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}</math> + <math>\frac{x}{11}</math> = <math>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}</math> + <math>\frac{x}{11}</math> = <math>N</math>. Simplifying we get <math>\frac{18x}{77} = N</math> | ||
− | 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. | + | 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</math> = <math>\boxed{\text{(B) }18}</math> is our final answer. |
==Video Solution 1== | ==Video Solution 1== |
Revision as of 17:18, 31 March 2021
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.