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

(New page: ==Problem== Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club? <math>(\mathrm {A}) 297</m...)
 
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==Problem==
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== Problem ==
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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>
  
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Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club?
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== Solution ==
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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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<math>(\mathrm {A}) 297</math>       <math>(\mathrm {B}) 330</math>       <math>(\mathrm {C}) 363</math>       <math>(\mathrm {D}) 396</math>       <math>(\mathrm {E}) 726</math>
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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>
  
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==Solution==
 
  
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There are 22-12+1=11 rows reserved for the youth club, and at 33 seats per row, there are 33*11 seats reserved.
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== Video Solution 1==
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https://youtu.be/6rkc-C9wllA
  
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33*11=363
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~Education, the Study of Everything
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== See Also ==
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{{AMC12 box|year=2004|ab=B|before=First Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:21, 22 October 2022

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?

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

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 $48 \cdot \frac{1}{2^4} = \boxed{\mathrm{(A)}\ 3}$


Video Solution 1

https://youtu.be/6rkc-C9wllA

~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. AMC logo.png

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