Difference between revisions of "2004 AMC 12B Problems/Problem 1"
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==Solution== | ==Solution== | ||
Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made <math>\frac{1}{2} \cdot 48 = 24</math> free throws, on the third <math>12</math>, on the second <math>6</math>, and on the first <math>3 \Rightarrow \mathrm{(A)}</math>. | Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made <math>\frac{1}{2} \cdot 48 = 24</math> free throws, on the third <math>12</math>, on the second <math>6</math>, and on the first <math>3 \Rightarrow \mathrm{(A)}</math>. | ||
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| + | Because there are five days, or four transformations between days (day 1 <math>\rightarrow</math> day 3 <math>\rightarrow</math> day 4 <math>\rightarrow</math> day 5), she makes <math>48 \cdot \frac{1}{2^4} = \boxed{(A) \qquad 3}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 19:57, 25 December 2011
Problem
At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 
free throws. How many free throws did she make at the first practice?
Solution
Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made 
free throws, on the third 
, on the second 
, and on the first 
.
Because there are five days, or four transformations between days (day 1 
day 3 day 4 day 5), she makes 
See Also
| 2004 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Question |
Followed by Problem 2 |
| 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 | |
