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Two containers A and B, contained different masses of peanuts at first. 4/5  of the mass of peanuts in A was poured into B. Then 1/2 of the mass of peanuts in B was poured into A. After these two transfers of peanuts between A and B, the ratio of the mass of peanuts in A to the mass of peanuts in B is 7:5. Find the ratio of the mass of peunuts in A to the mass of peanuts in B at first. 

 Sep 10, 2023
 #2
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Call the original mass in A = A

Call the original mass in B = B

 

When (4/5)  of the mass in A is  poured into B

A now contains (1/5)A

B contains B + (4/5)A

 

When (1/2)  of the mass of B is then poured into A

A  now contains (1/5)A  + [ B + (4/5)A ] / 2  =   (3/5)A + (1/2)B

B now cointains (B + 4/5)A)  / 2  =  (1/2)B + ( 2/5)A

 

And we have that 

 

[ (3/5)A + (1/2)B ]               7

______________  =        __              cross-multiply

[ (1/2)B + (2/5)A ]              5

 

5 [ (3/5)A + (1/2)B ] =  7 [ (1/2)B + (2/5)A ]

 

3A + (5/2)B =  (7/2)B + (14/5)A

 

3A - (14/5)A = (7/2)B - (5/2)B

(1/5)A = B

 

A / B  =   5 / 1    (the original ratio of their masses )

 

 

cool cool cool

 Sep 10, 2023

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