Number of eggs laid on each day = Number of hens in the poultry farm = 60.
Out of the eggs laid on each day, the number of eggs t h a t got rotten is either 2 or 3 or 4.
Out of the eggs laid on each day, the number of eggs that got broken is either 4 or 5 or 6.
Maximum possible number of eggs taken to the market for sale on day 1 = 60 – (2 + 4)
= 54.
Minimum possible number of eggs taken to the market for sale on day 1 = 60 – (4 + 6)
= 50.
The minimum number of eggs that are left unsold each day must be 5, as the number of eggs that are rotten and broken among them needs to be an integer. It can be at max 10, since number of egg left unsold on any day is lessthan 20% ofthe number of eggs laid on each day, i.e. 20% of 60 = 12.
So, the number of eggs that are sold on day 1 ranges from (50 – 10 = 40) to (54 – 5 = 49), (both inclusive).
On the next day again 60 eggs are laid, so from the above logic the range of number of eggs sold should again come out to be from 42 to 49 (both inclusive), but there are eggs that remain unsold at the end of the previous day.
Minimum possible number of eggs that are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 5 – (40% of 5) – (2% of 5) = 2.
Maximum possible number of eggsthat are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 10 – (40% of 10) – (20% of 10) = 4.
So, the range of number of eggs that are sold on day 2 varies from (40 + 2 = 42) to (49 + 4 = 53) (both inclusive) and this holds true for day 3, day 4 and day 5 also,
Number of eggs that got rotten and broken is maximum possible.
So, assume that on each day 10 eggs remain unsold at the end of each day.
So, out of these 10 eggs that remain unsold at the end of each day, the number of eggs that get rotten and broken on the next day is 4 and 2 respectively.
The maximum number of eggs that got rotten and broken from among the eggs laid each day is 4 and 6 respectively.
So, the number of eggs that got broken across all the five days = 6 × 5 + 2 × 4
= 38.
Number of eggs that got rotten across all the five days = 4 × 5 + 4 × 4 = 36.
Required difference = 38 – 36 = 2.