Practice Area of Garden

Jan 15, 2020
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GMAT WORD PROBLEMS

A rectangular garden is surrounded by a 3 ft. wide concrete sidewalk. If the length of the garden is 4 ft more than its width, and if the area of the sidewalk is 60 sq ft more than the area of the garden, then what is the length of the garden?

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MarkFL

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Let \(L\) be the length of the garden and \(W\) be the width.

\(\displaystyle A_G=LW\) area of garden

\(\displaystyle A_S=(L+6)(W+6)-LW=6(L+W+6)\) area of sidewalk

\(\displaystyle A_S=A_G+60\)

\(\displaystyle 6(L+W+6)=LW+60\)

\(\displaystyle 6(L+W)=LW+24\)

\(\displaystyle W=L-4\)

\(\displaystyle 6(L+L-4)=L(L-4)+24\)

\(\displaystyle L^2-16L+48=0\)

\(\displaystyle (L-4)(L-12)=0\)

We discard \(L=4\) since that results in \(W=0\), and we are left with \(L=12\).
 
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Jan 15, 2020
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Let \(L\) be the length of the garden and \(W\) be the width.

\(\displaystyle A_G=LW\) area of garden

\(\displaystyle A_S=(L+6)(W+6)-LW=6(L+W+6)\) area of sidewalk

\(\displaystyle A_S=A_G+60\)

\(\displaystyle 6(L+W+6)=LW+60\)

\(\displaystyle 6(L+W)=LW+24\)

\(\displaystyle W=L-4\)

\(\displaystyle 6(L+L-4)=L(L-4)+24\)

\(\displaystyle L^2-16L+48=0\)

\(\displaystyle (L-4)(L-12)=0\)

We discard \(L=4\) since that results in \(W=0\), and we are left with \(L=12\).
Can you explain why my reasoning here is not ok?
 

MarkFL

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Area is the product of length and width. Among other things, you simply added 60 to the width and said that is the area of the concrete.
 
Jan 15, 2020
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Bronx, NY
Area is the product of length and width. Among other things, you simply added 60 to the width and said that is the area of the concrete.
Oh boy. What agony for me. I just cannot come up with the right equation set up. This means I cannot now or ever work as a math tutor, which is a great second part-time job.
 

MarkFL

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Your diagram is good, and your definitions of length and width are good. So, as your next step, you want to express the area of the concrete and the area of the garden in terms of your variable \(x\).

The area of the garden will be:

\(\displaystyle A_G=x(x+4)\)

Now the area of the concrete will be the area of the outer rectangle formed by the concrete minus the area of the inner rectangle formed by the concrete, which is also the area of the garden. Since the width of the concrete is 3 ft., we can write:

\(\displaystyle A_C=(x+3+3)(x+4+3+3)-x(x+4)=(x+6)(x+10)-x(x+4)=12(x+5)\)

Okay, now we look at the fact that the problem tells us "the area of the sidewalk is 60 sq ft more than the area of the garden." And so we may write:

\(\displaystyle A_C=A_G+60\)

\(\displaystyle 12(x+5)=x(x+4)+60\)

Now, since we are asked for length, and you made your variable \(x\) the width, Let's let \(L=x+4\) and write

\(\displaystyle 12(L+1)=(L-4)L+60\)

\(\displaystyle 12L+12=L^2-4L+60\)

\(\displaystyle L^2-16L+48=0\)

This is the same equation I got before.
 
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Your diagram is good, and your definitions of length and width are good. So, as your next step, you want to express the area of the concrete and the area of the garden in terms of your variable \(x\).

The area of the garden will be:

\(\displaystyle A_G=x(x+4)\)

Now the area of the concrete will be the area of the outer rectangle formed by the concrete minus the area of the inner rectangle formed by the concrete, which is also the area of the garden. Since the width of the concrete is 3 ft., we can write:

\(\displaystyle A_C=(x+3+3)(x+4+3+3)-x(x+4)=(x+6)(x+10)-x(x+4)=12(x+5)\)

Okay, now we look at the fact that the problem tells us "the area of the sidewalk is 60 sq ft more than the area of the garden." And so we may write:

\(\displaystyle A_C=A_G+60\)

\(\displaystyle 12(x+5)=x(x+4)+60\)

Now, since we are asked for length, and you made your variable \(x\) the width, Let's let \(L=x+4\) and write

\(\displaystyle 12(L+1)=(L-4)L+60\)

\(\displaystyle 12L+12=L^2-4L+60\)

\(\displaystyle L^2-16L+48=0\)

This is the same equation I got before.
Excellent break down. Nothing could be more clear. Note: I will be posting lots of applications with the goal in mind to create the proper set up.