Problem Solving

In a farmer sells 15 of his chickens, his stock of feed will last for 4 more days than planned, but if he buys 20 more chickens, he will run out of feed 3 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?

A. 12
B. 24
C. 48
D. 55
E. 60

The OA is E.

I don't have clear this PS question. I appreciate if any expert explain it for me. Thank you so much.

IF....we sell 15 of the chickens, then there will be EXACTLY 4 more days of food than are needed. That's an INTERESTING piece of info - exactly 4 more days of food (not 3.999, not 3.5, not 2.7) - an INTEGER.

IF...we buy 20 more chickens, there there will be EXACTLY 3 fewer days of food than are needed. Again, that's INTERESTING - it's an INTEGER.

The ONLY way for those integers to appear is if the current number of chickens is a MULTIPLE of BOTH 15 and 20.

Can you clarify your reasoning here?
It seems to be based on the contention that the current number of chickens must be a multiple of the number of chickens bought or sold -- a contention that does not seem quite right.
Consider the following alternate version of the prompt:

If a farmer sells 20 of his chickens, his stock of feed will last for 10 more days than planned, but if he buys 20 more chickens, he will run out of feed 2 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?

A. 25
B. 30
C. 40
D. 60
E. 80

Total amount of feed = (number of chickens)(number of days).
The number of chickens is INVERSELY PROPORTIONAL to the number of days.
If the number of chickens DOUBLES, then the feed will last for 1/2 the number of days.
If the number of chickens TRIPLES, then the feed will last for 1/3 the number of days.

Rephrasing the equation above, we get:
Number of days = (total amount of feed)/(number of chickens).

We can PLUG IN THE ANSWERS, which represent the actual number of chickens.
When the correct answer is plugged in:
20 fewer chickens versus 20 additional chickens will yield the following time ratio:
(10 more days)/(2 fewer days) = 10/2.

Answer choice B: 30 chickens
In this case:
If 20 chickens are sold, the remaining number of chickens = 30-20 = 10.
If 20 chickens are purchased, the new number of chickens = 30+20 = 50.

Let the total amount of feed = the LCM of 30, 10 and 50 = 150 pounds.

Number of days for 30 chickens = (amount of feed)/(number of chickens) = 150/30 = 5 days.
Number of days for 10 chickens = (amount of feed)/(number of chickens) = 150/10 = 15 days, implying 10 more days of feed.
Number of days for 50 chickens = (amount of feed)/(number of chickens) = 150/50 = 3 days, implying 2 fewer days of feed.
Success!
The time ratio in green aligns with the blue portion above.

The correct answer is B.

In this version of the problem, 20 chickens are bought or sold.
The three answer choices that are divisible by the number of chickens bought or sold -- C, D, and E -- are incorrect.
The correct answer -- 30 -- is NOT divisible by the number of chickens bought or sold.