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

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==Problem==
 
==Problem==
 
Harry and Terry are each told to calculate <math>8-(2+5)</math>. Harry gets the correct answer. Terry ignores the parentheses and calculates <math>8-2+5</math>. If Harry's answer is <math>H</math> and Terry's answer is <math>T</math>, what is <math>H-T</math>?
 
Harry and Terry are each told to calculate <math>8-(2+5)</math>. Harry gets the correct answer. Terry ignores the parentheses and calculates <math>8-2+5</math>. If Harry's answer is <math>H</math> and Terry's answer is <math>T</math>, what is <math>H-T</math>?
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<math>\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10</math>
 
<math>\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10</math>
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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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 +
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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>.
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==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/dljBE3_G85E
 +
 +
~Education, the Study of Everything
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 +
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==Video Solution==
 +
https://youtu.be/6TRHPxBveh0
 +
 +
~savannahsolver
  
 
==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}}

Latest revision as of 00:57, 15 September 2024

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 $\boxed{\textbf{(A)}-10}$.



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 $\boxed{\textbf{(A)}-10}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/dljBE3_G85E

~Education, the Study of Everything


Video Solution

https://youtu.be/6TRHPxBveh0

~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

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