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M06-12 23 Jan 2025, 22:18
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I like the solution - it’s helpful.
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M06-12 26 Apr 2025, 07:05
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M06-12 25 May 2025, 04:55
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M06-12 25 May 2025, 12:01
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I did not quite understand the solution. .
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Bunuel
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M06-12 25 May 2025, 12:09
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Jeringas
I did not quite understand the solution. .
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This isn't an easy question, so careful study is recommended. It's a good idea to review the full discussion, especially the posts that walk through the reasoning in detail. There are also a few alternative solutions shared in the thread, and you can find more HERE.

If anything is still unclear, please ask a specific question so we can better address your doubt.
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M06-12 06 Jul 2025, 01:05
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M06-12 26 Jul 2025, 10:48
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M06-12 13 Aug 2025, 02:57
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This is a great question that’s helpful for learning and I like the solution - it’s helpful.
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Hi Bunuel, I assumed variables for cow=3x, chicken=3y and sheep=x+y. While I understood the forums' discussions, I wanted to know why I cannot reach the solution by assuming these variables? I have filled more than 8 pages still I am stuck with these variables
Thanks
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Vineet1102
Hi Bunuel, I assumed variables for cow=3x, chicken=3y and sheep=x+y. While I understood the forums' discussions, I wanted to know why I cannot reach the solution by assuming these variables? I have filled more than 8 pages still I am stuck with these variables
Thanks
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You can proceed EXACTLY the same way as shown in the official solution:

(3x + 3y) + (12x + 6y) = 100

5(3x) + 3(3y) = 100

So 3x = 17 and 3y = 5, OR 3x = 14 and 3y = 10.

Then you apply the same logic as in the official solution. Not sure why you’d set it up this way though, since there’s no need for x and y here.
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M06-12 26 Aug 2025, 01:47
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I didn't use multiples method--chicken=x,cow=y,sheep=z
x+y=3z 2x+4y=100-x-y
so 3x+5y=100-----9z+2y=100,
then A: z=5, NO
B: z=8 72+2y=100---y=14, and y>x,z
so it's B
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M06-12 17 Sep 2025, 20:41
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I don’t quite agree with the solution. What if there were 0 chickens, 15 cows and 5 sheep. Wouldn't it satisfy all requirements?
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M06-12 17 Sep 2025, 22:11
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Your solution satisfies the condition - 3×(chicken+sheep)=cow
3×(0+5)=15

but doesn't satisfy the second condition - 5×cow+3×chicken=100
5×15+3×0≠100 -> 75≠100

agnikhil
I don’t quite agree with the solution. What if there were 0 chickens, 15 cows and 5 sheep. Wouldn't it satisfy all requirements?
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M06-12 07 Oct 2025, 05:12
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M06-12 28 Oct 2025, 13:04
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This is a great question that’s helpful for learning and I like the solution - it’s helpful.
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M06-12 15 Nov 2025, 02:25
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M06-12 15 Nov 2025, 10:36
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