Difference between revisions of "2004 AMC 12B Problems/Problem 1"
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== Problem == | == 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 48 free throws. How many free throws did she make at the first practice? | |
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+ | <math>(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</math> | ||
== Solution == | == Solution == | ||
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Because there are five days, or four transformations between days (day 1 <math>\rightarrow</math> day 2 <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{\mathrm{(A)}\ 3}</math> | Because there are five days, or four transformations between days (day 1 <math>\rightarrow</math> day 2 <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{\mathrm{(A)}\ 3}</math> | ||
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+ | == Video Solution 1== | ||
+ | https://youtu.be/6rkc-C9wllA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== See Also == | == See Also == |
Latest revision as of 19:21, 22 October 2022
Contents
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 48 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 2
day 3
day 4
day 5), she makes
Video Solution 1
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.