MHB Find the amount of sand in each bag

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The total amount of sand in Bags A, B, and C is 360 g. After a series of transfers, each bag ends up with an equal amount of sand. By working backwards from the final equal amounts, it is determined that Bag C contained 150 g before the last transfer. Consequently, Bag A had 90 g and Bag B had 120 g initially. The problem illustrates a systematic approach to solving for the initial quantities through reverse calculations.
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Bags $A,\,B$ and $C$ contained 360 g of sand in total. At first, $\dfrac{1}{6}$ of sand in Bag $A$ was poured into Bag $B$. Then, $\dfrac{1}{3}$ of sand in Bag $B$ is poured into Bag $C$. Lastly, $\dfrac{1}{5}$ of sand in Bag $C$ is poured into Bag $A$. In the end, there was equal amount of sand in each bag.

Find the amount of sand in each bag at first.

I haven't been able to solve this problem without setting up equations and solve for the unknowns yet. I'll keep trying, and at the same time want to share this good problem here at MHB.
 
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Instead of going forward I would go backwards.

Just for notation not algebra let sand in A be a , in B be b, in C be c,

at the end because all are equal we have

a = 120, b = 120, c = 120

In the last step 1/5 of sand in C was poured into A. So in C remained was 4/5 .

now 4/5 is 120 so it was 120 * 5 / 4 or 150 before .

because it was 30 poured from C to A before last step we have

a = 90, b = 120, c = 150

similarly you can calculate other 2 steps.
 
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