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Difference between revisions of "2009 AMC 8 Problems/Problem 1"
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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>.
 
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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 +
==Video Solution==
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https://www.youtube.com/watch?v=USVVURBLaAc
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|before=First Problem|num-a=2}}
 
{{AMC8 box|year=2009|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:50, 28 January 2021

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

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 $7\cdot 2$, giving us the answer $\boxed{\textbf{(E)}\ 14}$.

Video Solution

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

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

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