AMC 12B Problems/Problem 9|Solution]]
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<cmath>m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3</cmath>
  
Since <math>f</math> is a linear function with slope <math>m</math>,
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<cmath>f(12) - f(2) = m \Delta x = 3(12 - 2) = 30 \Rightarrow \text (D)</cmath>
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==Solution 2==
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Since <math>f</math> is linear, we can easily guess and check to confirm that <math>f(x)=3x</math>. Indeed, <math>f(6)-f(2)=3(6-2)=12</math>. So, we have <math>f(12)-f(2)=3(12-2)=30 \Rightarrow \text (D).</math>
  
<cmath>m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3</cmath>
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Solution by franzliszt
  
<cmath>f(12) - f(2) = m \Delta x = 4(12 - 2) = 30 \Rightarrow \text (D)</cmath>
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==See Also==
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{{AMC12 box|year=2003|ab=B|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 18:05, 7 July 2020

Problem

Let $f$ be a linear function for which $f(6) - f(2) = 12.$ What is $f(12) - f(2)?$

$\text {(A) } 12 \qquad \text {(B) } 18 \qquad \text {(C) } 24 \qquad \text {(D) } 30 \qquad \text {(E) } 36$

Solution 1

Since $f$ is a linear function with slope $m$,

\[m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3\]

\[f(12) - f(2) = m \Delta x = 3(12 - 2) = 30 \Rightarrow \text (D)\]

Solution 2

Since $f$ is linear, we can easily guess and check to confirm that $f(x)=3x$. Indeed, $f(6)-f(2)=3(6-2)=12$. So, we have $f(12)-f(2)=3(12-2)=30 \Rightarrow \text (D).$

Solution by franzliszt

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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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