SOLUTION: Marvin decided to rearrange his stamps in his stamp books, Book A, Book B and Book C. First, he moved 2/9 of the stamps from Book B to Book C. Then, he moved 3/5 of the remainin

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Marvin decided to rearrange his stamps in his stamp books, Book A, Book B and Book C. First, he moved 2/9 of the stamps from Book B to Book C. Then, he moved 3/5 of the remainin      Log On


   



Question 1184480: Marvin decided to rearrange his stamps in his stamp books, Book A, Book B and Book
C. First, he moved 2/9 of the stamps from Book B to Book C. Then, he moved 3/5 of the
remaining stamps from Book B to Book A. Next, he moved 1/3 of his stamps from Book
A to Book C. In the end, there were 240 fewer stamps in Book B than in Book A. Book
C had 460 more stamps than Book B and there were 600 stamps left in Book C. How
many more stamps were there in Book B than Book A at first?

Found 4 solutions by JBnovelwriter, MathTherapy, ikleyn, 54929:
Answer by JBnovelwriter(34) About Me  (Show Source):
You can put this solution on YOUR website!
ABC
Make sure to understand that the original numbers for A, B, and C stay constant. any changes to the group you must include the constant.
.
First, he moved 2/9 of the stamps from Book B to Book C.
%282%2F9%29B%2BC =updated C book amount of stamps
Then, he moved 3/5 of the remaining stamps from Book B to Book A. (the remaining ould be what's left after subtracting 2/9ths of B.
%283%2F5%29%28B-%282B%2F9%29%29%2BA
%283%2F5%29%287B%2F9%29%2BA = updated A book amount of stamps
he moved 1/3 of his stamps from Book A to Book C
%28%281%2F3%29%28A%2B%28%283%2F5%29%287B%2F9%29%29%29%29%2B%28%282B%2F9%29%2BC%29 = current C book amount of stamps
%282%2F3%29%28A%2B%28%283%2F5%29%287B%2F9%29%29%29 is current A book
(In the end) this is very important terminology this now means everything following is the end results for each book not the original constants.
there were 240 fewer stamps in Book B than in Book A. 240 fewer than B = current state of B+240=current state of A.

current Book C had 460 more stamps than current Book B. took current state of book C and set it equal to current B plus 460 more stamps.
%28%289B%2F9%29-%282B%2F9%29%29=%287B%2F9%29
%287B%2F9%29-%283%2F5%29%287B%2F9%29 = Bs current book amount

current C book=600 stamps left
%28%281%2F3%29%28A%2B%283%2F5%29%287B%2F9%29%29%29%2B%28%282B%2F9%29%2BC%29=600
so this means that current B book is = to 140=(600-460)
600=%287B%2F9%29-%283%2F5%29%287B%2F9%29%2B460 solve for B, subtract 460 from each side
140=%287B%2F9%29-%283%2F5%29%287B%2F9%29multiply the fraction in
140=%287B%2F9%29-%2821B%2F45%29 im not a fan of fractions so multiply both sides by 9
6300=%2835B%29-%2821B%29
6300=14B
highlight%28450=B%29 We finally Found what B originally equaled. now we can use it in other equations to solve for the other books.
.
there were 240 fewer stamps in Book B than in Book A. 240 fewer than current A = current B +240= 140+240=380 is current A and current C is 600.
so now we have all the currents:
cA=380
cB=140
cC=600
so above is all the calculations you get from reading the word problem
.
you can now set each to the equations from above and use B=450 to find what C and A were at the start
.
cA=%282%2F3%29%28A%2B%28%283%2F5%29%287B%2F9%29%29%29
cC=%28%281%2F3%29%28A%2B%28%283%2F5%29%287B%2F9%29%29%29%29%2B%28%282B%2F9%29%2BC%29
.
%282%2F3%29%28A%2B%28%283%2F5%29%287B%2F9%29%29%29=380
B=450
solve for A
%282%2F3%29%28A%2B%28%283%2F5%29%287B%2F9%29%29%29=380 multiply by 3
2A%2B%28%283%2F5%29%287B%2F9%29%29=1140 multiply accross fractions
2A%2B%2835B%2F45%29=1140 multiply by 45
45%282A%29%2B%2835B%29=51300 substitute in the B=450
90A%2B%2835%28450%29%29=51300
90A%2B%2815750%29=51300
90A=35550
highlight%28A=395%29
A=395
B=450
How many more stamps were there in Book B than Book A at first?
B-A=x
450-395=x
highlight%2855%29=x




