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

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==Problem==
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== 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 <math>48</math> free throws. How many free throws did she make at the first practice?
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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?
  
<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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<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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== 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>.
  
 
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>
  
==See Also==
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== Video Solution 1==
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https://youtu.be/6rkc-C9wllA
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~Education, the Study of Everything
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== See Also ==
 
{{AMC12 box|year=2004|ab=B|before=First Question|num-a=2}}
 
{{AMC12 box|year=2004|ab=B|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{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 (ProblemsAnswer KeyResources)
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

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