A woman's "measurements" are three numbers that measure, in order, the inches around her bust, waist and hips.
Now, a certain beauty contest winner had a 36-23-34 figure.
Although no two contestants had exactly the same measurements, the two runners-up differed by less than an inch in each measurement from the winner and the waist of each was two-thirds the hips of the other.
If the sum of the three measurements was the same for all three girls, what were the vital statistics of the two runners-up (the tape is accurate to only a quarter of an inch)?
The measurements of the two runners up are B1-W1-H1 and B2-W2-H2. It is given that W1=(2/3)*H2 and W2=(2/3)*H1. Since the tape is accurate only to a quarter inch, H1 and H2 must be multiples of 3/4. The winners hips are 34=136/4. The only measurements that are multiples of 3/4 and are within one inch of this value are 135/4 (33 3/4) and 138/4 (34 1/2). Since we are not differentiating between the two runners up, let H1=135/4 and H2=138/4. Then
W1=(2/3)*138/4=92/4=23 and
W2=(2/3)*135/4=90/4=22 1/2.
Since B1+H1+W1=B2+W2+H2=36+23+34=93
B1=93-23-33 3/4=36 1/4
B2=93-22 1/2-34 1/2=36
To sum up, the measurements of the two runners up are 36 1/4-23-33 3/4 and 36-22 1/2-34 1/2.
|
|
Posted by Bryan
on 2003-05-28 09:51:38 |
Solution
