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,
The maximum number of eggs that were sold on day1 can be 49, in a scenario when 5 eggs were left unsold.
In this case, the maximum number of eggs that were sold on day 2 can be calculated as 60 (laid on day 2) –2 (minimum rotten out of 60) – 4 (minimum broken out of 60) + 5 (previous days unsold eggs) – 2(rotten out of 5) – 1(broken out of 5) – 5(minimum unsold) = 51 Since, on day 2 also. 5 eggs were left unsold the maximum number eggs sold on day 3 will again be 51.
The same holds true for day 4 and day 5.
So, the maximum number of eggs that can be sold in the entire week can be 49 + 51 × 4 = 253.
Note: Most of the students will make a mistake of considering 53 eggs for any two consecutive days but this is not possible on any two consecutive days simultaneously.