Lesson Solving more complicated problems by the backward method

Solving more complicated problems by the backward method

Problem 1

I will solve the problem  from the ending state to the starting step. 


I will organize my logic, my solution and my calculations using this table below.


    Name           starting       consumed         ending
   (step)           amount         amount          amount


  H (Heart)           x           0.5x+0.5        h = x - (0.5x+0.5) = 0.5x-0.5     (1)


  D (diamond)         h           0.5h+0.5        d = h - (0.5h+0.5) = 0.5h-0.5     (2)


  C (club)            d           0.5d+0.5        c = d - (0.5d+0.5) = 0.5d-0.5     (3)


  S (spade)           c = 1                                                         (4)



        So, I have 4 equations  (1),  (2),  (3)  and  (4)  for  4  unknowns  x, h,  d  and  c.

        Now I will move backward from (4) to (1) to determine the unknowns one after another.



The last equation  c = 1  (4)  in the table gives me the opportunity to find "d" from (3)

    1 = 0.5d-0.5  --->  1 + 0.5 = 0.5d  --->  1.5 = 0.5d  --->  d = 1.5/0.5 = 3;  d = 3.       (5)



Now, knowing d= 3, I substitute it into equation (2) and find h

    3 = 0.5h-0.5  --->  3 + 0.5 = 0.5h  --->  3.5 = 0.5h  --->  h = 3.5/0.5 = 7;  h = 7.       (6)



Finally, knowing h= 7, I substitute it into equation (1) and find x

    7 = 0.5x-0.5  --->  7 + 0.5 = 0.5x  --->  7.5 = 0.5x  --->  x = 7.5/0.5 = 15;  x = 15.     (7)



Thus the problem is just solved, and the ANSWER is


      +-----------------------------------------+
      |   The Knave of Hearts stole 15 tarts.   |
      +-----------------------------------------+

Let the original number of tarts be T.


1st time around: Ate  of the tarts, and  more gives us a remainder of 

     =  .



2nd time around: Ate  of REMAINING tarts, plus  more gives us a remainder of

     =  = .


3rd time around: Ate  of REMAINING tarts, plus  more gives us a remainder of

     =  = 



After 3rd time around, 1 tart remained;  so we get

     =  1.


Multiplying by 8, we obtain

    T - 7 = 8.


It gives the final answer for the original number of tarts

    T = 8 + 7 = 15.

Problem 2

First, let x be the remainder money amount after buying a number of DVDs.


Then for this amount of money you have this equation

    x -  = 24.     (1)


To solve it, multiply everything in equation (1) by 5  to get

    5x - (3x + 30) = 24*5

    5x - 3x        = 24*5 + 30

       2x          = 120 + 30 = 150

        x                     = 150/2 = 75.


Thus you found that after buying a number of DVDs, the remainder was 75 dollars.


OK.     It gives you  NEXT EQUATION from the first part of the problem 

    M -  = 75    (2)


where M is the unknown money amount Henry had at first.


From this equation, you get

     = 75 + 3 = 78,

        M    =  = 26*4 = 104.


ANSWER.  At first,  Henry had $104.

Problem 3

    1.  At the start, the total number of toys was 4850.

    2.  Ben gave 3/7 of his toys to Gilian.

    3.  Gilian then gave 1/3 of her toys to Ben.

    4.  After that, the ratio of Gilian's toys to Ben toys was 2:3.

Notice that, althought Gilian and Ben exchanged toys, the total number of toys remained 4850, with no change.

So, at the final step 4, there were totaly 4850 toys in the ratio Gilian to Ben as 2:3.


It means, there were 5 equal parts of toys, each part counted  4850/5 = 970 toys,

and Gilian had 2 such parts, or 2*970 = 1940 toys, while Ben had 3 such parts, or 3*970 = 2910 toys.


In particular, we just know the ANSWER to question (a):  Gilian had 1940 toys at the end.


Next, the last action that Gilian did was giving 1/3 of her toys to Ben.

    HENCE, her final number of toys, 1940, is 2/3 of what she had before this giving.

    So, we conclude that before giving to Ben, Gilian had  (3/2)*1940 = 2910 toys.

    Repeating it again, immediately before step 3, Gilian had 2910 toys.

    Hence, immediately before step 3, Ben had the rest  4850 - 2910 = 1940 toys.


Thus we analysed the final stage and then restored from the condition, that immediately before step 3,
Gilian had 2910 toys, while Ben had 1940 toys.


Next, moving back, we know that after step 2, Ben had 1940 toys; Gilian had 2910 toys.

    In other words, after giving Gilian 3/7 of his toys at the step 2, Ben left with 1940 toys; it is 4/7 what he had before his giving.

    HENCE, before this giving, Ben had 7/4*1940 = 3395 toys and Gilian had the rest 4850-3395 = 1455 toys.


    Again, immediately before step 2, Ben had 3395 toys anad Gilian had the rest 1455 toys.


It gives the ANSWER to question (a): at first, Ben had 3395 toys.

Problem 4

So, I will accurately divide the entire process by steps.


    (1)  First, he moved 2/9 of the stamps from Book B to Book C. 

    (2)  Then, he moved 3/5 of the remaining stamps from Book B to Book A. 

    (3)  Next, he moved 1/3 of his stamps from Book A to Book C. 

