| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 465.1 | What No Banana? | SYSENG::NELSON | | Thu Apr 10 1986 12:07 | 3 |
| A variation of Note 122. for 122. I get 5.75 inches.
SN
|
| 465.2 | Is not 5.75 inches. | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Thu Apr 10 1986 14:12 | 6 |
| re. .1
No, 5.75 inches is not it.
Kostas G.
|
| 465.3 | | ALIEN::POSTPISCHIL | | Thu Apr 10 1986 15:05 | 4 |
| Two feet.
-- edp
|
| 465.4 | Answering 122. | SYSENG::NELSON | | Thu Apr 10 1986 15:15 | 6 |
| Re .2:
The banana isn't 5.75 inches for 122. ? Or are you talking about
this problem.
SN
|
| 465.5 | I was talking about 465 not 122 | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Thu Apr 10 1986 16:53 | 7 |
| re. .4
I was talking about this problem. But 5.75 is near to the correct
answer.
Kostas G.
|
| 465.6 | | ALIEN::POSTPISCHIL | | Thu Apr 10 1986 17:10 | 7 |
| Re .5:
If 5.75 inches is near the correct answer, I believe the problem
is worded incorrectly.
-- edp
|
| 465.7 | Apples and Oranges | SYSENG::NELSON | | Thu Apr 10 1986 19:41 | 5 |
| I'd like to clarify my responses.
For problem 122. I have 5.75 inches for the banana.
For this problem I have 4.8 feet for the rope.
SN
|
| 465.8 | Elaboration on the unit | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri Apr 11 1986 00:17 | 10 |
| re. .6
The number may be near the correct answer but not the units (i.e.
not the inches part ). the problem is worded correctly.
Sorry for tthe confusion.
Kostas
|
| 465.9 | Three Feet | BEING::POSTPISCHIL | Always mount a scratch monkey. | Fri Apr 11 1986 09:56 | 137 |
| Re .0:
Please find the mistake, if any.
> Suppose there was a rope hanging over a pulley with a weight on one end
> of the rope, and at the other end a monkey the same weight as the
> weight.
Let W = weight of monkey = weight of weight.
> Now suppose the rope weighted 4 ounces for every foot, . . .
Let length of rope = L. Let weight of rope = w.
w = L * (4 ounces/feet).
> . . . and the age of the monkey and the monkey's mother was 4 years,
> . . .
Let age of monkey at time t = a(t). Let age of mother at time t
= A(t). Clearly, a(t) + d = A(t), where d is the difference between
the age of the mother and the age of the monkey.
Let the current time be t0. Then a(t0)+A(t0) = 4 years.
> . . . and the weight of the monkey was as many pounds as the mother
> was years old.
W = A(t0) * pounds/years.
> The monkey's mother is twice as old as the monkey was . . .
Note that the narrative switches here from the past tense to the
present tense, which is confusing. However, this seems to be a
statement that the mother at the current time (t0) is twice as old
as the monkey was at some other time. Let this other time be t1.
Then A(t0) = 2 a(t1).
> . . .when the monkey's mother was half as old as the monkey was . . .
At time t1, the monkey's mother was half as old as the monkey at
some other time, say t2. So A(t1) = 1/2 a(t2).
> . . . when the monkey's mother was 3 times as old as the monkey, . . .
At time t2, the monkey's mother was three times as old as the monkey,
also at time t2. So A(t2) = 3 a(t2).
We can now express a(t0) in terms of d. Let's start with
A(t2) = 3 a(t2).
Substitute a(t2)+d for A(t2).
a(t2)+d = 3 a(t2).
Subtract a(t2) from both sides.
d = 2 a (t2).
Divide by 2 and change sides.
a(t2) = d/2.
Substitute this value of a(t2) in A(t1) = 1/2 a(t2).
A(t1) = 1/2 (d/2) = d/4.
Substitute a(t1)+d for A(t1).
a(t1)+d = d/4.
Subtract d from both sides.
a(t1) = -3/4 d.
Note that the age of the monkey at time t1 is negative if the mother is
older than the monkey, a flaw in the problem.
Substitute this value of a(t1) in A(t0) = 2 a(t1).
A(t0) = 2 (-3/4 d) = -3/2 d.
Substitute a(t0)+d for A(t0).
a(t0)+d = -3/2 d.
Subtract d from both sides.
a(t0) = -5/2 d.
Substitute these values for a(t0) and A(t0) in a(t0)+A(t0) = 4 years.
(-5/2 d) + (-3/2 d) = 4 years.
-4 d = 4 years.
d = - one year.
The monkey's mother is one year younger than the monkey, a definite
flaw in the problem.
Substitute - one year for d in A(t0) = -3/2 d.
A(t0) = -3/2 (- one year) = 3/2 years.
Substitute this value for A(t0) in W = A(t0) * pounds/years.
W = (3/2 years) * pounds/years = 3/2 pounds.
