Difference between revisions of "2007 AMC 10B Problems/Problem 2"

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Substitute and simplify.
 
Substitute and simplify.
<cmath>(3+5)5 - (5+3)3 = (3+5)2 = (8)2 = \boxed{\textbf{(E) }16}</cmath>
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<cmath>(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{\textbf{(E) }16}</cmath>
  
 
==Solution 2==
 
==Solution 2==
 
Note that
 
Note that
<math>(a \star b) - (b \star a) = (a+b)b - (b+a)a
+
<math>(a \star b) - (b \star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2</math>. We can substitute <math>a=3</math> and <math>b=5</math> to get <math>5^2 - 3^2 = \boxed{\textbf{(E) }16}</math>.
= (a+b)(b-a)
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= b^2 - a^2
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~MathFun1000 (Minor Edits)
= 5^2 - 3^2 = \boxed{\textbf{(E) }16}</math>
 
Note numbers were only plugged in at the end.
 
  
 
==See Also==
 
==See Also==

Latest revision as of 09:33, 7 March 2022

Problem

Define the operation $\star$ by $a \star b = (a+b)b.$ What is $(3 \star 5) - (5 \star 3)?$

$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16$

Solution 1

Substitute and simplify. \[(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{\textbf{(E) }16}\]

Solution 2

Note that $(a \star b) - (b \star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2$. We can substitute $a=3$ and $b=5$ to get $5^2 - 3^2 = \boxed{\textbf{(E) }16}$.

~MathFun1000 (Minor Edits)

See Also

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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