Three farmers were selling chickens in the local market One farmer had 10 chickens to sell another had 16 chickens to sell the last had 26 chickens to sell In order not to compete with each other they agreed to sell the all chickens at the same price But by lunch time they decided to lower their price to the same lower price point By the end of the day they sold sold all their chickens It turned out they all collected the same amount of money $35 from the day''s chicken sales What was the price the chickens before lunch time and after lunch time?
Courtesy of chartgpt- To solve this problem, we need to determine the price of the chickens before lunch and the price after lunch. Let's denote the price before lunch as \( P_{\text{before}} \) and the price after lunch as \( P_{\text{after}} \). 1. **Total number of chickens sold:** - Farmer 1: 10 chickens - Farmer 2: 16 chickens - Farmer 3: 26 chickens - Total: \( 10 + 16 + 26 = 52 \) chickens 2. **Total money collected by each farmer:** - Each farmer collected $35 by the end of the day. Since they sold all the chickens and collected the same amount of money, we can set up equations for each farmer: Let \( x \) be the number of chickens sold before lunch and \( y \) be the number of chickens sold after lunch. The equations for each farmer would be: \[ 10P_{\text{before}} + (10 - x)P_{\text{after}} = 35 \] \[ 16P_{\text{before}} + (16 - y)P_{\text{after}} = 35 \] \[ 26P_{\text{before}} + (26 - z)P_{\text{after}} = 35 \] However, solving these equations directly might be complicated, so let's use the fact that the total money collected by all farmers is: \[ 35 \times 3 = 105 \text{ dollars} \] To simplify: \[ 10P_{\text{before}} + 16P_{\text{before}} + 26P_{\text{before}} + (10 - x)P_{\text{after}} + (16 - y)P_{\text{after}} + (26 - z)P_{\text{after}} = 105 \] \[ 52P_{\text{before}} + (52 - (x + y + z))P_{\text{after}} = 105 \] We know all chickens were sold, so: \[ x + y + z = 52 \] This reduces to: \[ 52P_{\text{before}} + 0P_{\text{after}} = 105 \] \[ 52P_{\text{before}} = 105 \] \[ P_{\text{before}} = \frac{105}{52} = 2.019 \] Each farmer sold chickens at a lower price after lunch. Since they collected $35 each, we can assume that they sold all their remaining chickens for the same total amount: If we assume the remaining chickens were sold for \( P_{\text{after}} \), then: \[ P_{\text{after}} = \frac{35 - (number\ of\ chickens\ sold\ before\ lunch) \times P_{\text{before}}}{number\ of\ chickens\ sold\ after\ lunch} \] However, for simplicity and equal distribution, it's logical to assume they dropped their price to $1 per chicken after lunch to sell them all equally for $35. So, \[ P_{\text{after}} = 1 \text{ dollar} \] Therefore, the price before lunch was approximately $2.019 per chicken, and the price after lunch was $1 per chicken.
chatgpt的回答有问题，如下： 1）说Let \( x \) be the number of chickens sold before lunch and \( y \) be the number of chickens sold after lunch. 并没有定义z 2）但是紧接着这段里面用到了z The equations for each farmer would be: \[ 10P_{\text{before}} + (10 - x)P_{\text{after}} = 35 \] 如果x是上午卖的数量，y是下午卖的数量，那么x是上午三家卖的数量总和，为什么将x 放在第一家的方程中，x是三家共有的 \[ 16P_{\text{before}} + (16 - y)P_{\text{after}} = 35 \] \[ 26P_{\text{before}} + (26 - z)P_{\text{after}} = 35 \] 3） We know all chickens were sold, so: \[ x + y + z = 52 \] 前面说10+16+26=52 不明白xyz的定义到底是什么 只看 \[ x + y + z = 52 \] 的话，x是第一家总共卖了多少只，y是第二家总共卖了多少只，z是第三家总共卖了多少只，但是这不是已知条件吗
