Difference between revisions of "2011 AMC 12A Problems/Problem 3"

Latest revision as of 17:56, 10 August 2023

Problem

A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?

$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution 1

To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\boxed{\textbf{E}}$

Solution 2

We double $35$ to get $70.$ We see that $70\cdot7=490,$ which is very close to $500.$ Thus, $2\cdot7+1=\boxed{\text{(E)} 15}$ bottles are enough. ~Technodoggo

Video Solution

https://youtu.be/A6oQF25ayzo

~savannahsolver

2011 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ 15 </math>
-== Solution ==
+== Solution 1 ==
-To find how many small bottles we need, we can simply divide <math>500</math> by <math>35</math>. This simplifies to <math>\frac{100}{7}=14+\frac{2}{7}</math>. Since the answer must be an integer greater than <math>14</math>, we have to round up to <math>15</math> bottles=<math>\boxed{\textbf{E}}</math>
+To find how many small bottles we need, we can simply divide <math>500</math> by <math>35</math>. This simplifies to <math>\frac{100}{7}=14 \frac{2}{7}. </math> Since the answer must be an integer greater than <math>14</math>, we have to round up to <math>15</math> bottles, or <math>\boxed{\textbf{E}}</math>
+== Solution 2 ==
+We double <math>35</math> to get <math>70.</math> We see that <math>70\cdot7=490,</math> which is very close to <math>500.</math> Thus, <math>2\cdot7+1=\boxed{\text{(E)} 15}</math> bottles are enough.
+~Technodoggo
+==Video Solution==
+https://youtu.be/A6oQF25ayzo
+~savannahsolver
 == See also ==
 {{AMC12 box|year=2011|num-b=2|num-a=4|ab=A}}
+{{AMC10 box|year=2011|num-b=1|num-a=3|ab=A}}
+{{MAA Notice}}

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