Difference between revisions of "2018 AMC 8 Problems/Problem 1"

(Solution)
(Solution 1(Good Ol' Ratios))
 
(59 intermediate revisions by 31 users not shown)
Line 1: Line 1:
==Problem 1==
+
==Problem==
An amusement park has a collection of scale models, with ratio <math>1 : 20</math>, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number?
+
An amusement park has a collection of scale models, with a ratio of <math> 1: 20</math>, of buildings and other sights from around the country. The height of the United States Capitol is <math>289</math> feet. What is the height in feet of its duplicate to the nearest whole number?
  
 
<math>\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math>
 
<math>\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math>
  
==Solution==
+
==Solution 1(Good Ol' Ratios)==
  
  
You can set up a ratio: <math>\frac{1}{20}=\frac{x}{289}</math>. Cross multiplying, you get <math>20x=289</math>. You divide by 20 to get <math>x=14.45</math>. The closest integer is <math>\boxed{\textbf{(A)}14}</math>
+
You can see that since the ratio of real building's heights to the model building's height is <math>1:20</math>. We also know that the U.S Capitol is <math>289</math> feet in real life, so to find the height of the model, we divide by 20. That gives us <math>14.45</math> which rounds to 14. Therefore, to the nearest whole number, the duplicate is <math>\boxed{\textbf{(A) }14}</math>
  
==See Also==
+
~avamarora, Nivaar..
 +
 
 +
==Solution 2==
 +
 
 +
 
 +
We can compute <math>\frac{289}{20}</math> and round our answer to get <math>\boxed{\textbf{(A) }14}</math>.
 +
It is basically Solution 1 without the ratio calculation. However, Solution 1 is referring further to the problem.
 +
 
 +
==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>
 +
 
 +
== Video Solution (CRITICAL THINKING!!!)==
 +
https://youtu.be/zF5PqNpI0RM
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/rRaMWpifJJE
 +
 
 +
~savannahsolver
 +
 
 +
==See also==
 
{{AMC8 box|year=2018|before=First Problem|num-a=2}}
 
{{AMC8 box|year=2018|before=First Problem|num-a=2}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:05, 8 July 2024

Problem

An amusement park has a collection of scale models, with a ratio of $1: 20$, of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?

$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

Solution 1(Good Ol' Ratios)

You can see that since the ratio of real building's heights to the model building's height is $1:20$. We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\boxed{\textbf{(A) }14}$.

~avamarora, Nivaar..

Solution 2

We can compute $\frac{289}{20}$ and round our answer to get $\boxed{\textbf{(A) }14}$. It is basically Solution 1 without the ratio calculation. However, Solution 1 is referring further to the problem.

Solution 3

We know that $20 \cdot 14 = 280 ,$ and that $20 \cdot 15 = 300 .$ These are the multiples of $20$ around $289 ,$ and the closest one of those is $280.$ Therefore, the answer is $\dfrac {280} {20} = \boxed{\textbf{(A) }14} .$

Video Solution (CRITICAL THINKING!!!)

https://youtu.be/zF5PqNpI0RM

~Education, the Study of Everything

Video Solution

https://youtu.be/rRaMWpifJJE

~savannahsolver

See also

2018 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png