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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

465.0. "<><><> The "Monkey on a rope" problem. <><><>" by KEEPER::KOSTAS (Kostas G. Gavrielidis &lt;o.o&gt; ) Sun Apr 06 1986 22:00

    Hello,
    
         here is the "Monkey on a rope" problem:
    
    _____________________________________________________________________
                                            
       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. Now suppose the rope weighted 4 ounces for
    every foot, and the age of the monkey and the monkey's mother was
    4 years, and the weight of the monkey was as many pounds as the
    mother was years old. The monkey's mother is twice as old as the
    monkey was when the monkey's mother was half as old as the monkey
    was when the monkey's mother was 3 times as old as the monkey, 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. 
    
    What is the length of the rope?
                                                                         
    _____________________________________________________________________
    
    Enjoy,
    
    Kostas G.
    <><><><><>
    
T.RTitleUserPersonal
Name
DateLines
465.1What No Banana?SYSENG::NELSONThu Apr 10 1986 15:073
    A variation of Note 122. for 122. I get 5.75 inches.
    
    SN
465.2Is not 5.75 inches.KEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Thu Apr 10 1986 17:126
    re. .1
    
         No, 5.75 inches is not it.
    
    Kostas G.
    
465.3ALIEN::POSTPISCHILThu Apr 10 1986 18:054
    Two feet.
    
    
    				-- edp
465.4Answering 122.SYSENG::NELSONThu Apr 10 1986 18:156
    Re .2:
    
    The banana isn't 5.75 inches for 122. ? Or are you talking about
    this problem.
    
    SN
465.5I was talking about 465 not 122KEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Thu Apr 10 1986 19:537
    re. .4
    
    I was talking about this problem. But 5.75 is near to the correct
    answer.
    
    Kostas G.
    
465.6ALIEN::POSTPISCHILThu Apr 10 1986 20:107
    Re .5:
    
    If 5.75 inches is near the correct answer, I believe the problem
    is worded incorrectly.
    
    
    				-- edp
465.7Apples and OrangesSYSENG::NELSONThu Apr 10 1986 22:415
    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
465.8Elaboration on the unitKEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Fri Apr 11 1986 03:1710
    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  
    
465.9Three FeetBEING::POSTPISCHILAlways mount a scratch monkey.Fri Apr 11 1986 12:56137
    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
465.10Does This Make Sense ?SYSENG::NELSONFri Apr 11 1986 14:4313
    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
465.114.8 feetBEING::POSTPISCHILAlways mount a scratch monkey.Fri Apr 11 1986 15:14133
    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
465.12Note: Use of the word "difference"KEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Fri Apr 11 1986 18:1723
    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.
    
465.13Some notation and ...KEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Fri Apr 11 1986 18:2626
    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.
    
    
    
465.14BEING::POSTPISCHILAlways mount a scratch monkey.Fri Apr 11 1986 19:287
    Re .12:
    
    ((W+W) - W) = |W - (W+W)| unless the monkey has inhaled a lot of
    helium.
    
    
    				-- edp
465.15direct translation is whatI was looking forKEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Fri Apr 11 1986 20:5723
    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.
    
465.16BEING::POSTPISCHILAlways mount a scratch monkey.Mon Apr 14 1986 13:169
    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
465.17Here is one correct solution by KGGKEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Mon Apr 14 1986 14:2738
    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.
    <><><><><>
    
465.18Matter of Interpretation?SYSENG::NELSONMon Apr 14 1986 16:5014
    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
465.19BEING::POSTPISCHILAlways mount a scratch monkey.Mon Apr 14 1986 17:2115
    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