Hey, this one is different. Can you someone solve it for me?
A farmer buys 100 animals for $100.
Cow=$10
Sheep=$5
Chicken=$3
Rabbit=$0.50
He baught atleast 1 of each animal.
How many of each animal did he buy?
Hey, this one is different. Can you someone solve it for me?
A farmer buys 100 animals for $100.
Cow=$10
Sheep=$5
Chicken=$3
Rabbit=$0.50
He baught atleast 1 of each animal.
How many of each animal did he buy?
No, sorry. As you read in the the Read Before Posting thread that's stickied at the top of every subforum, that's not the purpose of the forum. As our name suggests, we are here to provide help to students struggling with their homework, but we do not give fully worked answers. Please re-read the Read Before Posting thread and comply with the rules therein. In particular, please share with any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.
As a big hint to help you get started, if you're completely stuck at the beginning and don't even know how to start is to remember that the 100 animals is the stricter of the two conditions. The farmer can buy 10 cows, but that leaves him no money to buy any of the other animals. If he buys 9 cows, he has $10 left over. Can he buy the required 91 other animals with this money? What if the farmer buys only 8 cows? And so on.
I agree with ksdhart2. You need to try some stuff. Experimenting often leads to insight. Trial and error is like experimenting.
You might also want to consider how the values of different animals are related. That is, it might be handy, as you refine your guesses doing trial and error, to have a listing like:
1 cow = 20 rabbits (that is, 1 times $10 = 20 times $0.50)
1 cow = 2 sheep
1 chicken = 6 rabbits, and so on.
Do you see what I'm doing?
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
I thought I could use equations like C+S+c+R=100
and 10C+5S+3c+0.5R= 100
buts that’s 2 equations and so have to identify 4 variables. But that’s won’t work
The next step I did was assuming we have 90 rabbits = $45
We have left $55 and 10 animals to work with (cow,sheep and chicken). I tried different combinations until I came to an answer: it’s must be 3cows = $30 2sheep = $10 5chiken = $15
30+10+15+45= $100 And 90+3+2+5 = 100
or am I missing something here? This was just me guessing after hours...
Is there a simpler way?
Btw thanks for your tips!! 🙏🏼
Agree. In general (but there are some exceptions), to solve a system of equations you need the same number of equations as variables in the system.
There might be guessing strategies that are more efficient, but you'd have to spend time experimenting to find them anyway. Good job!… I tried different combinations until I came to an answer: it’s must be 3cows = $30 2sheep = $10 5chiken = $15 [and 90rabbit = $45]
… This was just me guessing after hours … Is there a simpler way?
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
You can find answers (or hints) for this classic kind of problem here, including both algebraic and non-algebraic solutions:
How Many Mice, Cats, and Dogs?
100 Animals, 100 Dollars, No Algebra
Buying Cows, Pigs, and Chickens
If you search the site for the phrase "100 animals", you'll find yet more examples.
It's nice, to have an experienced doctor in the house.
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
As it turns out, there's four unique solutions to this problem. You identified one such solution, which should suffice (although I have seen teachers refuse to accept any answer except their "correct" answer, even if the other solution(s) fully satisfy the problem's criteria... ). I found another solution very quickly using a bit of trial and error. I began by using the method I outlined in my original post and worked down until I realized that if the farmer bought 5 cows, he could not buy the required other 95 animals while sticking to the budget and the requirement that he buy at least one of each animal.
Then I looked at what would happen if he bought 4 cows and 1 of each other animal. That leaves $51.50 to buy 93 animals. I figured the best way to go from here was to buy as many rabbits as possible because they're cheapest. Buying 94 rabbits would give the desired total of 100 animals, but also put him 5 dollars over budget. At this point, I realized that if he bought a chicken instead of a sheep, he'd "earn back" $0.50 but spend $3, for a net increase of $2.50 to his expenditures. Perfect! If he exchanges two rabbits for sheep, that means an increase of $5 which is exactly what we needed. To recap, if the farmer buys 4 cows, 1 sheep, 3 chickens, and 92 rabbits, it satisfies all of the requirements. and now just to double check algebraically: c + s + h [for cHickens] + r = 4 + 1 + 3 + 92 = 100, and 10c + 5s + 3h + 0.5r = 40 + 5 + 9 + 46 = 100.
Similar such experimentation will uncover one solution where the farmer buys 4 cows, one where he buys 3 cows, one where he buys 2 cows, and one where he buys 1 cow. WolframAlpha verifies that these are the only four integer solutions given the constraints.
Bookmarks