> . . . and the weight of the weight and the weight of the rope was
> half as much again as the difference between the weight of the weight
> and the weight of the weight and the weight of the monkey.
The narrative seems to shift back to the past tense even though it is
referring to time t0.
W+w = 150% ((W+W) - W).
Simplify.
W+w = 3/2 W.
w = 1/2 W.
Substitute 3/2 pounds for W.
w = 1/2 (3/2 pounds) = 3/4 pound = 3/4 (16 ounces) = 12 ounces.
Substitute this value for w in w = L * (4 ounces/feet).
12 ounces = L * (4 ounces/feet).
Multiply both sides by (feet/4 ounces).
12 ounces * (feet/4 ounces) = L.
3 feet = L.
-- edp
|
| 465.10 | Does This Make Sense ? | SYSENG::NELSON | | Fri Apr 11 1986 11:43 | 13 |
| Re .9:
I had the same difficulty with the use of the past tenses. At first
I came up with the monkey having to be older than the mother and
this must be a flaw in the problem. After rereading several times
carefully and mulling it over, I decided this is what had to be
meant:
"when the monkey's mother was 3 times as old as the monkey,"
^
is now!
That seemed to make all the difference. SN
|
| 465.11 | 4.8 feet | BEING::POSTPISCHIL | Always mount a scratch monkey. | Fri Apr 11 1986 12:14 | 133 |
| Re .10:
Thank you. Here is a corrected solution.
> Suppose there was a rope hanging over a pulley with a weight on one end
> of the rope, and at the other end a monkey the same weight as the
> weight.
Let W = weight of monkey = weight of weight.
> Now suppose the rope weighted 4 ounces for every foot, . . .
Let length of rope = L. Let weight of rope = w.
w = L * (4 ounces/feet).
> . . . and the age of the monkey and the monkey's mother was 4 years,
> . . .
Let age of monkey at time t = a(t). Let age of mother at time t
= A(t). Clearly, a(t) + d = A(t), where d is the difference between
the age of the mother and the age of the monkey.
Let the current time be t0. Then a(t0)+A(t0) = 4 years.
> . . . and the weight of the monkey was as many pounds as the mother
> was years old.
W = A(t0) * pounds/years.
> The monkey's mother is twice as old as the monkey was . . .
Note that the narrative switches here from the past tense to the
present tense, which is confusing. However, this seems to be a
statement that the mother at the current time (t0) is twice as old
as the monkey was at some other time. Let this other time be t1.
Then A(t0) = 2 a(t1).
> . . .when the monkey's mother was half as old as the monkey was . . .
At time t1, the monkey's mother was half as old as the monkey at
some other time, say t2. So A(t1) = 1/2 a(t2).
> . . . when the monkey's mother was 3 times as old as the monkey, . . .
At time t2, the monkey's mother was three times as old as the monkey
is now. So A(t2) = 3 a(t0).
We can now express a(t0) in terms of d. Let's start with
A(t2) = 3 a(t0).
Substitute a(t2)+d for A(t2).
a(t2)+d = 3 a(0).
Subtract d from both sides.
a(t2) = 3 a(t0) - d.
Substitute this value of a(t2) in A(t1) = 1/2 a(t2).
A(t1) = 1/2 (3 a(t0) - d) = 3/2 a(t0) - d/2.
Substitute a(t1)+d for A(t1).
a(t1)+d = 3/2 a(t0) - d/2.
Subtract d from both sides.
a(t1) = 3/2 a(t0) - 3/2 d.
Substitute this value of a(t1) in A(t0) = 2 a(t1).
A(t0) = 2 (3/2 a(t0) - 3/2 d) = 3 a(t0) - 3d.
Substitute a(t0)+d for A(t0).
a(t0)+d = 3 a(t0) - 3d.
Add 3d-a(t0) to both sides.
4d = 2 a(t0).
Divide by two and exchange sides.
a(t0) = 2d.
Recall that a(t0)+A(t0) = 4 years. Substitute a(t0)+d for A(t0).
a(t0)+a(t0)+d = 4 years.
Substitude 2d for a(t0).
4d+d = 4 years.
d = 4/5 years.
Now A(t0) = a(t0) + d = 2d+d = 3d = 12/5 years.
Substitute this value for A(t0) in W = A(t0) * pounds/years.
W = (12/5 years) * pounds/years = 12/5 pounds.
> . . . and the weight of the weight and the weight of the rope was
> half as much again as the difference between the weight of the weight
> and the weight of the weight and the weight of the monkey.
The narrative seems to shift back to the past tense even though it is
referring to time t0.
W+w = 150% ((W+W) - W).
Simplify.
W+w = 3/2 W.
w = 1/2 W.
Substitute 12/5 pounds for W.
w = 1/2 (12/5 pounds) = 6/5 pound = 6/5 (16 ounces) = 96/5 ounces.
Substitute this value for w in w = L * (4 ounces/feet).
96/5 ounces = L * (4 ounces/feet).
Multiply both sides by (feet/4 ounces).
