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{(B)}}</math>
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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 $\frac{a}{b} - c$, where $a, b,$ and $c$ 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 $\frac{a}{b} - c$?

$\textbf{(A)}\ \text{She rounds all three numbers up.}\\ \qquad\textbf{(B)}\ \text{She rounds } a \text{ and } b \text{ up, and she rounds } c \text{down.}\\ \qquad\textbf{(C)}\ \text{She rounds } a \text{ and } c \text{ up, and she rounds } b \text{down.} \\ \qquad\textbf{(D)}}\ \text{She rounds } a \text{ up, and she rounds } b \text{ and } c \text{down.}\\ \qquad\textbf{(E)}\ \text{She rounds } c \text{ up, and she rounds } a \text{ and } b \text{down.}$ (Error compiling LaTeX. Unknown error_msg)

Solution

To maximize our estimate, we want to maximize $\frac{a}{b}$ and minimize $c$, because both terms are positive values. Therefore we round $c$ down. To maximize $\frac{a}{b}$, round $a$ up and $b$ down. $\Rightarrow \boxed{\textbf{(D)}}$

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
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