Difference between revisions of "1999 AMC 8 Problems/Problem 17"
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<math>1</math> pan requires <math>2</math> eggs, so <math>15</math> pans require <math>2\cdot 15 = 30</math> eggs. Since there are <math>6</math> eggs in a half dozen, we need <math>\frac{30}{6} = 5</math> half-dozens of eggs, and the answer is <math>\boxed{C}</math> | <math>1</math> pan requires <math>2</math> eggs, so <math>15</math> pans require <math>2\cdot 15 = 30</math> eggs. Since there are <math>6</math> eggs in a half dozen, we need <math>\frac{30}{6} = 5</math> half-dozens of eggs, and the answer is <math>\boxed{C}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/NraP1LG6zoU Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|num-b=16|num-a=18}} | {{AMC8 box|year=1999|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:00, 21 December 2022
Contents
Problem
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: 
cups of flour, 
eggs, 
tablespoons butter, 
cups sugar, and 
package of chocolate drops. They will make only full recipes, not partial recipes.
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)
Solution
If 
students eat cookies on average, there will need to be 
cookies. There are 
cookies per pan, meaning there needs to be 
pans. However, since half-recipes are forbidden, we need to round up and make 
pans.
pan requires eggs, so pans require 
eggs. Since there are 
eggs in a half dozen, we need 
half-dozens of eggs, and the answer is 
Video Solution
https://youtu.be/NraP1LG6zoU Soo, DRMS, NM
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
