Difference between revisions of "2020 AMC 10A Problems/Problem 9"
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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> | 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{\ | + | 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. |
==Video Solution 1== | ==Video Solution 1== | ||
Revision as of 19:33, 13 June 2022
Contents
Problem
A single bench section at a school event can hold either 
adults or 
children. When 
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 
Solution 1
The least common multiple of and is 
. Therefore, there must be adults and children. The total number of benches is 
.~taarunganesh
Solution 2
Let 
denote how many adults there are. Since the number of adults is equal to the number of children we can write as 
. Simplifying we get 
Since both 
and have to be positive integers, has to equal . Therefore, 
is our final answer.
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~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. 
