1997 AJHSME Problems/Problem 10

Problem 10

What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale.

[asy] unitsize(8); fill((0,0)--(6,0)--(6,6)--(0,6)--cycle,black); fill((0,0)--(5,0)--(5,5)--(0,5)--cycle,white); fill((0,0)--(4,0)--(4,4)--(0,4)--cycle,black); fill((0,0)--(3,0)--(3,3)--(0,3)--cycle,white); fill((0,0)--(2,0)--(2,2)--(0,2)--cycle,black); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,white); draw((0,6)--(0,0)--(6,0)); [/asy]

$\text{(A)}\ \dfrac{5}{12} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{7}{12} \qquad \text{(D)}\ \dfrac{2}{3} \qquad \text{(E)}\ \dfrac{5}{6}$


Solution 1

Fill in grid lines to make the shape look like a small 6x6 checkerboard.

There are $6\times6 = 36$ squares total. The smallest shaded region has $3$ squares. The medium shaded region has $7$ squares. The largest shaded region has $11$ squares. The total shaded area is $3 + 7 + 11 = 21$ squares.

Thus, $\frac{21}{36} = \frac{7}{12}$ of the square is shaded, and the answer is $\boxed{C}$

Solution 2

Similar to the first solution, fill in the gridlines to make $36$ squares total.

Instead of counting shaded squares, count unshaded squares. There are $1 + 5 + 9 = 15$ unshaded squares.

Thus $\frac{15}{36} = \frac{5}{12}$ of the square is unshaded, and $1 - \frac{5}{12} = \frac{7}{12}$ of the square is shaded.

See also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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