Difference between revisions of "2020 AMC 8 Problems/Problem 24"
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==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 <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> | 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> | ||
− | ~ junaidmansuri | + | ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri] |
==Solution 3== | ==Solution 3== |
Revision as of 18:38, 18 November 2020
A large square region is paved with gray square tiles, each measuring inches on a side. A border inches wide surrounds each tile. The figure below shows the case for . When , the gray tiles cover of the area of the large square region. What is the ratio for this larger value of
Contents
Solution 1
WLOG, let . Then, the total area of the squares of side is , of the area of the large square, which would be , making the side of the large square . Then, borders have a total length of . Since if is the value we're asked to find, the answer is .
Solution 2
When , we see that the total height of the large square is . Similarly, when , the total height of the large square is . The total area of the gray tiles is and the area of the large white square is . We are given that the ratio of the gray area to the area of the large square is . Thus, our equation becomes . Square rooting both sides, we get . Cross multiplying, we get . Combining like terms, we get , which implies that .
~junaidmansuri
Solution 3
The area of the shaded region is . The total area of the square is because for each side of the square there one extra row/column of the boarder. Our equation is . Taking the square root of both sides gives . Cross multiplying and rearranging gives .
-franzliszt
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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