Difference between revisions of "2004 AMC 10B Problems/Problem 3"

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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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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>
 
<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==
  
At the fourth practice she made <math>48/2=24</math> throws, at the third one it was <math>24/2=12</math>, then we get <math>12/2=6</math> throws for the second practice, and finally <math>6/2=\boxed{3}</math> throws at the first one.
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At the fourth practice she made <math>48/2=24</math> throws, at the third one it was <math>24/2=12</math>, then we get <math>12/2=6</math> throws for the second practice, and finally <math>6/2=3\Rightarrow\boxed{\mathrm{(A)}\ 3}</math> throws at the first one.
  
 
== See also ==
 
== See also ==
  
 
{{AMC10 box|year=2004|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2004|ab=B|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 20:02, 22 July 2014

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

At the fourth practice she made $48/2=24$ throws, at the third one it was $24/2=12$, then we get $12/2=6$ throws for the second practice, and finally $6/2=3\Rightarrow\boxed{\mathrm{(A)}\ 3}$ throws at the first one.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 10 Problems and Solutions

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