Difference between revisions of "2003 AMC 12B Problems/Problem 9"
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− | == Problem | + | == 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> | ||
+ | ==Solution 1== | ||
+ | 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>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 | ||
+ | |||
+ | ==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> (NOTE: <math>b</math> cancels) | ||
+ | <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}} |
Latest revision as of 20:51, 11 September 2025
Problem
Let be a linear function for which
What is
Solution 1
Since is a linear function with slope
,
Solution 2
Since is linear, we can easily guess and check to confirm that
. Indeed,
. So, we have
Solution by franzliszt
Solution 3
Since is linear,
=
, we can use a system of equations to solve for
.
Now subtracting:
Analyze the target equation to get:
(NOTE:
cancels)
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.