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

m (SOLUTION 2)
(Solution)
 
(2 intermediate revisions by the same user not shown)
Line 10: Line 10:
  
  
==SOLUTION 2== PEMDAS
 
  
 +
==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>.
 
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>.
  

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png