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

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==Problem 24==
 
A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math>
 
A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math>
  
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filldraw((9,1)--(12,1)--(12,4)--(9,4)--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((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);
−
filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);
+
filldraw((12,12)--(12,9)--(9,9)--(9,12)--cycle, mediumgray);
 
</asy>
 
</asy>
  
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==Solution 1==
 
==Solution 1==
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Let <math>s=1</math>. Then, the total area of the squares of side <math>s</math> is <math>576</math>, <math>64\%</math> of the area of the large square, which would be <math>900</math>, making the side of the large square <math>30</math>. Then, <math>25</math> borders have a total length of <math>30-24=6</math>. Since <math>\frac{d}{s}=d</math> if <math>s=1</math> is the value we're asked to find, the answer is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>.
+
The area of the shaded region is <math>(24s)^2</math>. To find the area of the large square, we note that there is a <math>d</math>-inch border between each of the <math>23</math> pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of <math>23+2 = 25</math> times the length of the border, i.e. <math>25d</math>. Adding this to the total length of the consecutive squares, which is <math>24s</math>, the side length of the large square is <math>(24s+25d)</math>, yielding the equation <math>\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}</math>. Taking the square root of both sides (and using the fact that lengths are non-negative) gives <math>\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}</math>, and cross-multiplying now gives <math>120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}</math>.
 +
 
 +
 
 +
Note: Once we obtain <math>\tfrac{24s}{24s+25d} = \tfrac{4}{5},</math> to ease computation, we may take the reciprocal of both sides to yield <math>\tfrac{24s+25d}{24s} = 1 + \tfrac{25d}{24s} = \tfrac{5}{4},</math> so <math>\tfrac{25d}{24s} = \tfrac{1}{4}.</math> Multiplying both sides by <math>\tfrac{24}{25}</math> yields the same answer as before. ~peace09
  
 
==Solution 2==
 
==Solution 2==
−
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{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}=\boxed{\textbf{(A)}\frac{6}{25}}</math>. ~jmansuri
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Without loss of generality, we may let <math>s=1</math> (since <math>d</math> will be determined by the scale of <math>s</math>, and we are only interested in the ratio <math>\frac{d}{s}</math>). Then, as the total area of the <math>576</math> gray tiles is simply <math>576</math>, the large square has area <math>\frac{576}{0.64} = 900</math>, making the side of the large square <math>\sqrt{900}=30</math>. As in Solution 1, the side length of the large square consists of the total length of the gray tiles and <math>25</math> lots of the border, so the length of the border is <math>d = \frac{30-24}{25} = \frac{6}{25}</math>. Since <math>\frac{d}{s}=d</math> if <math>s=1</math>, the answer is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>.
 +
Nice.
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 +
==Solution 3 (using answer choices)==
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As in Solution 2, we let <math>s = 1</math> without loss of generality. For sufficiently large <math>n</math>, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown:
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<asy>
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draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);
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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);
 +
 
 +
for(int i = 1; i <= 13; i += 4){
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draw((1,i)--(13,i), red);
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draw((i,1)--(i,13), red);
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}
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</asy>
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Each red square has side length <math>(1+d)</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of the total area covered by the gray tiles will be slightly less than <math>\frac{1}{(1+d)^2}</math>, which implies <math>\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}</math>. Hence <math>d</math> (and thus <math>\frac{d}{s}</math>, since we are assuming <math>s=1</math>) is less than <math>\frac{1}{4}</math>, and the only choice that satisfies this is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>.
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=Qs8zS0YEfg1iP-Y2&t=5584
 +
 
 +
~Math-X
 +
 
 +
==Video Solution(🚀 Just 2 m)==
 +
 
 +
https://www.youtube.com/watch?v=Vnk73Kd8t4o&list=PL73YVYWi-yG8Exr884k6y3eq8VBFMZRIF&index=24
 +
 
 +
<i>~Education, the Study of Everything </i>
 +
 
 +
==Video Solution==
 +
https://youtu.be/t8MVmKEyUhw
 +
 
 +
Please like and subscribe!
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/UpCURw5Moig?t=31
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/NHYB0VI3dcY
 +
 
 +
~savannahsolver
 +
 
 +
[edit: false link]
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=1515
 +
 
 +
~Interstigation
 +
 
 +
==Video Solution by STEMbreezy==
 +
https://youtu.be/wq8EUCe5oQU?t=353
 +
 
 +
~STEMbreezy
  
 
==See also==
 
==See also==

Latest revision as of 14:38, 21 January 2024

Problem 24

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((12,12)--(12,9)--(9,9)--(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

The area of the shaded region is $(24s)^2$. To find the area of the large square, we note that there is a $d$-inch border between each of the $23$ pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of $23+2 = 25$ times the length of the border, i.e. $25d$. Adding this to the total length of the consecutive squares, which is $24s$, the side length of the large square is $(24s+25d)$, yielding the equation $\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}$. Taking the square root of both sides (and using the fact that lengths are non-negative) gives $\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}$, and cross-multiplying now gives $120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}$.


Note: Once we obtain $\tfrac{24s}{24s+25d} = \tfrac{4}{5},$ to ease computation, we may take the reciprocal of both sides to yield $\tfrac{24s+25d}{24s} = 1 + \tfrac{25d}{24s} = \tfrac{5}{4},$ so $\tfrac{25d}{24s} = \tfrac{1}{4}.$ Multiplying both sides by $\tfrac{24}{25}$ yields the same answer as before. ~peace09

Solution 2

Without loss of generality, we may let $s=1$ (since $d$ will be determined by the scale of $s$, and we are only interested in the ratio $\frac{d}{s}$). Then, as the total area of the $576$ gray tiles is simply $576$, the large square has area $\frac{576}{0.64} = 900$, making the side of the large square $\sqrt{900}=30$. As in Solution 1, the side length of the large square consists of the total length of the gray tiles and $25$ lots of the border, so the length of the border is $d = \frac{30-24}{25} = \frac{6}{25}$. Since $\frac{d}{s}=d$ if $s=1$, the answer is $\boxed{\textbf{(A) }\frac{6}{25}}$. Nice.

Solution 3 (using answer choices)

As in Solution 2, we let $s = 1$ without loss of generality. For sufficiently large $n$, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: [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); for(int i = 1; i <= 13; i += 4){ draw((1,i)--(13,i), red); draw((i,1)--(i,13), red); } [/asy] Each red square has side length $(1+d)$, so by solving $\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}$, we obtain $d = \frac{1}{4}$. The actual fraction of the total area covered by the gray tiles will be slightly less than $\frac{1}{(1+d)^2}$, which implies $\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}$. Hence $d$ (and thus $\frac{d}{s}$, since we are assuming $s=1$) is less than $\frac{1}{4}$, and the only choice that satisfies this is $\boxed{\textbf{(A) }\frac{6}{25}}$.

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=Qs8zS0YEfg1iP-Y2&t=5584

~Math-X

Video Solution(🚀 Just 2 m)

https://www.youtube.com/watch?v=Vnk73Kd8t4o&list=PL73YVYWi-yG8Exr884k6y3eq8VBFMZRIF&index=24

~Education, the Study of Everything

Video Solution

https://youtu.be/t8MVmKEyUhw

Please like and subscribe!

Video Solution by OmegaLearn

https://youtu.be/UpCURw5Moig?t=31

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/NHYB0VI3dcY

~savannahsolver

[edit: false link]

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=1515

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/wq8EUCe5oQU?t=353

~STEMbreezy

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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