Let x be the original cost of 1 dozen bananas , for Rs. 600 we  gets y dozens.  
xy =600 .................... {1}
\(y={600\over x}\) 
By increasing the cost of 1 dozen of bananas by Rs. 10 we get 3 dozen less bananas
(x +10)(y - 3)=600  ..........................{2}
Substituting the y value in (2), we get 
\((x+10)({600\over x}-3)=600\) 
\((x+10)({(600-3x)\over x})=600\)
\(6000-30x-3x^2=0\) 
\(3(x^2+10x-2000)=0\) 
\((x^2+10x-2000)=0\) 
(x+50)(x-40)=0
x=-50 or 40
Since cost of bananas cannot be negative, x = 40. So, the original cost of one dozen of bananas is Rs.40