Bag A contains X coins of 50 paise and Y coins of 25 paise. Bag B contains X coins of 25 paise and Y coins of 50 paise. If two coins of 50 paise from Bag A are put into Bag B and three coins of 25 paise from Bag B is put into Bag A, the value of money in both the bags become equal. What may be the minimum value of the toal amount in both the bags ?

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Bag A contains X coins of 50 paise and Y coins of 25 paise. Bag B contains X coins of 25 paise and Y coins of 50 paise. If two coins of 50 paise from Bag A are put into Bag B and three coins of 25 paise from Bag B is put into Bag A, the value of money in both the bags become equal. What may be the minimum value of the toal amount in both the bags ?

Mathematics
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A : 0.5x + 0.25y B : 0.25x + 0.5y ---------------------- \(0.5(x-2) + 0.25(y+3) = 0.25(x-3) + 0.5(y+2)\) simplifying gives : \(x-y=2\)
I could solve till this step. After this?
Notice that \(x\ge 3\) therefore the minimum values of \(x\) and \(y\) that satisfy above quation are \(x = 3, y = 1\)

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Answer is Rs.3 @ganeshie8
Yes, do you get why \(x=3,y=1\) produce \(3\) as the sum of the sum of money in the bags ?
what are the minimum positive integers, \(x,y\) that satisfy the equation \(x-y=2\) ?
3 and 1. But how would that be 3 rupees? @ganeshie8
A : 0.5x + 0.25y B : 0.25x + 0.5y ---------------------- A+B = ?
Got it. Thank you @ganeshie8

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