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Difference between revisions of "2020 AMC 8 Problems/Problem 24"

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==Solution 2==
 
==Solution 2==
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When <math>n=3</math>, we see that the total height of the large square is <math>3s+4d</math>. Similarly, when <math>n=24</math>, the total height of the large square is <math>24s+25d</math>. The total area of the 576 gray tiles is <math>576s^2</math> and the area of the large white square is <math>(24s+25d)^2</math>. We are given that the ratio of the gray area to the area of the large square is <math>\frac{64}{100}=\frac{16}{25}</math>. Thus, our equation becomes <math>\frac{576s^2}{(24s+25d)^2}=\frac{16}{25}</math>. Square rooting both sides, we get <math>\frac{24s}{24s+25d}=\frac{4}{5}</math>. Cross multiplying, we get <math>120s=96s+100d</math>. Combining like terms, we get <math>24s=100d</math>, which implies that <math>\frac{d}{s}=\frac{24}{100}=\frac{6}{25}\implies\boxed{\textbf{(A) }\frac{6}{25}}</math>.<br>
+
When <math>n=3</math>, we see that the total height of the large square is <math>3s+4d</math>. Similarly, when <math>n=24</math>, the total height of the large square is <math>24s+25d</math>. The total area of the <math>576</math> gray tiles is <math>576s^2</math> and the area of the large white square is <math>(24s+25d)^2</math>. We are given that the ratio of the gray area to the area of the large square is <math>\frac{64}{100}=\frac{16}{25}</math>. Thus, our equation becomes <math>\frac{576s^2}{(24s+25d)^2}=\frac{16}{25}</math>. Square rooting both sides, we get <math>\frac{24s}{24s+25d}=\frac{4}{5}</math>. Cross multiplying, we get <math>120s=96s+100d</math>. Combining like terms, we get <math>24s=100d</math>, which implies that <math>\frac{d}{s}=\frac{24}{100}=\frac{6}{25}\implies\boxed{\textbf{(A) }\frac{6}{25}}</math>.<br>
 
~jmansuri
 
~jmansuri
  

Revision as of 04:48, 18 November 2020

A large square region is paved with $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$. When $n=24$, the $576$ gray tiles cover $64\%$ of the area of the large square region. What is the ratio $\frac{d}{s}$ for this larger value of $n?$

[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); [/asy]

$\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}$

Solution 1

Let $s=1$. Then, the total area of the squares of side $s$ is $576$, $64\%$ of the area of the large square, which would be $900$, making the side of the large square $30$. Then, $25$ borders have a total length of $30-24=6$. Since $\frac{d}{s}=d$ if $s=1$ is the value we're asked to find, the answer is $\boxed{\textbf{(A) }\frac{6}{25}}$.

Solution 2

When $n=3$, we see that the total height of the large square is $3s+4d$. Similarly, when $n=24$, the total height of the large square is $24s+25d$. The total area of the $576$ gray tiles is $576s^2$ and the area of the large white square is $(24s+25d)^2$. We are given that the ratio of the gray area to the area of the large square is $\frac{64}{100}=\frac{16}{25}$. Thus, our equation becomes $\frac{576s^2}{(24s+25d)^2}=\frac{16}{25}$. Square rooting both sides, we get $\frac{24s}{24s+25d}=\frac{4}{5}$. Cross multiplying, we get $120s=96s+100d$. Combining like terms, we get $24s=100d$, which implies that $\frac{d}{s}=\frac{24}{100}=\frac{6}{25}\implies\boxed{\textbf{(A) }\frac{6}{25}}$.
~jmansuri

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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

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