T.R | Title | User | Personal Name | Date | Lines |
---|
122.1 | | HARE::STAN | | Fri Aug 10 1984 03:05 | 3 |
| This classic problem was submitted by Douglas Williams to the DEC
publication "Systems and Clusters Engineering News" (as filler)
and appeared on page 5 of the June 1984 issue (issue #140).
|
122.2 | How ripe is the banana? | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Wed Jun 22 1988 13:11 | 72 |
| Spoiler follows:
Let
L = length of rope (inches)
D = density of rope (ounces per inch)
D = 1/3 (pounds per foot) * 16 (ounces per pound) * 1/12 (feet per inch)
D = 4/9
W = weight of monkey (ounces) = weight of widget (ounces)
d = density of banana (ounces per inch)
d = 2
A = age of monkey (years)
L/12 = A
M = age of monkey's mother (years)
W = M
A + M = 30
l = length of banana (inches)
ld + W/2 = (W + LD)/4
T = twice the monkey's age (years)
T = 2A
U = monkey's mother's age when she was one-half as old as ...
U = T/2
V = monkey's age when he is three times as old as his mother was ...
V = 3U
M = V/2
So we must solve for l in:
D = 4/9
d = 2
L/12 = A
W = M
A + M = 30
ld + W/2 = (W + LD)/4
T = 2A
U = T/2
V = 3U
M = V/2
Try to express each variable in terms of A:
L = 12A
T = 2A
U = T/2 = A
V = 3U = 3A
M = V/2 = 3A/2
W = M = 3A/2
Now solve for A:
A + M = 30
A + 3A/2 = 30
5A/2 = 30
A = 12
Now solve for l:
L = 12A = 144
W = 3A/2 = 18
ld + W/2 = (W + LD)/4
2l + 18/2 = (18 + (4/9)144)/4
2l + 9 = 20 1/2
2l = 11 1/2
l = ...
l = 5 3/4
The banana is 5 3/4 inches long.
Dan
|
122.3 | This appeared just a couple of months ago in "Parade" magazine | SEURAT::NEWMAN | Chuck Newman, 297-5499, MRO4-1/H16, Pole J13 | Thu Sep 06 1990 13:25 | 1 |
| It was in the "Ask Marilyn" column.
|