    (4)  In the end, there were 240 fewer stamps in Book B than in Book A, 

                     and  Book C had 460 more stamps than Book B and there were 600 stamps left in Book C. 


First of all, based on description of final state  (4),  we can easily find the number of stamps in each book A, B and C:

          = 600;    = 600 - 460 = 140;    = 140 + 240 = 380.


Next,  I will introduce  arrays of numbers  , ,     ( distribution of stamps immediately before step 1; 
                                                                same as the initial distribution of stamps )

                                            , ,     ( distribution of stamps immediately before step 2) 

                                            , ,     ( distribution of stamps immediately before step 3)

                                            , ,     ( distribution of stamps immediately before step 4)


We just found out the values , , : they are   = 380;   = 140;   = 600.



We will now move from step (4) to step (3).  Marvin's action from step (3) to step (4) was  "he moved 1/3 of the stamps from Book A to Book C". 
                                             So, we can write

     = ,   = ,   =  + .


First of these equations gives  =  =  = 570;  second gives   = 140;  third gives   = 600 -  = 410.


Thus we can fill the row  3 :    = 570;   = 140;   = 410.




We will now move from step (3) to step (2).  Marvin's action from step (2) to step (3) was  "he moved 3/5 of the stamps from Book B to Book A."
                                             So, we can write

     = ,   = ,   = .


First of these equations gives 570 = ;  second gives  140 = ;  third gives   = 410.

It implies   =  = 350;  570 =  =  + 210;  hence,   = 570-210 = 360.


Thus we can fill the row  2 :   = 360;   = 350;   = 410.




We will now move from step (2) to step (1).  Marvin's action from step (1) to step (2) was  "he moved 2/9 of the stamps from Book B to Book C."
                                             So, we can write

     = ,   = ,   = .


First of these equations gives  = 360;  second gives   =  = 450;  third gives   =  = 310.


Thus we can fill the row  1 :   = 360;   = 450;   = 310.


So the answer to the problem question is   -  = 450-360 = 90  stamps.      ANSWER

Problem 5

Let x be what Derrick had finally.

Then Cindy had (x+20) dollars at the end;  Edmund had (x+30) dollars at the end.


The total is $410;  it did not change after these interior changes.  So


    x + (x+20) + (x+30) = 410

           3x + 50 = 410

           3x = 410 - 50 = 360

            x = 360/3 = 120  dollars.


Thus at final, Derrick had  120 dollars;  Cindy had 120+20 = 140 dollars  and  Edmund had 120+30 = 150 dollars.



Now we will make one step back, from the end to the beginning.



    As the problem says, Derrick gave 2/5 of his money to Cindy.

    Hence, Derrick left with only 3/5 of his money, and this 3/5 of his original money is 120 dollars.

    It means that originally Derrick had   = 5*40 = 200 dollars.


       HENCE, Derrick gave  2/5  of $200,  or 80 dollars to Cindy.


   
    The problem says: "Cindy gave 1/3 of her money to Edmund".

    So, Cindy left with 2/3 of her original money, PLUS 80 dollars she got from Derrick.


    So, we write this equation  for the Cindy original amount "C"

         = 140  dollars.


    We solve this equation easily and get   = 140-80 = 60;  hence  C =  = 90.


ANSWER.  Cindy had 90 dollars, at first.

Problem 6

In course of mutual exchanging, the total amount of dollars does not change: 
at the end, the total amount is the same $540, as at the beginning.

Therefore, at the end, Stephen and Siew Mei have 540/2 = 270 dollars each.



Let x be the amount which Stephan had at first.

Then Siew Mei had (540-x) dollars at first.


When Stephen gave 1/7 of his money to Siew Mei, Stephen left with  dollars;

Siew Mei had .


When Siew Mei gave 1/4 of the total amount she had to Stephen, 

Stephen had finally  +  dollars.


This final amounts of Stephen's money is 270 dollars, so we write this equation

     = 270.


Simplify this equation and find x. First multiply both sides by 4*7 = 28
    
    3*7*((540-x) + (1/7)*x) = 270*28

    21*(540-x) + 3x = 270*28

    21*540 - 21x + 3x = 270*28

    21*540 - 270*28 =  18x

    3780 = 18x

    x =  = 210.


ANSWER.  Stephen had initially 210 dollars;  hence, Siew Mei had initially 540-210 = 330 dollars.


CHECK.  After first giving, Stephen had 210-30 = 180 dollars; Siew Mei had 330+30 = 360 dollars.

        After second giving, Stephen had 180 + 90 = 270 dollars; Siew Mei had 360-90 = 270 dollars.   ! Equal amounts, correct !

In course of mutual exchanging, the total amount of dollars does not change: 
at the end, the total amount is the same $540, as at the beginning.


Therefore, at the end, Stephen and Siew Mei have 540/2 = 270 dollars each.


According to the problem, at the last exchange Siew Mei gave 1/4 of her money to Stephen.

Since Siew Mei left after that with 270 dollars, it means that 270 dollars is 3/4 of the amount she had 
immediately before this transaction; so, she had   = 360 dollars before this transaction.

Hence, Siew Mei gave 360-270 = 90 dollars to Stephen at this last transaction.


From it, we conclude, that Stephen had 270-90 = 180 dollars before the last transaction.


These 180 dollars are  of what he had at the beginning; hence, at the beginning Stephen had  = 210 dollars,


                   and the problem is just SOLVED and completed.