Difference between revisions of "2015 AMC 12A Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | To maximize our estimate, we want to maximize <math>\frac{a}{b}</math> and minimize <math>c</math>, because both terms are positive values. Therefore we round <math>c</math> down. To maximize <math>\frac{a}{b}</math>, round <math>a</math> up and <math>b</math> down. <math>\Rightarrow \boxed{\textbf{( | + | To maximize our estimate, we want to maximize <math>\frac{a}{b}</math> and minimize <math>c</math>, because both terms are positive values. Therefore we round <math>c</math> down. To maximize <math>\frac{a}{b}</math>, round <math>a</math> up and <math>b</math> down. <math>\Rightarrow \boxed{\textbf{(D)}}</math> |
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2015|ab=A|num-b=4|num-a=6}} |
Revision as of 20:41, 4 February 2015
Problem
Amelia needs to estimate the quantity , where and are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of ?
$\textbf{(A)}\ \text{She rounds all three numbers up.}\\ \qquad\textbf{(B)}\ \text{She rounds } a \text{ and } b \text{ up, and she rounds } c \text{down.}\\ \qquad\textbf{(C)}\ \text{She rounds } a \text{ and } c \text{ up, and she rounds } b \text{down.} \\ \qquad\textbf{(D)}}\ \text{She rounds } a \text{ up, and she rounds } b \text{ and } c \text{down.}\\ \qquad\textbf{(E)}\ \text{She rounds } c \text{ up, and she rounds } a \text{ and } b \text{down.}$ (Error compiling LaTeX. Unknown error_msg)
Solution
To maximize our estimate, we want to maximize and minimize , because both terms are positive values. Therefore we round down. To maximize , round up and down.
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 |