Difference between revisions of "2013 AMC 12A Problems/Problem 5"
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==Problem== | ==Problem== | ||
| + | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $<math>105</math>, Dorothy paid $<math>125</math>, and Sammy paid $<math>175</math>. In order to share the costs equally, Tom gave Sammy <math>t</math> dollars, and Dorothy gave Sammy <math>d</math> dollars. What is <math>t-d</math>? | ||
| + | |||
| + | <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 </math> | ||
| + | |||
| + | ==Solution== | ||
Add up the amounts that Tom, Dorothy, and Sammy paid to get $<math>405</math>, and divide by 3 to get $<math>135</math>, the amount that each should have paid. | Add up the amounts that Tom, Dorothy, and Sammy paid to get $<math>405</math>, and divide by 3 to get $<math>135</math>, the amount that each should have paid. | ||
| − | |||
Tom, having paid $<math>105</math>, owes Sammy $<math>30</math>, and Dorothy, having paid $<math>125</math>, owes Sammy $<math>10</math>. | Tom, having paid $<math>105</math>, owes Sammy $<math>30</math>, and Dorothy, having paid $<math>125</math>, owes Sammy $<math>10</math>. | ||
Thus, <math>t - d = 30 - 10 = 20</math>, which is <math>\boxed{\textbf{(B)}}</math> | Thus, <math>t - d = 30 - 10 = 20</math>, which is <math>\boxed{\textbf{(B)}}</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=A|num-b=4|num-a=6}} | ||
Revision as of 18:33, 22 February 2013
Problem
Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $
, Dorothy paid $
, and Sammy paid $
. In order to share the costs equally, Tom gave Sammy 
dollars, and Dorothy gave Sammy 
dollars. What is 
?
Solution
Add up the amounts that Tom, Dorothy, and Sammy paid to get $
, and divide by 3 to get $
, the amount that each should have paid.
Tom, having paid $, owes Sammy $
, and Dorothy, having paid $, owes Sammy $
.
Thus, 
, which is 
See also
| 2013 AMC 12A (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 12 Problems and Solutions | |
