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 ?

Question

ganeshie8 · Answer

A : 0.5x + 0.25y 
B : 0.25x + 0.5y
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$0.5(x-2) + 0.25(y+3) = 0.25(x-3) + 0.5(y+2)$

simplifying gives :
$x-y=2$

anonymous · Answer

I could solve till this step. After this?

ganeshie8 · Answer

Notice that $x\ge 3$
therefore the minimum values of $x$ and $y$ that satisfy above quation are 
$x = 3, y = 1$

anonymous · Answer

Answer is Rs.3 @ganeshie8

ganeshie8 · Answer

Yes, do you get why $x=3,y=1$ produce $3$ as the sum of the sum of money in the bags ?

anonymous · Answer

How? @ganeshie8

ganeshie8 · Answer

what are the minimum positive integers, $x,y$  that satisfy the equation
$x-y=2$ ?

anonymous · Answer

3 and 1. But how would that be 3 rupees? @ganeshie8

ganeshie8 · Answer

A : 0.5x + 0.25y 
B : 0.25x + 0.5y
----------------------

A+B = ?

anonymous · Answer

Got it. Thank you @ganeshie8