Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 8 Problems/Problem 10"
(→‎Solution)
Line 6: Line 6:
 
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>
  
 +
==Solution 2==
 +
Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for 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 the equation <math>5(3)-x = 5</math> we get <math>-9+x=1<cmath></cmath>x=\boxed{\textbf{(D)}  \,        10}</math>
 
{{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 16:50, 12 May 2020

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

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 the equation $3(2)-y=1$, we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for the equation $5(3)-x = 5$ we get $-9+x=1<cmath></cmath>x=\boxed{\textbf{(D)} \, 10}$

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_10&oldid=122330"