Difference between revisions of "2011 AMC 12A Problems/Problem 3"
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\textbf{(E)}\ 15 </math> | \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, or <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== | ==Video Solution== | ||
Revision as of 17:56, 10 August 2023
Problem
A small bottle of shampoo can hold 
milliliters of shampoo, whereas a large bottle can hold 
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?
Solution 1
To find how many small bottles we need, we can simply divide by . This simplifies to 
Since the answer must be an integer greater than 
, we have to round up to 
bottles, or 
Solution 2
We double to get 
We see that 
which is very close to 
Thus, $2\cdot7+1=\boxed{\text{(E)} 15$ (Error compiling LaTeX. Unknown error_msg) bottles are enough.
~Technodoggo
Video Solution
~savannahsolver
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 | |
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
