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 "2020 AMC 10A Problems/Problem 9"

m (→‎Solution 1)
(→‎Solution 2)
 
(17 intermediate revisions by 6 users not shown)
Line 6: Line 6:
 
== Solution 1 ==  
 
== Solution 1 ==  
  
−
The least common multiple of <math>7</math> and <math>11</math> is <math>77</math>. Therefore, there must be <math>77</math> adults and <math>77</math> children. The total number of benches is <math>\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}</math>.
+
The least common multiple of <math>7</math> and <math>11</math> is <math>77</math>. Therefore, there must be <math>77</math> adults and <math>77</math> children. The total number of benches is <math>\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\textbf{(B) }18}</math>.~taarunganesh
−
-taarunganesh
 
  
 
== Solution 2 ==
 
== Solution 2 ==
  
−
This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both <math>7</math> and <math>11</math> are relatively prime, their LCM must be their product. So the answer would be <math>7 + 11 = \boxed{\text{(B) } 18}</math>.     ~Baolan
+
Let <math>x</math> denote how many adults there are. Since the number of adults is equal to the number of children we can write <math>N</math> as <math>\frac{x}{7}+\frac{x}{11}=N</math>. Simplifying we get <math>\frac{18x}{77} = N</math>
 +
Since both <math>n</math> and <math>x</math> have to be positive integers, <math>x</math> has to equal <math>77</math>. Therefore, <math>N=\boxed{\textbf{(B) }18}</math> is our final answer.
 +
 
 +
==Solution 3==
 +
We have 1 Bench = 7 Adults = 11 Children, or
 +
<math>A = \frac{11}{7}C</math>.
 +
 
 +
The minimum number of benches is <math>N = x(A + C)</math>.
 +
 
 +
This gives <math>\frac{77}{18}N = x</math>, so that the minimum value for x is the integer 18, or (B).
 +
 
 +
~PeterDoesPhysics
  
 
==Video Solution 1==
 
==Video Solution 1==
Line 28: Line 38:
  
 
~savannahsolver
 
~savannahsolver
 +
 +
== Video Solution 4==
 +
https://youtu.be/ZhAZ1oPe5Ds?t=1616
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 02:27, 2 September 2024

Problem

A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$

$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

Solution 1

The least common multiple of $7$ and $11$ is $77$. Therefore, there must be $77$ adults and $77$ children. The total number of benches is $\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\textbf{(B) }18}$.~taarunganesh

Solution 2

Let $x$ denote how many adults there are. Since the number of adults is equal to the number of children we can write $N$ as $\frac{x}{7}+\frac{x}{11}=N$. Simplifying we get $\frac{18x}{77} = N$ Since both $n$ and $x$ have to be positive integers, $x$ has to equal $77$. Therefore, $N=\boxed{\textbf{(B) }18}$ is our final answer.

Solution 3

We have 1 Bench = 7 Adults = 11 Children, or $A = \frac{11}{7}C$.

The minimum number of benches is $N = x(A + C)$.

This gives $\frac{77}{18}N = x$, so that the minimum value for x is the integer 18, or (B).

~PeterDoesPhysics

Video Solution 1

Education, The Study of Everything

https://youtu.be/GKTQO99CKPM

Video Solution 2

https://youtu.be/JEjib74EmiY

~IceMatrix

Video Solution 3

https://youtu.be/w2_H96-yzk8

~savannahsolver

Video Solution 4

https://youtu.be/ZhAZ1oPe5Ds?t=1616

~ pi_is_3.14

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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=2020_AMC_10A_Problems/Problem_9&oldid=226199"