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

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. Then 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. ~ ellenmom

Video Solution 1

Education, The Study of Everything

~IceMatrix

~savannahsolver

~ pi_is_3.14

See Also

 2020 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 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.

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