M08-13

Official Solution:

A farmer discovers that if he sells 75 of his chickens, his stock of feed will last 20 days longer than planned. On the other hand, if he buys 100 more chickens, the feed will run out 15 days earlier than planned. If no chickens are sold or bought, the farmer's stock of feed will align precisely with the intended schedule. How many chickens does the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300


Let the number of chickens be \(c\) and the number of days the stock of feed is intended to last be \(d\). We can also assume that the amount of feed each chicken consumes per day is 1 unit (since it will be proportionally reduced anyway, the exact amount does not matter).

If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned. This implies that 75 chickens in \(d\) days would have consumed the same amount of food as \((c - 75)\) chickens in 20 days (since both represent the amount of food saved). Thus, we can write the equation: \(75d = (c - 75)*20\).

If the farmer buys 100 more chickens, he will run out of feed 15 days earlier than planned. This implies that 100 chickens in \(d-15\) days would have consumed the same amount of food as \(c\) chickens in 15 days (since both represent the additional amount of food consumed). We can write another equation from this situation: \(100(d-15) = 15c\).

Solving the system of equations, \(75d = (c - 75) *20\) and \(100(d-15) = 15c\), gives us \(c = 300\).

Therefore, the farmer has 300 chickens.


Answer: E
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I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with the explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Hey Bunuel , can you please elaborate the solution, I am having trouble understanding the solution.

Can you please tell me which part is unclear in the solution? Thank you!

P.S. Else you can check alternative solutions HERE.
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Hey Bunuel the logic you have used i.e the increase in feed with reduced chicken consumption part. I know, even though you have clearly mentioned the logic, I still have some trouble understanding it. Can you please elaborate on the reasoning you have used, if that's possible.
Thanks in advance

1. The problem stipulates that if the farmer sells 75 chickens, his feed would last 20 more days than intended. This suggests that the amount of feed these 75 chickens would have eaten over the originally planned duration (d days) is equal to the amount of feed the remaining chickens (c - 75) would consume in the additional 20 days. We can equate these two quantities and form our first equation: 75d = (c - 75) * 20.

2. In the second scenario, the farmer buys 100 more chickens, leading to the feed running out 15 days earlier than initially planned. This implies that these additional 100 chickens' feed consumption over the adjusted period (d - 15 days) is equivalent to the amount of feed the original quantity of chickens (c) would consume in these 15 days saved. By equating these two quantities, we can form our second equation: 100(d - 15) = 15c.

Does this make sense?
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Yes, It makes sense now. Thanks a lot Bunuel

This one seems easy but it has taken me a while to understand the reasoning. Any alternative method of solving it?

You can check alternative solutions in the following post If the farmer sells 75 of his chickens.
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