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

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==Solution==
 
==Solution==
  
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We can work backwards to solve the problem. Bridget had 7 apples before she gave Cassie 3 apples. These 7 apples were half of Bridget’s 14 original apples. So the answer is <math>\boxed{\bold{\text{E}}}</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
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==See Also==
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{{AMC8 box|year=2009|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 12:37, 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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