Art of Problem Solving

2018 AMC 8 Problems/Problem 1: Difference between revisions

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==Solution 2==
==Solution 2==


Alternatively, you can note that <math>20\cdot14</math> is <math>280</math>, and that <math>20\cdot15</math> is <math>300</math>. Since <math>289</math> is closer to <math>280</math>, the answer is <math>\boxed{\textbf{(A)}14}</math> ~ Mathscienceclass
Also, you can note that <math>20\cdot14</math> is <math>280</math>, and that <math>20\cdot15</math> is <math>300</math>. Since <math>289</math> is closer to <math>280</math>, the answer is <math>\boxed{\textbf{(A)}14}</math> ~ Mathscienceclass
(Vishal Vinjamuri)


==See Also==
==See Also==

Revision as of 23:36, 18 June 2019

Problem 1

An amusement park has a collection of scale models, with ratio $1 : 20$, 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 replica 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

You can set up a ratio: $\frac{1}{20}=\frac{x}{289}$. Cross multiplying, you get $20x=289$. You divide by 20 to get $x=14.45$. The closest integer is $\boxed{\textbf{(A)}14}$

Solution 2

Also, you can note that $20\cdot14$ is $280$, and that $20\cdot15$ is $300$. Since $289$ is closer to $280$, the answer is $\boxed{\textbf{(A)}14}$ ~ Mathscienceclass (Vishal Vinjamuri)

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

These problems are copyrighted © by the Mathematical Association of America.