Difference between revisions of "1999 AMC 8 Problems/Problem 18"
Talkinaway (talk | contribs) (Solutions 1 and 2) |
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<math>\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math> | <math>\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math> | ||
| − | + | ==Solution== | |
| − | ==Solution 1== | + | ===Solution 1=== |
If <math>108</math> students eat <math>2</math> cookies on average, there will need to be <math>108\cdot 2 = 216</math> cookies. But with the smaller attendance, you will only need <math>100\% - 25\% = 75\%</math> of these cookies, or <math>75\% \cdot 216 = 0.75\cdot 216 = 162</math> cookies. | If <math>108</math> students eat <math>2</math> cookies on average, there will need to be <math>108\cdot 2 = 216</math> cookies. But with the smaller attendance, you will only need <math>100\% - 25\% = 75\%</math> of these cookies, or <math>75\% \cdot 216 = 0.75\cdot 216 = 162</math> cookies. | ||
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<math>162</math> cookies requires <math>\frac{162}{15} = 10.8</math> batches. However, since half-batches are forbidden, we must round up to get <math>\lceil \frac{162}{15} \rceil = 11</math> batches, and the correct answer is <math>\boxed{E}</math>. | <math>162</math> cookies requires <math>\frac{162}{15} = 10.8</math> batches. However, since half-batches are forbidden, we must round up to get <math>\lceil \frac{162}{15} \rceil = 11</math> batches, and the correct answer is <math>\boxed{E}</math>. | ||
| − | ==Solution 2== | + | ===Solution 2=== |
If there were <math>108</math> students before, with the <math>25\%</math> of students missing, there will be <math>75\%</math> of <math>108</math> students left. This is <math>75\% \cdot 108 = 0.75 \cdot 108 = 81</math> students. These students eat <math>81 \cdot 2 = 162</math> cookies. Follow the logic of the second paragraph above to find that there needs to be <math>11</math> batches, and the correct answer is <math>\boxed{E}</math>. | If there were <math>108</math> students before, with the <math>25\%</math> of students missing, there will be <math>75\%</math> of <math>108</math> students left. This is <math>75\% \cdot 108 = 0.75 \cdot 108 = 81</math> students. These students eat <math>81 \cdot 2 = 162</math> cookies. Follow the logic of the second paragraph above to find that there needs to be <math>11</math> batches, and the correct answer is <math>\boxed{E}</math>. | ||
| − | ==See | + | ==See Also== |
{{AMC8 box|year=1999|num-b=17|num-a=19}} | {{AMC8 box|year=1999|num-b=17|num-a=19}} | ||
Revision as of 12:59, 23 December 2012
Problem
At Central Middle School the 108 students who take the AMC8 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 flour, 
eggs, 
tablespoons butter, 
cups sugar, and 
package of chocolate drops. They will make only full recipes, not partial recipes.
They learn that a big concert is scheduled for the same night and attendance will be down 25%. How many recipes of cookies should they make for their smaller party?
Solution
Solution 1
If 
students eat cookies on average, there will need to be 
cookies. But with the smaller attendance, you will only need 
of these cookies, or 
cookies.
cookies requires 
batches. However, since half-batches are forbidden, we must round up to get 
batches, and the correct answer is 
.
Solution 2
If there were students before, with the 
of students missing, there will be 
of students left. This is 
students. These students eat 
cookies. Follow the logic of the second paragraph above to find that there needs to be 
batches, and the correct answer is .
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
