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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2012 AMC 10A Problems/Problem 6"

(Fixed brackets)
(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 23: Line 23:
 
The sum would be <math>\frac{3}{2}+6</math> = <math>\boxed{\textbf{(D)}\ \frac{15}{2}}</math>
 
The sum would be <math>\frac{3}{2}+6</math> = <math>\boxed{\textbf{(D)}\ \frac{15}{2}}</math>
  
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/jyA5_tjDOjc
 +
 +
~Education, the Study of Everything
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2012|ab=A|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 12:55, 1 July 2023

Problem

The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the reciprocal of the other number. What is the sum of the two numbers?

$\textbf{(A)}\ \frac{10}{3}\qquad\textbf{(B)}\ \frac{20}{3}\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ \frac{15}{2}\qquad\textbf{(E)}\ 8$

Solution

Let the two numbers equal $x$ and $y$. From the information given in the problem, two equations can be written:

$xy=9$

$\frac{1}{x}=4 \left( \frac{1}{y} \right)$

Therefore, $4x=y$

Replacing $y$ with $4x$ in the equation,

$4x^2=9$

So $x=\frac{3}{2}$ and $y$ would then be $4 \times$ $\frac{3}{2}=6$

The sum would be $\frac{3}{2}+6$ = $\boxed{\textbf{(D)}\ \frac{15}{2}}$

Video Solution (CREATIVE THINKING)

https://youtu.be/jyA5_tjDOjc

~Education, the Study of Everything

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_6&oldid=195023"