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.

jakeviamcknight24 Sep 10, 2023

#2**+1 **

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 )

CPhill Sep 10, 2023