Raju and Sarita play a number game. First, each one of them chooses a positive integer independently. Separately, they both multiply their chosen integers by 2, and then subtract 20 from their resultant numbers. Now, each of them has a new number. Then, they divide their respective new numbers by 5. Finally, they added their results and found that the sum is 16. What can be the maximum possible difference between the positive integers chosen by Raju and Sarita?

Let the positive integers Raju & Sarita chose be 'r' & 's'.

After doing the given operations, the final numbers they get are $$\frac{2r-20}{5}\ \&\frac{2s-20}{5}$$

Adding results we get, $$\frac{2r-20}{5}\ +\frac{2s-20}{5}\ =\ 16$$

$$2r-20\ +2s-20\ =\ 80$$

$$2r+2s\ =\ 80+40+40=120$$

$$r+s\ =60$$

For maximum value of |r-s|, one of 'r' & 's' has to be maximum and other has to be minimum.

As r & s are positive integers minimum value they can take is "1".

If $$r=1,\ then\ s=59$$

$$\left|r-s\right|=\left|1-59\right|=58$$.

If $$s=1,\ then\ r=59$$

$$\left|r-s\right|=\left|59-1\right|=58$$

Hence maximum value of $$\left|r-s\right|$$ is 58. Option B is correct.