Difference between revisions of "2005 AMC 10A Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | While eating out, Mike and Joe each tipped their server <math>2</math> | + | While eating out, Mike and Joe each tipped their server <math>\$2</math>. Mike tipped <math>10\%</math> of his bill and Joe tipped <math>20\%</math> of his bill. What was the difference, in dollars, between their bills? |
| − | <math> \ | + | <math> |
| + | \textbf{(A) } 2\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 10\qquad \textbf{(E) } 20 | ||
| + | </math> | ||
==Solution== | ==Solution== | ||
| − | Let <math>m</math> be Mike's bill and <math>j</math> be Joe's bill. | + | Let <math>m</math> be Mike's bill and <math>j</math> be Joe's bill (both in dollars). We then have <math>\frac{10}{100}m = 2 \iff m = 20</math> and <math>\frac{20}{100}j = 2 \iff j = 10</math>, so the desired difference is <math>\left\lvert m-j\right\rvert = 20-10 = \boxed{\textbf{(D) } 10}</math>. |
| − | + | ==Video Solution 1== | |
| + | https://youtu.be/OT47ZnF5MPY | ||
| − | + | ==Video Solution 2== | |
| + | https://youtu.be/mApPRP9v_XE | ||
| − | + | ~Charles3829 | |
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| − | |||
| − | |||
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==See Also== | ==See Also== | ||
| − | |||
| − | + | {{AMC10 box|year=2005|ab=A|before=First Problem|num-a=2}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:58, 1 July 2025
Problem
While eating out, Mike and Joe each tipped their server 
. Mike tipped 
of his bill and Joe tipped 
of his bill. What was the difference, in dollars, between their bills?
Solution
Let 
be Mike's bill and 
be Joe's bill (both in dollars). We then have 
and 
, so the desired difference is 
.
Video Solution 1
Video Solution 2
~Charles3829
See Also
| 2005 AMC 10A (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 AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
