Difference between revisions of "2014 AMC 8 Problems/Problem 1"

(Solution)
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==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)}-10}</math>.
+
, so our answer is <math>\boxed{\textbf{(A)}-11}</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 $8-(2+5)$. Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$. If Harry’s answer is $H$ and Terry’s answer is $T$, what is $H-T$?


$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$


Solution

We have $H=8-7=1$ and $T=8-2+5=11$. Clearly $1-11=-10$ , so our answer is $\boxed{\textbf{(A)}-11}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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