SOLUTION: Andy, Berlin and Cheryl had a total of 6750 stamps. At first, Andy have 50% of his stamps to Berlin. Berlin then gave 1/3 of her stamps to Cheryl. Finally, Cheryl gave 1/6 of her

Andy, Berlin and Cheryl had a total of 6750 stamps. At first, Andy have 50% of his stamps to Berlin. Berlin then gave 1/3 of her
stamps to Cheryl. Finally, Cheryl gave 1/6 of her stamps to Andy. In the end, the ratio of the number of Andy’s stamps to the number
of Berlin’s stamps became 4:5. Cheryl had twice the total number of stamps that Andy and Berlin had. How many stamps did Berlin and
Cheryl have in total at first?

Whatever/Whomever it is that responded, is WRONG!!

Let initial amounts Andy, Berlin, and Cheryl had, be A, B, and C, respectively

A gave B 50% of his, so A’s remainder was .5A
B received 50% of A’s, or .5A, after which B had B + .5A
B, after giving  to C, had  left
C, after receiving  from B, then had: 
C, after giving  to A, then had:  remaining
A, after receiving  of C’s, then had: 

A:B = 4:5 ====>  =====> 
                      =====>  
                      =====>  
                      =====>  
                                4(6A + 12B) = 5(9.5A + B + 3C) ----- Cross-multiplying
                                  24A + 48B = 47.5A + 5B + 15C
               47.5A - 24A + 5B - 48B + 15C = 0
                          23.5A - 43B + 15C = 0 ---- eq (i)

Also, C = 2(A + B) ====> C = 2A + 2B ====> 2A + 2B - C = 0 
        
 2(9.5A + B + 3C) + 6(2A + 4B) - (2.5A + 5B + 15C) = 0 ----- Multiplying by LCD, 18
       19A + 2B + 6C + 12A + 24B - 2.5A - 5B - 15C = 0
       19A + 12A - 2.5A + 2B + 24B - 5B + 6C - 15C = 0
                                  28.5A + 21B - 9C = 0 
                                    ----- Dividing by 3 
                                    9.5A + 7B - 3C = 0 ----- eq (ii)

         They ALL started with 6,750, so A + B + C = 6,750 ---- eq (iii)

                                 23.5A - 43B + 15C = 0 ------- eq (i)
                                  9.5A +  7B -  3C = 0 ------- eq (ii)
                                     A +   B +   C = 6,750 --- eq (iii)
                                    3A +  3B +  3C = 20,250 ---- Multiplying eq (iii) by 3 ---- eq (iv) 
                                       12.5A + 10B = 20,250 ---- Adding eqs (ii) & (iv) 
                                        ---- Dividing by 10 
                                         1.25A + B = 2,025
                                                 B = 2,025 - 1.25A ------ eq (v)
                                   15A + 15B + 15C = 101,250 ----- Multiplying eq (iii) by 15 ---- eq (vi)
                                        8.5A - 58B = - 101.250 --- Subtracting eq (vi) from eq (i) ---- eq (vii)
                          8.5A - 58(2,025 - 1.25A) = - 101,250 --- Substituting 2,025 - 1.25A for B in eq (vii)
                            8.5A - 117,450 + 72.5A = - 101,250
                                               81A = - 101,250 + 117,450
                                               81A = 16,200
                  Amount Andy had, initially, or A =  = 200

With ALL 3 starting with a total of 6,750 stamps, and A starting with 200, amount B and C started with = 6,750 - 200 = 6,550

You can do the CHECK!!

             (a)  Analyzing the ending situation from the problem



The stamps were circulated inside the team and did not go out or go into from outside.

Therefore, as they started with 6750 stamps at the beginning, so they ended with the same 
6750 stamps at the end.


      From it, we can analyze the ending situation from the problem.


The problem says that at the end 

    "Cheryl had twice the total number of stamps that Andy and Berlin had."


It means that at the end Cheryl had 2/3 of 6750, i.e. 4500 stamps, while Andy and Berlin
had 1/3 of 6750, or  2250 stamps, together.


Also, the problem says that at the end 

    "the ratio of the number of Andy’s stamps to the number of Berlin’s stamps became 4:5".


It means that at the end Andy had 4/9 of 2250, or 1000 stamps,
while Berlin had at the end 5/9 of 2250, or 1250 stamps.


So, at the end, Andy had 1000 stamps;  Berlin had 1250 stamps, and Cheryl had 4500 stamps.



             (b)  Making steps back from the end to the beginning



                     (b32)  From step (3) to step (2)



At step (3), Cheryl gave 1/6 of her stamps to Andy. 

So, we can write this equation

    C -  = 4500,  or   = 4500,  giving  C =  = 6*900 = 5400.


So, immediately before step (3), Cheryl had 5400 stamps.

Next, at step 3, Cheryl gave  = 900 stamps to Andy; so, immediately before step (3), Andy had 1000-900 = 100 stamps.


Thus, immediately after step 2, Andy had 100 stamps;  Berlin had 1250 stamps; Cheryl had 5400 stamps.



                     (b21)  From step (2) to step (1)



At step (2), Berlin gave 1/3 of her stamps to Cheryl.

So, we can write this equation

    B -  = 1250,  or   = 1250,  giving  B =  = 1875.


So, immediately before step (2), Berlin had 1875 stamps.

Next, at step 2, Berlin gave  = 625 stamps to Cheryl; so, immediately before step (2), Cheryl had 5400-625 = 4775 stamps.


Thus, immediately after step 1, Andy had 100 stamps;  Berlin had 1875 stamps; Cheryl had 4775 stamps.



                     (b10)  From step (1) to the beginning



At step (1), Andy gave 1/2 of his stamps to Berlin. 

So, we can write this equation

    A -  = 100,  or   = 100,  giving A = 200.


So, at the beginning, Andy had 200 stamps.

Next, at step (1), Andy gave  = 100 stamps to Berlin; so, at the beginning, Berlin had 1875-100 = 1775 stamps.


Thus, at the beginning, Andy had 200 stamps; Berlin had 1775 stamps, and Cheryl had 4775 stamps.


Finally, the problem asks "How many stamps did Berlin and Cheryl have in total at first ?"


The ANSWER  is  1775 + 4775 = 6550.