Difference between revisions of "2012 AMC 12B Problems/Problem 6"
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== Solution == | == Solution == | ||
The original expression <math>x-y</math> now becomes <math>(x+k) - (y-k)=(x-y)+2k>x-y</math>, where <math>k</math> is a positive constant, hence the answer is <math>\boxed{\textbf{(A)}}</math>. | The original expression <math>x-y</math> now becomes <math>(x+k) - (y-k)=(x-y)+2k>x-y</math>, where <math>k</math> is a positive constant, hence the answer is <math>\boxed{\textbf{(A)}}</math>. | ||
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+ | == Solution 2 == | ||
+ | The problem never says what <math>x</math> and <math>y</math> are, so we can decide what they are. Let <math>x = 1.6</math> and <math>y = 1.4</math>. We round <math>x</math> to <math>2</math> and <math>y</math> to <math>1</math>. Then the new <math>x - y = 1</math>, while the original <math>x - y = 0.2</math>. Thus, the new <math>x - y</math> is greater than the original <math>x - y</math>. The answer is <math>\boxed{\textbf{(A)}}</math>. ~Extremelysupercooldude | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=B|num-b=5|num-a=7}} | {{AMC12 box|year=2012|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:32, 29 June 2023
Contents
[hide]Problem
In order to estimate the value of where and are real numbers with , Xiaoli rounded up by a small amount, rounded down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct?
Solution
The original expression now becomes , where is a positive constant, hence the answer is .
Solution 2
The problem never says what and are, so we can decide what they are. Let and . We round to and to . Then the new , while the original . Thus, the new is greater than the original . The answer is . ~Extremelysupercooldude
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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.