Difference between revisions of "2018 AMC 8 Problems/Problem 1"
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==Solution 3== | ==Solution 3== | ||
We know that <math> 20 \cdot 14 = 280 ,</math> and that <math> 20 \cdot 15 = 300 .</math> These are the multiples of <math>20</math> around <math>289 ,</math> and the closest one of those is <math>280.</math> Therefore, the answer is <math> \dfrac {280} {20} = \boxed{\textbf{(A) }14} .</math> | We know that <math> 20 \cdot 14 = 280 ,</math> and that <math> 20 \cdot 15 = 300 .</math> These are the multiples of <math>20</math> around <math>289 ,</math> and the closest one of those is <math>280.</math> Therefore, the answer is <math> \dfrac {280} {20} = \boxed{\textbf{(A) }14} .</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/rRaMWpifJJE | ||
| + | |||
| + | ~savannahsolver | ||
==See also== | ==See also== | ||
Revision as of 10:03, 18 February 2022
Problem
An amusement park has a collection of scale models, with a ratio of 
, of buildings and other sights from around the country. The height of the United States Capitol is 
feet. What is the height in feet of its duplicate to the nearest whole number?
Solution 1
You can see that since the ratio of real building's heights to the model building's height is 
. We also know that the U.S Capitol is feet in real life, so to find the height of the model, we divide by 20. That gives us 
which rounds to 14. Therefore, to the nearest whole number, the duplicate is 
. ~avamarora.
Solution 2
We can compute 
and round our answer to get 
.
It is basically Solution 1 without the ratio calculation. However, Solution 1 is referring further to the problem.
Solution 3
We know that 
and that 
These are the multiples of 
around 
and the closest one of those is 
Therefore, the answer is 
Video Solution
~savannahsolver
See also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