sorry about all the updates after the post, realized I made a few mistakes and had to correct them. this was not an easy problem. It was quite difficult to keep everything straight. I got hung up on when it took a third of the current A book and forgot to change the current state of A to two-thirds of the updated A book. but it's all fixed now. :-)


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Marvin decided to rearrange his stamps in his stamp books, Book A, Book B and Book C. First, he moved 2/9 of the stamps from Book B to Book C. Then, he moved 3/5 of the remaining stamps from Book B to Book A. Next, he moved 1/3 of his stamps from Book A to Book C. In the end, there were 240 fewer stamps in Book B than in Book A. Book C had 460 more stamps than Book B and there were 600 stamps left in Book C. How many more stamps were there in Book B than Book A at first?
The other person is PARTLY-WRONG, but that should be expected seeing that the problem's setup and calculations
are difficult enough to easily make mistakes and end up with one or more INCORRECT answers. 

Let number of stamps in Books A, B, and C, be a, b, and c, respectively
After moving 2%2F9 from Book B to Book C, Book B then had matrix%281%2C5%2C+%281+-+2%2F9%29%2C+of%2C+b%2C+or%2C+7b%2F9%29 left, and Book C had: matrix%281%2C5%2C+c+%2B+2%2F9%2C+of%2C+b%2C+%22=%22%2C+c+%2B+2b%2F9%29  
After moving 3%2F5 of remainder from Book B to Book A, Book B then had  left.
Book A then had: matrix%281%2C5%2C+a+%2B+3%2F5%2C+of%2C+7b%2F9%2C+%22=%22%2C+a+%2B+7b%2F15%29  

After moving 1%2F3 from Book A to Book C, Book A then had  left
Book C then had: 
As there were 240 fewer stamps in Book B than in Book A, in the end, we get: 
                                                                               2a = 3(240) ------- Cross-multiplying
                                                       
                                                Original  number in Book A, or 

As there were 460 more stamps in Book C than in Book B, in the end, we get: matrix%281%2C3%2C+c+%2B+2b%2F9+%2B+a%2F3+%2B+7b%2F45+-+460%2C+%22=%22%2C+14b%2F45%29
  
                                                                              matrix%282%2C3%2C+600+-+460%2C+%22=%22%2C++14b%2F45%2C+140%2C+%22=%22%2C+14b%2F45%29 ---- Substituting 600 for c+%2B+2b%2F9+%2B+a%2F3+%2B+7b%2F45
                                                                              14b = 140(45) ---- Cross-multiplying

                                                Original  number in Book B, or 

               Difference between original number in Book A and Book B: highlight_green%28matrix%281%2C4%2C+450+-+360%2C+%22=%22%2C+90%2C+stamps%29%29

