Difference between revisions of "2011 AMC 12A Problems/Problem 3"
(Created page with '== Problem == == Solution == == See also == {{AMC12 box|year=2011|num-b=2|num-a=4|ab=A}}') |
m |
||
(12 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | A small bottle of shampoo can hold <math>35</math> milliliters of shampoo, whereas a large bottle can hold <math>500</math> 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? | ||
+ | |||
+ | <math> | ||
+ | \textbf{(A)}\ 11 \qquad | ||
+ | \textbf{(B)}\ 12 \qquad | ||
+ | \textbf{(C)}\ 13 \qquad | ||
+ | \textbf{(D)}\ 14 \qquad | ||
+ | \textbf{(E)}\ 15 </math> | ||
+ | |||
== Solution == | == Solution == | ||
+ | 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> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/A6oQF25ayzo | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=2|num-a=4|ab=A}} | {{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}} |
Latest revision as of 17:51, 22 December 2020
Contents
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
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
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.