The perimeter of a rectangular field is 3 times its length. If the perimeter is $240 \mathrm{~m}$, the width of the field is
(A) $80 \mathrm{~m}$
(B) $40 \mathrm{~m}$
(C) $20 \mathrm{~m}$
(D) $30 \mathrm{~m}$
(E) $120 \mathrm{~m}$
Solution 1

Since the perimeter of the field is 3 times the length and the perimeter is $240 \mathrm{~m}$, then the length of the field is $80 \mathrm{~m}$.

Since the perimeter of a rectangle is two times the length plus two times the width, then the length accounts for $160 \mathrm{~m}$ of the perimeter, leaving $80 \mathrm{~m}$ for two times the width.

Therefore, the width of the field is $40 \mathrm{~m}$.

## Solution 2

Let the perimeter of the field be $P$, the length be $l$, and the width $w$.

We are given that $P=3 l$ and $P=240$, so $l=\frac{1}{3}(240)=80$.

Since $P=240$ and $P=2 l+2 w$, we have $240=2(80)+2 w$ or $w=40$.

ANSwer: (B)