Difference between revisions of "2005 AMC 10A Problems/Problem 5"
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The store's offer means that every <math>5</math>th window is free. | The store's offer means that every <math>5</math>th window is free. | ||
− | Dave would get <math>\lfloor\frac{7}{5}\rfloor=1</math> free window. | + | Dave would get <math>\left\lfloor\frac{7}{5}\right\rfloor=1</math> free window. |
− | Doug would get <math>\lfloor\frac{8}{5}\rfloor=1</math> free window. | + | Doug would get <math>\left\lfloor\frac{8}{5}\right\rfloor=1</math> free window. |
This is a total of <math>2</math> free windows. | This is a total of <math>2</math> free windows. | ||
− | Together, they would get <math>\lfloor\frac{8+7}{5}\rfloor = \lfloor\frac{15}{5}\rfloor=3</math> free windows. | + | Together, they would get <math>\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3</math> free windows. |
So they get <math>3-2=1</math> additional window if they purchase the windows together. | So they get <math>3-2=1</math> additional window if they purchase the windows together. | ||
− | Therefore they save <math>1\cdot100= | + | Therefore they save <math>1\cdot100=\left(\text{A}\right)100</math>. |
− | == | + | ==Video Solution== |
− | + | CHECK OUT Video Solution: https://youtu.be/ECBXQ1cE2SI | |
− | + | ==See also== | |
+ | {{AMC10 box|year=2005|ab=A|num-b=4|num-a=6}} | ||
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{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:01, 30 October 2020
Contents
Problem
A store normally sells windows at each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?
Solution
The store's offer means that every th window is free.
Dave would get free window.
Doug would get free window.
This is a total of free windows.
Together, they would get free windows.
So they get additional window if they purchase the windows together.
Therefore they save .
Video Solution
CHECK OUT Video Solution: https://youtu.be/ECBXQ1cE2SI
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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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