Difference between revisions of "2014 AMC 8 Problems/Problem 1"
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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)}-10}</math>. | , so our answer is <math>\boxed{\textbf{(A)}-10}</math>. | ||
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| + | ==Solution 2== | ||
| + | We use PEMDAS, 8-(2+5) is the same as 8-7 which is 1, 8-2+5 is the same as 6+5 because 8-2 comes first, and the sum is 11. H-T is 1-11 which is <math>\boxed{\textbf{(A)}-10}</math>. | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
Latest revision as of 00:57, 15 September 2024
Contents
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 
.
Solution 2
We use PEMDAS, 8-(2+5) is the same as 8-7 which is 1, 8-2+5 is the same as 6+5 because 8-2 comes first, and the sum is 11. H-T is 1-11 which is .
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
