Difference between revisions of "2003 AMC 12B Problems/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> | ||
| + | |||
| + | <math> | ||
| + | \text {(A) } 12 \qquad \text {(B) } 18 \qquad \text {(C) } 24 \qquad \text {(D) } 30 \qquad \text {(E) } 36 | ||
| + | </math> | ||
| + | |||
| + | ==Solution 1== | ||
Since <math>f</math> is a linear function with slope <math>m</math>, | Since <math>f</math> is a linear function with slope <math>m</math>, | ||
<cmath>m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3</cmath> | <cmath>m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3</cmath> | ||
| − | <cmath>f(12) - f(2) = m \Delta x = | + | <cmath>f(12) - f(2) = m \Delta x = 3(12 - 2) = 30 \Rightarrow \text (D)</cmath> |
| + | |||
| + | ==Solution 2== | ||
| + | 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> | ||
| + | |||
| + | Solution by franzliszt | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2003|ab=B|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:05, 7 July 2020
Contents
Problem
Let 
be a linear function for which 
What is 
Solution 1
Since is a linear function with slope 
,
![\[m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3\]](http://latex.artofproblemsolving.com/a/a/2/aa203aedcd175b965d80f840ac73c7f9356e3456.png)
![\[f(12) - f(2) = m \Delta x = 3(12 - 2) = 30 \Rightarrow \text (D)\]](http://latex.artofproblemsolving.com/1/f/d/1fd57e2924ffbf1a5fdb30a0312b68a900db6039.png)
Solution 2
Since is linear, we can easily guess and check to confirm that 
. Indeed, 
. So, we have 
Solution by franzliszt
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