Just thought of another way by assuming: (1) the first farmer sold (10-a) chickens before lunch, and (a) chickens after lunch (2) the second farmer sold (16-b) chickens before lunch, and (b) chickens after lunch (3) the third farmer sold (26-c) chickens before lunch, and (c) chickens after lunch and (4) the price in the morning is as k times as expensive as that in the afternoon Then （10-a) * k + a = (16-b) k + b ==> (10-a -16+b)* k = (b-a) ==> (6+a-b) * k = (a-b) k = (a-b)/(6+a-b) = 1 - 6 / (6 + a-b) > 1, then 6 + a - b < 0 ==> b - a > 6 （16-b)*k + b = (26-c)* k + c ==> (16-b-26+c)*k = (c - b) ===> (10+b-c)* k = (b-c) k = (b-c)/(10+b-c) = 1 - 10 / (10 + b-c) > 1, then 10 + b - c < 0 ==> c - b > 10 Thus, a-b/(6+a-b) = (b-c)/(10+b-c) ==> (a-b)/6 = (b-c)/10 ==> (a-b)/3 = (c-b)/5 then the only solution is c - b = 15, b -a = 9, and thus (a, b, c) = (a, a+9, a+24) thus k = 3 is derived, which means the price in the morning is as three times as expensive as that in the afternoon. Let the price in the morning is 3y, and thus y in the afternoon, per chicken. Then 3 * (10-a) * y + a * y = 35 3 * (16-b) * y + b * y = 35 3 * (26-c) * y + c * y = 35 all the above equalization are equivalent to each other due to the previous process, and thus reduced to (30-2a)* y = 35 That is, y = 35/(30-2a) as (a, b, c) = (a, a+9, a+24), where a +24 <=26 , then potential values of (a,b,c) are (0, 9, 24), (1,10,25), (2, 11,26) Most likely the (1, 10, 25) is the right candidate, then y = 35/(30-2*1)= 5/4, and thus the price in the morning is 15/4 per chicken the first farmer sold (9,1) respectively in the morning and afternoon, a total of (9*3+1)* 5/4 = 28* 5/4 = 35 the second farmer sold (6,10) respectively in the morning and afternoon, a total of (6*3+10)* 5/4 = 28* 5/4 = 35 the third farmer sold (1,25) respectively in the morning and afternoon, a total of (1*3+25)* 5/4 = 28* 5/4 = 35 However，I still wonder what is wrong with the previous solving process attache above, and can not figure it out.
One farmer had 10 chickens to sell another had 16 chickens to sell the last had 26 chickens to sell In order not to compete with each other they agreed to sell the all chickens at the same price
But by lunch time they decided to lower their price to the same lower price point By the end of the day they sold sold all their chickens It turned out they all collected the same amount of money $35 from the day''s chicken sales
What was the price the chickens before lunch time and after lunch time?
然后列方程，解方程，反复计算，还是求不出合理答案。
请教大家帮忙查看哪里出问题了，或者给出解法。
1. **Total number of chickens sold:** - Farmer 1: 10 chickens - Farmer 2: 16 chickens - Farmer 3: 26 chickens - Total: \( 10 + 16 + 26 = 52 \) chickens
2. **Total money collected by each farmer:** - Each farmer collected $35 by the end of the day.
Since they sold all the chickens and collected the same amount of money, we can set up equations for each farmer:
Let \( x \) be the number of chickens sold before lunch and \( y \) be the number of chickens sold after lunch.
The equations for each farmer would be: \[ 10P_{\text{before}} + (10 - x)P_{\text{after}} = 35 \] \[ 16P_{\text{before}} + (16 - y)P_{\text{after}} = 35 \] \[ 26P_{\text{before}} + (26 - z)P_{\text{after}} = 35 \]
However, solving these equations directly might be complicated, so let's use the fact that the total money collected by all farmers is: \[ 35 \times 3 = 105 \text{ dollars} \]
To simplify: \[ 10P_{\text{before}} + 16P_{\text{before}} + 26P_{\text{before}} + (10 - x)P_{\text{after}} + (16 - y)P_{\text{after}} + (26 - z)P_{\text{after}} = 105 \] \[ 52P_{\text{before}} + (52 - (x + y + z))P_{\text{after}} = 105 \]
We know all chickens were sold, so: \[ x + y + z = 52 \]
This reduces to: \[ 52P_{\text{before}} + 0P_{\text{after}} = 105 \] \[ 52P_{\text{before}} = 105 \] \[ P_{\text{before}} = \frac{105}{52} = 2.019 \]
Each farmer sold chickens at a lower price after lunch. Since they collected $35 each, we can assume that they sold all their remaining chickens for the same total amount:
If we assume the remaining chickens were sold for \( P_{\text{after}} \), then: \[ P_{\text{after}} = \frac{35 - (number\ of\ chickens\ sold\ before\ lunch) \times P_{\text{before}}}{number\ of\ chickens\ sold\ after\ lunch} \]
However, for simplicity and equal distribution, it's logical to assume they dropped their price to $1 per chicken after lunch to sell them all equally for $35.
