Each animal has 1 head so h c 24 Each chicken has 2 legs and each cow has 4 legs so 2 h 4 c 76 Solve for h and c. If you want to use algebra you can let h be the number of chickens and c be the number of cows. Thus you are left with 14 cows and 24 - 14 10 chickens. Here $x,y$ are the prices before and after lunch, $k,l,m$ are the number of sold chickens before lunch. With each of these pairs of legs you can turn a chicken into a cow. A full solution is not difficult although it can be a bit tedious to obtain, but one key is to derive the Diophantine condition $$5x - 8y 3z = 0,$$ in conjunction with the obvious constraint $10 \ge x > y > z \ge 0$. The only plausible qualitative difference between the solutions is that the latter corresponds to an exact amount to the penny the former does not (but why should this matter from a mathematical standpoint?). The difference is $35-10c$, so the number of chickens sold before lunchtime is $\frac.$$ It is not stated that all farmers sold at least one chicken prior to lunchtime, and the difference between the first and second solution (one chicken each) is not, in my opinion, subjectively large enough to justify saying that the farmers were unhappy with the rate of sales for the former but not the latter. With this, we can know how many chickens were sold at the higher price.įor the one who sold $10$ chickens, if he sold everything at the after-lunchtime price, his sales will be $10c$. Hence every chicken sold at the before-lunchtime price rather than the after-lunchtime price will incur an increase of sales of $d$. However, when he "switches" a chicken from the "after-lunchtime" price to the "before-lunchtime" price, there will be a price increase of $d$ (which is the difference between the after-lunchtime and before-lunchtime prices). Think of it this way: if the seller sold everything at $c$ after lunchtime, the total sales will be $kc$, where $k$ is the number of chickens. Now we attempt to solve for how many did each seller sell at each of the two prices. Let $c$ be the morning price and let $d$ be the difference between the morning and afternoon prices.
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