Difference between revisions of "2020 AMC 10A Problems/Problem 9"

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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77</math>
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77</math>
  
== Solution ==  
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== 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>.
 
  
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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>.~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
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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}</math> + <math>\frac{x}{11}</math> = <math>N</math>. Simplifying we get <math>\frac{18x}{77} = N</math>
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Since both <math>n</math> and <math>x</math> have to be integers, <math>x</math> has to equal <math>77</math>. Therefore, <math>N</math> = <math>\boxed{\text{(B) }18}</math> is our final answer.
  
==Video Solution==
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==Video Solution 1==
  
 
Education, The Study of Everything
 
Education, The Study of Everything
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https://youtu.be/GKTQO99CKPM
 
https://youtu.be/GKTQO99CKPM
  
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==Video Solution 2==
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  
 
~IceMatrix
 
~IceMatrix
  
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==Video Solution 3==
 
https://youtu.be/w2_H96-yzk8
 
https://youtu.be/w2_H96-yzk8
  
 
~savannahsolver
 
~savannahsolver
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== Video Solution 4==
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https://youtu.be/ZhAZ1oPe5Ds?t=1616
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 11:19, 2 February 2021

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 $\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}$.~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 integers, $x$ has to equal $77$. Therefore, $N$ = $\boxed{\text{(B) }18}$ is our final answer.

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 (ProblemsAnswer KeyResources)
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

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