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# M08-13

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Math Expert
Joined: 02 Sep 2009
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Math Expert
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General Discussion
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with the explanation.
Math Expert
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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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Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Hey Bunuel , can you please elaborate the solution, I am having trouble understanding the solution.
Math Expert
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rickyric395 wrote:
Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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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Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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.
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rickyric395 wrote:
Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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.

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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Bunuel wrote:
rickyric395 wrote:
Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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.

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?

Yes, It makes sense now. Thanks a lot Bunuel
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This one seems easy but it has taken me a while to understand the reasoning. Any alternative method of solving it?
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echo9000 wrote:
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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Hi bunuel,
I tried solving the question using this method:
let x be the number of actual chicken and y be the number of days of food for these chicken and lets assum that these each chicken eats 1 unit of food each day
So, xy = (x-75)(y+20) and xy = (x+100)(y-15) but solving these equations the answer is not coming out actually. Could you pls explain why this concept is wrong?
Math Expert
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tanishqgirotra wrote:
Hi bunuel,
I tried solving the question using this method:
let x be the number of actual chicken and y be the number of days of food for these chicken and lets assum that these each chicken eats 1 unit of food each day
So, xy = (x-75)(y+20) and xy = (x+100)(y-15) but solving these equations the answer is not coming out actually. Could you pls explain why this concept is wrong?

­xy = (x - 75)(y + 20) and xy = (x + 100)(y - 15) gives x = 300 and y = 60, which is correct.
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­c: number of chickens
d: number of days

c * d = (c - 75) * (d + 20)
= (c + 100) * (d - 15)

=> cd = cd - 75d + 20c - 1500 (i)
= cd + 100d - 15c - 1500 (ii)

=> - 75d + 20c = 100d - 15c
=> 35c = 175d
=> c = 5d

From (i) => 100d - 15c = 1500
=> 20d - 3c = 300
=> 4c - 3c = 300 (coz c = 5d)
=> c = 300­