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Difference between revisions of "2009 AMC 8 Problems/Problem 1"

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==Solution==
 
==Solution==
  
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Let us work backwards. We know that Cassie had 4 apples for herself at the end, and we know she gave away 3 apples before. Therefore, she had 7 apples before giving half of her original amount of apples to someone else. Since half of the amount of original apples is equal to seven, then the original amount of apples Bridget had is <math>7\cdot 2</math>, giving us the answer <math>\boxed{\textbf{(E)}\ 14}</math>.
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If we set up an equation, we find out <math>x=(3+4)\cdot 2</math> because 3 apples were left after giving half, then four away.  
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We can simplify the equations to <math>x=7\cdot 2=14.</math>
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The answer is <math>\text{(E) } 14.</math>
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~John0412
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==Solution 2==
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You can work backwards and add 3 apples that she gave to Cassie to the 4 she currently has, which results in 7, and then multiply by 2 since she gave half the apples to Ann, resulting in <math> \qquad\textbf{(E)}\ 14 </math>.  
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~Anabel.disher
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==Video Solution==
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https://www.youtube.com/watch?v=USVVURBLaAc
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==Video Solution 2==
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https://youtu.be/a2-76knCCEE
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|before=First Problem|num-a=2}}
 
{{AMC8 box|year=2009|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 11:37, 31 December 2023

Problem

Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 14$

Solution

If we set up an equation, we find out $x=(3+4)\cdot 2$ because 3 apples were left after giving half, then four away. We can simplify the equations to $x=7\cdot 2=14.$ The answer is $\text{(E) } 14.$

~John0412

Solution 2

You can work backwards and add 3 apples that she gave to Cassie to the 4 she currently has, which results in 7, and then multiply by 2 since she gave half the apples to Ann, resulting in $\qquad\textbf{(E)}\ 14$. ~Anabel.disher

Video Solution

https://www.youtube.com/watch?v=USVVURBLaAc

Video Solution 2

https://youtu.be/a2-76knCCEE

~savannahsolver

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
First Problem 		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 AJHSME/AMC 8 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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