2014 AMC 8 Problems/Problem 1

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

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Video Solution

https://youtu.be/6TRHPxBveh0

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2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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

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