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

Revision as of 20:50, 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: $f(12) - f(2) = 3(12) + b - (3(2) + b)$ $= 30 \Rightarrow \text(D)$

Solution by CYB3RFLARE7408

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 17: / Line 17: @@
 Solution by franzliszt
+==Solution 3==
+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>.
+<cmath>f(6) = a(6) + b </cmath>
+<cmath>f(2) = a(2) + b </cmath>
+Now subtracting:
+<cmath>f(6) - f(2) = 4a </cmath>
+<cmath>12 = 4a \Rightarrow a = 3 </cmath>
+Analyze the target equation to get:
+<cmath>f(12) - f(2) = 3(12) + b - (3(2) + b)</cmath>
+<cmath> = 30 \Rightarrow \text(D)</cmath>
+Solution by CYB3RFLARE7408
 ==See Also==
 {{AMC12 box|year=2003|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}

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