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