So, \[ P_{\text{after}} = 1 \text{ dollar} \]
Therefore, the price before lunch was approximately $2.019 per chicken, and the price after lunch was $1 per chicken.
3） We know all chickens were sold, so: \[ x + y + z = 52 \] 前面说10+16+26=52 不明白xyz的定义到底是什么 只看 \[ x + y + z = 52 \] 的话，x是第一家总共卖了多少只，y是第二家总共卖了多少只，z是第三家总共卖了多少只，但是这不是已知条件吗
Just thought of another way by assuming: (1) the first farmer sold (10-a) chickens before lunch, and (a) chickens after lunch (2) the second farmer sold (16-b) chickens before lunch, and (b) chickens after lunch (3) the third farmer sold (26-c) chickens before lunch, and (c) chickens after lunch and (4) the price in the morning is as k times as expensive as that in the afternoon
Then （10-a) * k + a = (16-b) k + b ==> (10-a -16+b)* k = (b-a) ==> (6+a-b) * k = (a-b) k = (a-b)/(6+a-b) = 1 - 6 / (6 + a-b) > 1, then 6 + a - b < 0 ==> b - a > 6
（16-b)*k + b = (26-c)* k + c ==> (16-b-26+c)*k = (c - b) ===> (10+b-c)* k = (b-c) k = (b-c)/(10+b-c) = 1 - 10 / (10 + b-c) > 1, then 10 + b - c < 0 ==> c - b > 10
Thus, a-b/(6+a-b) = (b-c)/(10+b-c) ==> (a-b)/6 = (b-c)/10 ==> (a-b)/3 = (c-b)/5 then the only solution is c - b = 15, b -a = 9, and thus (a, b, c) = (a, a+9, a+24) thus k = 3 is derived, which means the price in the morning is as three times as expensive as that in the afternoon.
Let the price in the morning is 3y, and thus y in the afternoon, per chicken. Then 3 * (10-a) * y + a * y = 35 3 * (16-b) * y + b * y = 35 3 * (26-c) * y + c * y = 35 all the above equalization are equivalent to each other due to the previous process, and thus reduced to (30-2a)* y = 35 That is, y = 35/(30-2a)
as (a, b, c) = (a, a+9, a+24), where a +24 <=26 , then potential values of (a,b,c) are (0, 9, 24), (1,10,25), (2, 11,26)
Most likely the (1, 10, 25) is the right candidate, then y = 35/(30-2*1)= 5/4, and thus the price in the morning is 15/4 per chicken
the first farmer sold (9,1) respectively in the morning and afternoon, a total of (9*3+1)* 5/4 = 28* 5/4 = 35 the second farmer sold (6,10) respectively in the morning and afternoon, a total of (6*3+10)* 5/4 = 28* 5/4 = 35 the third farmer sold (1,25) respectively in the morning and afternoon, a total of (1*3+25)* 5/4 = 28* 5/4 = 35
However，I still wonder what is wrong with the previous solving process attache above, and can not figure it out.
直接上答案吧，早上三块七毛五，三人分别卖9只，6只，1只鸡。 下午一块两毛五，分别卖1只，10只，25只。
也就是英文大家有quarter的概念， 翻译成中文题中国小孩得懵，啥年代了价格会到分。
我觉得三个答案都可以。8楼已经推出来了。
但是阅读理解似乎隐含了上下午每个人都卖了鸡。😂
我还是一直发现不了自己一开始三页为啥老做不出来答案。不知道哪里错了。
终于查出原来第三页错在哪里了，竟然6 - b + a = 6 -(a-b) 这样的错误，查了好几遍。 修正的第三页和以后的如下
先从原点画个斜率为 x 的直线；在这根直线上取 3 个点 A，B，C，在每个点画根斜率为 y （y < x) 的平行线；
然后 y 轴值为 35 的地方画根 X 轴的平行线，和 3 根平行线交与 3 个点，X 坐标分别为 10，16， 26
这样 C，B，A 的 横坐标 显然就 3 种整数选择 （分别是 3 人早上卖掉的鸡的数目），再求 x, y 就出来了
不明觉厉！🐮人！
我来想想你这个解法，为啥这么个思路？
因为每个人卖鸡所得的钱的数目 就是从原点开始，沿着两根斜率为 x 和 y 的直线所到达的高度
赞。“一针见脓”，想不起考啥好成语了，先放个place holder 在这里 😂