A man purchased 40 fruits: apples and oranges for Rs. 17. Had he purchased as many oranges as apples and as many apples as oranges, he would have paid Rs. 15/-. Find the cost of one pair of an apple and an orange.

Let us assume the number of apples purchased by the man to be x.

Then, the number of oranges will be (40-x).

Let the price of each apple be Rs. y and that of an orange be Rs. z

It is given that, xy+ (40-x)z= 17.

=> xy+40z-xz=17 ---------------(1)

Now, if the number of apples and oranges are interchanged, the apples will be (40-x) and the oranges will be x.

Given, the new total cost will be Rs. 15

So, (40-x)y + xz=15

=>40y-xy +xz= 15 ---------------(2)

Adding eqns 1 and 2, we get,

xy+40z-xz+40y-xy+xz=15+17

=> 40(y+z)=32

=>y+z= $$\ \frac{\ 32}{40}=\ \frac{\ 4}{5}$$

Or, the cost of one apple and one orange, x+y= Rs.$$\frac{\ 4}{5}=\ 80\ paise$$