Difference between revisions of "2004 AMC 12B Problems/Problem 1"

Revision as of 10:48, 6 October 2020

Jasmine had 2 paperclips on Monday, then she had 6 on Tuesday, and her number of paperclips proceeded to triple on each subsequent day. On what day of the week did she first have more than 100 paperclips?

Solution

Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\frac{1}{2} \cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \Rightarrow \mathrm{(A)}$ .

Because there are five days, or four transformations between days (day 1 $\rightarrow$ day 2 $\rightarrow$ day 3 $\rightarrow$ day 4 $\rightarrow$ day 5), she makes $48 \cdot \frac{1}{2^4} = \boxed{\mathrm{(A)}\ 3}$

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.

@@ Line 1: / Line 1: @@
-==Problem==
+Jasmine had 2 paperclips on Monday, then she had 6 on Tuesday, and her number of paperclips proceeded to triple on each subsequent day. On what day of the week did she first have more than 100 paperclips?
-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 <math>48</math> free throws. How many free throws did she make at the first practice?
-<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 6 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 12 \qquad \mathrm{(E)}\ 15</math>
 ==Solution==

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Difference between revisions of "2004 AMC 12B Problems/Problem 1"

Revision as of 10:48, 6 October 2020

Solution

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