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

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Solution by franzliszt
 
Solution by franzliszt
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==Solution 3==
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Since <math>f</math> is linear, <math>f(x)</math> = <math>ax+b</math>, we can use a system of equations to solve for <math>f</math>.
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<cmath>f(6) = a(6) + b </cmath>
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<cmath>f(2) = a(2) + b </cmath>
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Now subtracting:
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<cmath>f(6) - f(2) = 4a </cmath>
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<cmath>12 = 4a \Rightarrow a = 3 </cmath>
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Analyze the target equation to get:
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<cmath>f(12) - f(2) = 3(12) + b - (3(2) + b)</cmath> (NOTE: <math>b</math> cancels)
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<cmath> = 30 \Rightarrow \text(D)</cmath>
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Solution by CYB3RFLARE7408
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2003|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2003|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:51, 11 September 2025

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

Solution 3

Since $f$ is linear, $f(x)$ = $ax+b$, we can use a system of equations to solve for $f$.

\[f(6) = a(6) + b\] \[f(2) = a(2) + b\]


Now subtracting: \[f(6) - f(2) = 4a\] \[12 = 4a \Rightarrow a = 3\]


Analyze the target equation to get: \[f(12) - f(2) = 3(12) + b - (3(2) + b)\] (NOTE: $b$ cancels) \[= 30 \Rightarrow \text(D)\]

Solution by CYB3RFLARE7408



See Also

2003 AMC 12B (Problems • Answer Key • Resources)
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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