Difference between revisions of "2005 AMC 12A Problems/Problem 4"

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== Solution ==
 
== Solution ==
For <math>n</math> windows, the store offers a discount of <math>100 \cdot \lfloor\frac{n}{5}\rfloor</math> ([[floor function]]). Dave receives a discount of <math>100 \cdot \lfloor \frac{7}{5}\rfloor = 100</math> and Doug receives a discount of <math>100 \cdot \lfloor \frac{8}{5}\rfloor = 100</math>. These amount to <math>200</math> dollars in discounts. Together, they receive a discount of <math>100 \cdot \lfloor \frac{15}{5} \rfloor = 300</math>, so they save <math>300-200=100\ \mathrm{(A)}</math>.
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For <math>n</math> windows, the store offers a discount of <math>100 \cdot \left\lfloor\frac{n}{5}\right\rfloor</math> ([[floor function]]). Dave receives a discount of <math>100 \cdot \left\lfloor \frac{7}{5}\right \rfloor = 100</math> and Doug receives a discount of <math>100 \cdot \left\lfloor \frac{8}{5}\right\rfloor = 100</math>. These amount to <math>200</math> dollars in discounts. Together, they receive a discount of <math>100 \cdot \left\lfloor \frac{15}{5} \right\rfloor = 300</math>, so they save <math>300-200=100\ \mathrm{(A)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:57, 29 December 2008

Problem

A store normally sells windows at <dollar/>$100$ 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 much 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

For $n$ windows, the store offers a discount of $100 \cdot \left\lfloor\frac{n}{5}\right\rfloor$ (floor function). Dave receives a discount of $100 \cdot \left\lfloor \frac{7}{5}\right \rfloor = 100$ and Doug receives a discount of $100 \cdot \left\lfloor \frac{8}{5}\right\rfloor = 100$. These amount to $200$ dollars in discounts. Together, they receive a discount of $100 \cdot \left\lfloor \frac{15}{5} \right\rfloor = 300$, so they save $300-200=100\ \mathrm{(A)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 12 Problems and Solutions