Difference between revisions of "2013 AMC 10A Problems/Problem 5"
(→Solution: added slightly faster solution) |
|||
| Line 6: | Line 6: | ||
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 </math> | <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
The total amount paid is <math>105 + 125 + 175 = 405</math>. To get how much each should have paid, we do <math>405/3 = 135</math>. | The total amount paid is <math>105 + 125 + 175 = 405</math>. To get how much each should have paid, we do <math>405/3 = 135</math>. | ||
Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that <math>t-d = 30 - 10 = \boxed{\textbf{(B) }20}</math>. | Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that <math>t-d = 30 - 10 = \boxed{\textbf{(B) }20}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | The difference in the money that Tommy paid and Dorothy paid is <math>20</math>. In order for them both to have paid the same amount, Tommy must pay <math>20</math> more than Dorothy. The answer is <math>\boxed{\mathbb{B}}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 12:21, 27 January 2015
Contents
Problem
Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy 
dollars, and Dorothy gave Sammy 
dollars. What is 
?
Solution 1
The total amount paid is 
. To get how much each should have paid, we do 
.
Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that 
.
Solution 2
The difference in the money that Tommy paid and Dorothy paid is 
. In order for them both to have paid the same amount, Tommy must pay more than Dorothy. The answer is 
.
See Also
| 2013 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 | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
