Difference between revisions of "2012 AMC 10B Problems/Problem 6"
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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)} | + | |
+ | <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 \\ \qquad \textbf{(D) } \text{Her estimate equals } y-x \qquad \textbf{(E) } \text{Her estimate is } 0</math> | ||
== Solution == | == Solution == | ||
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<math>\left(x+z\right) - \left(y-z\right) = x+z-y+z = x-y+2z</math> | <math>\left(x+z\right) - \left(y-z\right) = x+z-y+z = x-y+2z</math> | ||
− | We can see that <math>x-y+2z</math> is greater than <math>x-y</math>, and the answer is <math>\textbf{(A)} \text{Her estimate is larger than} x-y</math> | + | We can see that <math>x-y+2z</math> is greater than <math>x-y</math>, and so the answer is <math>\boxed{\textbf{(A) } \text{Her estimate is larger than } x-y}</math>. |
Latest revision as of 13:47, 9 August 2024
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
Let's define as the amount rounded up by and down by.
The problem statement tells us that Xiaoli performed the following computation:
We can see that is greater than , and so the answer is .
See Also
2012 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.