Difference between revisions of "2012 AMC 12B Problems/Problem 6"
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| − | == Problem== | + | == Problem == |
In order to estimate the value of <math>x-y</math> where <math>x</math> and <math>y</math> are real numbers with <math>x>y>0</math>, Xiaoli rounded <math>x</math> up by a small amount, rounded <math>y</math> down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct? | In order to estimate the value of <math>x-y</math> where <math>x</math> and <math>y</math> are real numbers with <math>x>y>0</math>, Xiaoli rounded <math>x</math> up by a small amount, rounded <math>y</math> down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct? | ||
| − | + | <math>\textbf{(A) } \text{Her estimate is larger than } x - y \qquad \textbf{(B) } \text{Her estimate is smaller than } x - y \qquad \textbf{(C) } \text{Her estimate equals } x - y</math> | |
| + | <math>\textbf{(D) } \text{Her estimate equals } y - x \qquad \textbf{(E) } \text{Her estimate is } 0</math> | ||
| − | + | == 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>. | |
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| − | ==Solution== | ||
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| − | 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 | ||
== See Also == | == See Also == | ||
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{{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}} | ||
Revision as of 00:25, 19 October 2020
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 
.
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. 
