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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.  
  
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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.  

Revision as of 15:24, 21 October 2020

Problem

A store normally sells windows at $100$ 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?

$\mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$

Solution

The store's offer means that every $5$th window is free.

Dave would get $\left\lfloor\frac{7}{5}\right\rfloor=1$ free window.

Doug would get $\left\lfloor\frac{8}{5}\right\rfloor=1$ free window.

This is a total of $2$ free windows.

Together, they would get $\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3$ free windows.

So they get $3-2=1$ additional window if they purchase the windows together.

Therefore they save $1\cdot100=100\Rightarrow A$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AMC 10 Problems and Solutions

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