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

Revision as of 01:54, 7 January 2021

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?

$(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15$

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 <math>2</math> paperclips on Monday, then she had <math>6</math> 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 <math>100</math> paperclips?
+<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 01:54, 7 January 2021

Problem

Solution

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