Difference between revisions of "2006 AMC 8 Problems/Problem 1"

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== Solution ==
 
== Solution ==
The three prices round to <math> \textdollar 2 </math>, <math> \textdollar 5 </math>, and <math> \textdollar 10 </math>, which has a sum of <math> \boxed{\textbf{(D)}\ 17} </math>
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The three prices round to <math> \textdollar 2 </math>, <math> \textdollar 5 </math>, and <math> \textdollar 10 </math>, which have a sum of <math> \boxed{\textbf{(D)}\ 17} </math>.
  
{{AMC8 box|year=2006|num-b=First Problem|num-a=2}}
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We know that there will not be a rounding error, as the total amount rounded is clearly less than <math> \textdollar 0.50 </math>.
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==See Also==
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{{AMC8 box|year=2006|before=First <br />Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 13:14, 18 March 2020

Problem

Mindy made three purchases for $\textdollar 1.98$ dollars, $\textdollar 5.04$ dollars, and $\textdollar 9.89$ dollars. What was her total, to the nearest dollar?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

Solution

The three prices round to $\textdollar 2$, $\textdollar 5$, and $\textdollar 10$, which have a sum of $\boxed{\textbf{(D)}\ 17}$.

We know that there will not be a rounding error, as the total amount rounded is clearly less than $\textdollar 0.50$.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
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 AJHSME/AMC 8 Problems and Solutions

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