On a rectangular garden with dimensions 8 x 10 yards, Jack decides to

Question

On a rectangular garden with dimensions 8 x 10 yards, Jack decides to dig a square pit with dimension 3 x 3 yards. What is the probability that no part of the square pit shall extend beyond the garden?

A) \(\frac{1}{2}\)

B) \(\frac{1}{10}\)

C) \(\frac{7}{16}\)

D) \(\frac{9}{80}\)

E) \(\frac{3}{4}\)

A)  \frac{1}{2}

B)  \frac{1}{10}

C)  \frac{7}{16}

D)  \frac{9}{80}

E)  \frac{3}{4}

GMATGuruNY · Answer

For a favorable outcome, the pit must lie entirely within the rectangular garden. Consider the most extreme placements that satisfy this condition. The most southwest placement of the pit requires that the top left corner of the pit be placed at the green dot in the figure below: pit and garden 1.png The most northwest placement of the pit requires that the top left corner of the pit be placed at the second green dot in the figure below: pit and garden 2.png The most northeast placement of the pit requires that the top left corner of the pit be placed at the third green dot in the figure below: pit and garden 3.png The most southeast placement of the pit requires that the top left corner of the pit be placed at the fourth green dot in the figure below: pit and garden 4.png Connecting the four green dots, we can see that -- for the pit to lie entirely within the garden -- the top left corner of the pit must lie within the pink rectangle in the figure below: pit and garden 5.png Thus: P(pit lies entirely within the garden) = \frac{pink-area}{garden-area} = \frac{5*7}{8*10} = \frac{7}{16} C

JerryAtDreamScore · Answer

Breaking Down the Info:

Assuming the center of the pit can be anywhere within the rectangular garden, we need him to select a point near the center of the garden and not near the edges.

The center of the pit must be 1.5 yards away (vertically and horizontally) from any of the 4 edges. Then we effectively reduce the area of the garden by 1.5 on all sides.

Therefore the region that he can dig would be a (8 - 3)*(10 - 3) = 35 sq yard rectangle. The total area he is allowed to dig is 8*10 = 80 sq yards, so that is a  \frac{35}{80} = \frac{7}{16}  probability.