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Marvin decided to rearrange his stamps in his stamp books, Book A, Book B and Book C.
First, he moved 2/9 of the stamps from Book B to Book C.
Then, he moved 3/5 of the remaining stamps from Book B to Book A.
Next, he moved 1/3 of his stamps from Book A to Book C.
In the end, there were 240 fewer stamps in Book B than in Book A.
Book C had 460 more stamps than Book B and there were 600 stamps left in Book C.
How many more stamps were there in Book B than Book A at first?
~~~~~~~~~~~~~~~


            Looking into the solutions of the two other tutors,  I decided to try the backward method, hoping that the solution will be simpler.

            Let's look what I got  ( ! )


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:

         c%5B4%5D = 600;   b%5B4%5D = 600 - 460 = 140;   a%5B4%5D = 140 + 240 = 380.


Next,  I will introduce  arrays of numbers  a%5B1%5D, b%5B1%5D, c%5B1%5D    ( distribution of stamps immediately before step 1; 
                                                                same as the initial distribution of stamps )

                                            a%5B2%5D, b%5B2%5D, c%5B2%5D    ( distribution of stamps immediately before step 2) 

                                            a%5B3%5D, b%5B3%5D, c%5B3%5D    ( distribution of stamps immediately before step 3)

                                            a%5B4%5D, b%5B4%5D, c%5B4%5D    ( distribution of stamps immediately before step 4)


We just found out the values a%5B4%5D, b%5B4%5D, c%5B4%5D: they are  a%5B4%5D = 380;  b%5B4%5D = 140;  c%5B4%5D = 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

    a%5B4%5D = %282%2F3%29%2Aa%5B3%5D,  b%5B4%5D = b%5B3%5D,  c%5B4%5D = c%5B3%5D + %281%2F3%29%2Aa%5B3%5D.


First of these equations gives a%5B3%5D = %283%2F2%29%2Aa%5B4%5D = %283%2F2%29%2A380 = 570;  second gives  b%5B3%5D = 140;  third gives  c%5B3%5D = 600 - %281%2F3%29%2A570 = 410.


Thus we can fill the row  3 :   a%5B3%5D = 570;  b%5B3%5D = 140;  c%5B3%5D = 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

    a%5B3%5D = a%5B2%5D+%2B+%283%2F5%29%2Ab%5B2%5D,  b%5B3%5D = %282%2F5%29%2Ab%5B2%5D,  c%5B3%5D = c%5B2%5D.


First of these equations gives 570 = a%5B2%5D+%2B+%283%2F5%29%2Ab%5B2%5D;  second gives  140 = %282%2F5%29%2Ab%5B2%5D;  third gives  c%5B2%5D = 410.

It implies  b%5B2%5D = %285%2F2%29%2A140 = 350;  570 = a%5B2%5D+%2B+%283%2F5%29%2A350 = a%5B2%5D + 210;  hence,  a%5B2%5D = 570-210 = 360.


Thus we can fill the row  2 :  a%5B2%5D = 360;  b%5B2%5D = 350;  c%5B2%5D = 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

    a%5B2%5D = a%5B1%5D,  b%5B2%5D = %287%2F9%29%2Ab%5B1%5D,  c%5B2%5D = c%5B1%5D%2B%282%2F9%29%2Ab%5B1%5D.


First of these equations gives a%5B1%5D = 360;  second gives  b%5B1%5D = %289%2F7%29%2A350 = 450;  third gives  c%5B1%5D = 410-%282%2F9%29%2A450 = 310.


Thus we can fill the row  1 :  a%5B1%5D = 360;  b%5B1%5D = 450;  c%5B1%5D = 310.


So the answer to the problem question is  b%5B1%5D - a%5B1%5D = 450-360 = 90  stamps.      ANSWER

Solved.

After obtaining this  ANSWER,  I, naturally,  checked it,  walking through the steps from  (1)  to  (4).

What I got,  confirmed correctness of my calculations.



Answer by 54929(12) About Me  (Show Source):
You can put this solution on YOUR website!
Dim the first stamps of A, B, C
with a,b,c.
Step 1: A: a
B: 7%2F9b
C: 2%2F9b + c
Step 2: A 3/5 * 7/9 b + a = 21/45 b + a
B : 2/5 * 7/9b = 14/45 b [1]
Step 3: A : 2%2F3 (21%2F45b + a)
= 2%2F3a + 14%2F45b [2]
end: C : 600
B : 600 - 460 = 140 [1]
A: 140 + 240 = 380 [2]
=> [1] = [1] => a = 360
[2] = [2] => b = 450
highlight_green%28matrix%281%2C4%2C+b+-+a%2C+%22=%22%2C+90%2C+stamps%29%29