Difference between revisions of "2005 AMC 8 Problems/Problem 22"
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==Solution== | ==Solution== | ||
Suppose the small size costs <math>\textdollar 1</math> and the large size has <math>10</math> oz. The medium size then costs <math>\textdollar 1.50</math> and has <math>8</math> oz. The small size has <math>5</math> oz and the large size costs <math>\textdollar 1.95</math>. The small, medium, and large size cost respectively, <math>0.200, 0.188, 0.195</math> dollars per oz. The sizes from best to worst buy are <math>\boxed{\textbf{(E)}\ \text{MLS}}</math>. | Suppose the small size costs <math>\textdollar 1</math> and the large size has <math>10</math> oz. The medium size then costs <math>\textdollar 1.50</math> and has <math>8</math> oz. The small size has <math>5</math> oz and the large size costs <math>\textdollar 1.95</math>. The small, medium, and large size cost respectively, <math>0.200, 0.188, 0.195</math> dollars per oz. The sizes from best to worst buy are <math>\boxed{\textbf{(E)}\ \text{MLS}}</math>. | ||
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==Video Solution== | ==Video Solution== | ||
Revision as of 08:51, 17 June 2024
Contents
Problem
A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs 50% more than the small size and contains 20% less detergent than the large size. The large size contains twice as much detergent as the small size and costs 30% more than the medium size. Rank the three sizes from best to worst buy.
Solution
Suppose the small size costs 
and the large size has 
oz. The medium size then costs 
and has 
oz. The small size has 
oz and the large size costs 
. The small, medium, and large size cost respectively, 
dollars per oz. The sizes from best to worst buy are 
.
Video Solution
https://www.youtube.com/watch?v=65cfokoNyfA ~David
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
