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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>
  
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== Solution ==  
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== Solution 1 ==  
  
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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>.
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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{\textbf{(B) }18}</math>.~taarunganesh
  
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== Solution 2 ==
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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>
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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.
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==Solution 3==
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We have 1 Bench = 7 Adults = 11 Children, or
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<math>A = \frac{11}{7}C</math>.
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The minimum number of benches is <math>N = x(A + C)</math>.
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This gives <math>\frac{77}{18}N = x</math>, so that the minimum value for x is the integer 18, or (B).
  
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== Solution 2 ==
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~PeterDoesPhysics
  
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This is similar to Solution 1, with the same basic idea, but a little different. Since both <math>7</math> and <math>11</math> are prime, their LCM must be their product. However, we don't want to find the LCM, but rather the number of seats required to fit the children and adults. So the answer would be <math>7 + 11 = \boxed{\text{(B)} 18}</math>.
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==Video Solution 1==
  
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Education, The Study of Everything
  
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~Baolan
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https://youtu.be/GKTQO99CKPM
  
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==Video Solution==
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==Video Solution 2==
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution 3==
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https://youtu.be/w2_H96-yzk8
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~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==

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

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