- SOLUTION: If 15 oxen or 20 cows can eat the grass of a field in 80 days, then in how many days will 6 oxen and 2 cows eat the same grass?
- Cow grass eating problem
- contest math
SOLUTION: If 15 oxen or 20 cows can eat the grass of a field in 80 days, then in how many days will 6 oxen and 2 cows eat the same grass?
You can oxen eat ( 1 field ) / ( 80 days ) ox eats ( 1 field ) / ( 15*80 days ) ------------------------------------- cows eat ( 1 field ) / ( 80 days ) cow eats ( 1 field ) / ( 20*80 days ) ------------------------------------- Add their rates of eating to get their rate eating together Let their rate eating together = ( 1 field ) / ( d days ) --------------------------------------------- Multiply both sides by 6 oxen and 2 cows eat 1 field of grass in 160 days check: Multiply both sides by OK
Cow grass eating problem
I found a cool web page called mathproblems.info that has a lot of different story problems from different types of math. There are gaming questions, geometry questions, probability, calculus and brain teasers. Check it out. Five cows can eat 2 acres of grass in 10 days. Seven cows can eat 3 acres of grass in 30 days. Each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass? I have not yet looked at the solution so I am going to think about it here on this page. My first reaction to this problem is that it should be easy but there are three variables, cows, acres and days and only two equations to find the answer with. It seem reasonable to also think that this problem is linear somehow because the cows eat at the same rate so one cow is the same as another. 2 cows should eat twice as much OR twice as fast as one cow. The amount of grass does not change so one acre is the same as the next. If we place two fields next to each other as in the first situation we can find that 10 cows eat 4 acres in 10 days, or three fields together gives 15 cows eat 6 acres in 10 days. We need 16 cows and 7 acres. If 5 cows eat 2 acres in 10 days then one cow eats 2/5 of an acre in 10 days…. so 16 cows eat 6 2/5 acres in 10 days. Using the second sentence, 7 cows eat 3 acres in 30 days means that 14 cows eat 6 acres in 30 days. Wow this totally can’t make sense because we just said 15 cows e...
contest math
I encountered this problem about Newton's problem of cows and fields: In a field, 17 cows can finish the whole grass in the field for 30 days. 19 cows can finish in 24 days. If a group of cows eat the grass for 6 days, then 4 cows are sold, the remaining cows can finish the grass in 2 days. What is the initial number of cows before 4 of them are sold? I could do this problem using algebraic method but when I consulted another source, a different approach is used and I could not understand that approach. Here is my way: First we find the amount of grass that grows each day: $(17\times30-19\times24)\div(30-24)=9$ (the amount of grass is equivalent to the amount of grass that 9 cows could eat in a day). Then we find the initial amount of grass before any cow eats: $(17\times30-9\times30)=240$ (the amount of grass is equivalent to the amount of grass that 240 cows could eat in a day). Then we let the initial number of cows before 4 of them are sold be $x$. Then $(x-9)\times6+(x-4-9)\times2=240$. So $x=40$. Another approach that confuses me is $(240+8\times9+1\times2\times4)\div28=40$. The explanation is: Suppose the 4 cows are not sold, then we have $240+8\times9+2\times4=320$ amount of grass (where we use the same unit as above). Then we have $320\div(6+2)=40$ cows. My question is: when the cows are not sold, then we won't need 2 more days to finish the grass, why should we add $2\times4$? Basically I don't understand the reason for adding up $240+8\times9+2\times4$ and then ...
