Difference between revisions of "2014 AMC 8 Problems/Problem 1"
(→Solution) |
(→Solution) |
||
Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
We have <math>H=8-7=1</math> and <math>T=8-2+5=11</math>. Clearly <math>1-11=-10</math> | We have <math>H=8-7=1</math> and <math>T=8-2+5=11</math>. Clearly <math>1-11=-10</math> | ||
− | , so our answer is <math>\boxed{\textbf{(A)}- | + | , so our answer is <math>\boxed{\textbf{(A)}-10}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|before=First Problem|num-a=2}} | {{AMC8 box|year=2014|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:55, 10 November 2019
Problem
Harry and Terry are each told to calculate . Harry gets the correct answer. Terry ignores the parentheses and calculates . If Harry’s answer is and Terry’s answer is , what is ?
Solution
We have and . Clearly , so our answer is .
See Also
2014 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.