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Difference between revisions of "2016 AMC 8 Problems/Problem 10"
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<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math>
 
<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math>
  
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==Solutions==
 
  
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===Solution 1===
+
==Solution 1==
 
Let us plug in <math>(5 * x)=1</math> into <math>3a-b</math>. Thus it would be <math>3(5)-x</math>. Now we have <math>2*(15-x)=1</math>. Plugging <math>2*(15-x)</math> into <math>3a-b</math>, we have <math>6-15+x=1</math>. Solving for <math>x</math> we have <cmath>-9+x=1</cmath><cmath>x=\boxed{\textbf{(D)}  \,        10}</cmath>
 
Let us plug in <math>(5 * x)=1</math> into <math>3a-b</math>. Thus it would be <math>3(5)-x</math>. Now we have <math>2*(15-x)=1</math>. Plugging <math>2*(15-x)</math> into <math>3a-b</math>, we have <math>6-15+x=1</math>. Solving for <math>x</math> we have <cmath>-9+x=1</cmath><cmath>x=\boxed{\textbf{(D)}  \,        10}</cmath>
  
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===Solution 2===
+
==Solution 2==
 
Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for y in the equation <math>3(2)-y=1</math>, we see that y is equal to five. By substitution, we see that <math>5 * x</math> = 5. Solving for x in the equation <math>5(3)-x = 5</math> we get <cmath>x=\boxed{\textbf{(D)}  \,        10}</cmath>
 
Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for y in the equation <math>3(2)-y=1</math>, we see that y is equal to five. By substitution, we see that <math>5 * x</math> = 5. Solving for x in the equation <math>5(3)-x = 5</math> we get <cmath>x=\boxed{\textbf{(D)}  \,        10}</cmath>
 +
 +
==See Also==
 
{{AMC8 box|year=2016|num-b=9|num-a=11}}
 
{{AMC8 box|year=2016|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:31, 21 April 2021

Problem

Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$


Solution 1

Let us plug in $(5 * x)=1$ into $3a-b$. Thus it would be $3(5)-x$. Now we have $2*(15-x)=1$. Plugging $2*(15-x)$ into $3a-b$, we have $6-15+x=1$. Solving for $x$ we have \[-9+x=1\]\[x=\boxed{\textbf{(D)} \, 10}\]

Solution 2

Let us set a variable $y$ equal to $5 * x$. Solving for y in the equation $3(2)-y=1$, we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for x in the equation $5(3)-x = 5$ we get \[x=\boxed{\textbf{(D)} \, 10}\]

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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