Difference between revisions of "2012 AMC 10B Problems/Problem 6"
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| − | == Problem | + | == Problem == |
| − | In order to estimate the value of x-y where x and y are real numbers with x > y > 0, Xiaoli rounded x up by a small amount, rounded y 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? |
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| + | <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 } x-y \qquad \textbf{(E) } \text{Her estimate is } 0</math> | ||
| + | == Solution == | ||
| − | + | Let's define <math>z</math> as the amount rounded up by and down by. | |
| − | + | The problem statement tells us that Xiaoli performed the following computation: | |
| − | + | <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 so the answer is <math>\boxed{\textbf{(A) } \text{Her estimate is larger than } x-y}</math>. | |
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| − | + | ==See Also== | |
| − | + | {{AMC10 box|year=2012|ab=B|num-b=5|num-a=7}} | |
| − | + | {{MAA Notice}} | |
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Latest revision as of 16:07, 18 January 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
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. 
