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Difference between revisions of "2012 AMC 12B Problems/Problem 6"

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== Problem==
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== Problem ==
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In order to estimate the value of x-y where x and y are real numbers with x <math>\g</math> y
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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?
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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</math>
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<math>\textbf{(D) } \text{Her estimate equals } y - x \qquad \textbf{(E) } \text{Her estimate is } 0</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 <math>\boxed{\textbf{(A)}}</math>.
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== Solution 2 ==
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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
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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 06:32, 29 June 2023

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?

$\textbf{(A) } \text{Her estimate is larger than } x - y \qquad \textbf{(B) } \text{Her estimate is smaller than } x - y \qquad \textbf{(C) } \text{Her estimate equals } x - y$ $\textbf{(D) } \text{Her estimate equals } y - x \qquad \textbf{(E) } \text{Her estimate is } 0$

Solution

The original expression $x-y$ now becomes $(x+k) - (y-k)=(x-y)+2k>x-y$, where $k$ is a positive constant, hence the answer is $\boxed{\textbf{(A)}}$.

Solution 2

The problem never says what $x$ and $y$ are, so we can decide what they are. Let $x = 1.6$ and $y = 1.4$. We round $x$ to $2$ and $y$ to $1$. Then the new $x - y = 1$, while the original $x - y = 0.2$. Thus, the new $x - y$ is greater than the original $x - y$. The answer is $\boxed{\textbf{(A)}}$. ~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

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