Difference between revisions of "2009 AMC 8 Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | If we set up an equation, we find out x=(3+4) | + | 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. |
| − | We can simplify the equations to x=7 | + | We can simplify the equations to <math>x=7\cdot 2=14.</math> |
| − | The answer is (E) 14. | + | The answer is <math>\text{(E) } 14.</math> |
~John0412 | ~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 <math> \qquad\textbf{(E)}\ 14 </math>. | ||
| + | ~Anabel.disher | ||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=USVVURBLaAc | https://www.youtube.com/watch?v=USVVURBLaAc | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/a2-76knCCEE | ||
| + | |||
| + | ~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}} | ||
{{MAA Notice}} | {{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?
Solution
If we set up an equation, we find out 
because 3 apples were left after giving half, then four away.
We can simplify the equations to 
The answer is 
~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 
.
~Anabel.disher
Video Solution
https://www.youtube.com/watch?v=USVVURBLaAc
Video Solution 2
~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. 
