Price of 1 apple = price of 2 oranges.

Price of 2 apples and 16 oranges=> 2(2) oranges + 16 oranges = Price of 20 oranges.

Price of 3 apples and 14 oranges=> 2(3) oranges + 14 oranges = Price of 20 oranges.

Price of 6 apples and 8 oranges=>2(6) oranges + 10 oranges = Price of 20 oranges

Let price of an apple be ‘a’ and price of an orange be ‘b’. Let the number of apples and oranges be ‘m’ and ‘n’ respectively.

Given, price of an apple is twice the price of an orange. Thus a = 2b.

Taking 2 to the other side, we get b = a/2

Price of 20 oranges = 20a

In the question we have to find price of the combination of apples and oranges equaling the price of 20 oranges, i.e., 20a. So we get the following equation:

20a = ma + nb

Since we have ‘a’ on the LHS, which is the price of 1 orange, we will try to convert RHS in terms of ‘a’ so as to cancel out the term. Using, b = a/2 in the above equation, we get:

20a = ma + na/2

20a = a(m+n/2)

20 = m+n/2

20 = (number of apples + number of oranges/2)---------(1)

Now substituting the values of number of apples and oranges given in the options, we can find out the correct ones.

thanks