Difference between revisions of "2020 AMC 10A Problems/Problem 9"
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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>. | 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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+ | ==Video Solution== | ||
+ | https://youtu.be/JEjib74EmiY | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 23:13, 31 January 2020
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
The least common multiple of and
is
. Therefore, there must be
adults and
children. The total number of benches is
.
Video Solution
~IceMatrix
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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