96/5 ounces * (feet/4 ounces) = L.
24/5 feet = L.
L = 4.8 feet.
-- edp
|
| 465.12 | Note: Use of the word "difference" | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri Apr 11 1986 15:17 | 23 |
| re. .11
I believe you reasoning is correct up to the point where the statement:
W+w = 150% ((W+W) - W )
let be note something here
The word "difference" is quite unambiquous in arithmetic where the
magnitude of the numbers is known, but is ambiquous in algebra.
The difference between two unknown numbers a and b is expressed
either as |a - b| or as |b - a|, read as "absolute value of
(a - b)" or "absolute value of (b - a)".
Also the way I read the the last statement of the original problem
your W+w = 150% ((W+W) - W) should be W+w = 150% |W - (W+W)|
Enjoy,
Kostas G.
|
| 465.13 | Some notation and ... | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri Apr 11 1986 15:26 | 26 |
| Well,
to make this even simpler I have included some notation standards.
They are:
let
L = the length of the rope in feet
K = the weight of the rope in ounces
M and m be the ages of the mother and the monkey in years
n and w be the weights, in ounces, of the monkey and the
weight
x, y, z, be the time intervals in yeras to the different ages
in the problem as follows:
x = number of years ago when the monkey was ...
y = number od years later when the mokey will be ...
z = number of years ago when the mother was ...
given all of these what remains is the equations of the 9 statements
and solving for L which is the length of the rope in feet.
Enjoy,
Kostas G.
|
| 465.14 | | BEING::POSTPISCHIL | Always mount a scratch monkey. | Fri Apr 11 1986 16:28 | 7 |
| Re .12:
((W+W) - W) = |W - (W+W)| unless the monkey has inhaled a lot of
helium.
-- edp
|
| 465.15 | direct translation is whatI was looking for | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri Apr 11 1986 17:57 | 23 |
| re. .14:
Yes I agree that ((W+W) -W)) = |W - (W+W)| but the direct
translations form the statement given in the original problem does
not equal what you gave (i.e. ((W+W) - W)) ), and also
I do not use the character W for both the weight of the monkey and
for the weight of the weight.
In my notations I believe I used n for the weight of the monkey
and w for the weight of the weight
so that the nineth statement from the original problem will translate
to:
(9): w + K = (3/2) |w - (w+n)|
I also think the monkey may have inhaled a lot of helium why else
will it be on a rope? So it will not go up and up and up and up
...
kgg.
|
| 465.16 | | BEING::POSTPISCHIL | Always mount a scratch monkey. | Mon Apr 14 1986 10:16 | 9 |
| Re .15:
Regardless of whether or not the expression is a direct translation,
it is still a correct statement of the relationship between the
values, so why is my answer not correct? Or what do you think the
correct answer is and why?
-- edp
|
| 465.17 | Here is one correct solution by KGG | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Mon Apr 14 1986 11:27 | 38 |
| re. .16
Well here is how my solution goes:
The 9 statements in the original problem are given below in order:
(1): n = w
(2): K = 4*L
(3): m + M = 4
(4): n/16 = M
(5): M = 2 * (m - x)
(6): M - x = (m + y)/2
(7): m + y = 3(M - z)
(8): M - z = 3(m - z)
(9): w + K = (3/2) |w - (w+n)|
Subract (5) from (3) ---> m = 4 - 2m + 2x => m = (4+2x)/3
substitute into (5) ---> M = (8 + 4x)/3 - 2x => M = (8-2x)/3
substitute into (8) ---> (8-2x)/3 - z = 4 + 2x -3z
=> z = (2 + 4x)/3
Add (7) and twice (6) ---> 2M - 2x + m + y = m + y + 3M - 3z
substitute from M and z and solve for x, => x = 1/4
calculate m, M and z, ---> m = 3/2, M = 5/2, z = 1
from (4) n = 40
from (1) w = 40
from (9) K = 3n/2 - w = 20
from (2) L = K/4 = 5
So the length of the rope is 5 feet.
Enjoy,
Kostas G.
<><><><><>
|
| 465.18 | Matter of Interpretation? | SYSENG::NELSON | | Mon Apr 14 1986 13:50 | 14 |
| re .13
> x = number of years ago when the monkey was ...
> y = number od years later when the mokey will be ...
> z = number of years ago when the mother was ...
My question is how did you arrive at years "later" when the monkey
"will be" for y, from the wording and past tense of the original
problem.
This determination affects the equations starting with (6) in
re .17 where y is added and not subtracted.
SN
|
| 465.19 | | BEING::POSTPISCHIL | Always mount a scratch monkey. | Mon Apr 14 1986 14:21 | 15 |
| Re .17:
I do not see how equation (7) comes from the problem. It says the
monkey's age at some time was/is/will be three times the mother's
age at some other time. I do not see such a statement in .0.
Re .18:
As long as y is used consistently with the same sign, the sign does not
matter; the value of y will just come out positive or negative as
appropriate.
-- edp
|