Again, great work on the previous POTW. Please refer to the comments section of POTW#1 for the correct answer to it. Remember to provide solutions and explanations to your work. Simply writing "I think the answer is 1/512" is not sufficient.
POTW #2:
Mr. Milette owns a chicken farm. In order to keep his free-range chickens behaving properly he also has many dogs to act as security. So, the only animals on his farm are dogs and chickens. One day in March he wanted to count his animals for tax-return purposes. He needed to know exactly how many chickens and how many dogs he had. He got a bit scrambled in his counting (number sense tripped him up a bit). He ended up counting a total of 134 legs and 48 heads (again, total). Figure out how many chickens and how many dogs Mr. Milette has. Show your work. See if you can solve it in a variety of ways, or in ways that differ from your classmates. One sentence responses ARE NOT ENOUGH! For those of you that may have already done a problem like this in the past, I challenge you to use algebra in your solution.
There are 29 chickens and 19 dogs, which I labeled as x and y respectively:
ReplyDelete1) x+y=48
2) 2x+4y=134
2(x+2y)=134
x+2y=134/2
x+2y=67
x=67-2y
Isolate x:
x=48-y
x=67-2y
48-y=67-2y
-y+2y=67-48
y=19
So now I can plug the amount of hens into our first equation:
x+y=48
x+19=48
x=48-19
x=29
So the final answer is there are 29 chickens and 19 hens.
I don't really know how to solve this, can I have some help?
ReplyDeleteTry guess and check/trial and error. Start plugging in numbers of dogs and chickens and see how you can get to 134 legs (4 per dog, two per chicken) and 48 heads. Please look at other students' work for inspiration as well.
DeleteThings I know: 48 animals, 134 legs, chicken legs = 2, dog legs = 4.
ReplyDeleteFirst I paired a chicken and a dog together -> 2 legs (chicken) + 4 legs (dog) = 6 legs, 1 head (chicken) + 1 head (dog) = 2 heads.
Therefore 1 pair of a chicken and a dog has 2 heads and 6 legs.
I know that there are 48 heads and each pair has 2 heads so I divide 48 heads by 2 to get the number of groups. Which equals 24.
Next I multiplied 24 groups by 6 to get the amount of legs 24 groups have. Which equals 144 legs. Because there are only 134 legs in the question, I know that I am over the amount of legs.
To find the amount of legs I am over by, I subtracted 144 by 134 and got 10. Now I know that I am 10 legs over.
I found the difference between the number if legs a chicken has and the number of legs a dog has and got 2.
I take that 2 and divide 10 by it which gives me 5. I now know that I have to switch 5 dogs with 5 chickens.
So next I subtracted 5 from 24 (dogs) and added 5 to 24 (chickens) and got 19 (dogs) and 29 (chickens).
Now I know how many animals Mr. Milette has.
To check my work I put the number of dogs and chickens into an equation.
19 dogs + 29 chickens = 48 animals. Check.
(19 dogs * 4) + (29 chickens * 2) = 134. Check.
In conclusion, Mr. Milette has 19 dogs and 29 chickens on his farm.
2x48=96 134-96=38 38 divided by 2=19 dogs 48-19=29 chickens
ReplyDeleteFirst, I saw how many legs will be used if I give two legs to every head. Then I subtracted 96 from 134 to give me 38 which is the remainder of the legs left. I split the legs in pairs of two because each animal has either 2 or 4 legs. I got 19 pairs of legs so I added those pairs to 19 heads which means 19 heads/animals have four legs so 19 animals are dogs. This leaves 48-19=29 chickens. So Mr.Milette has 19 dogs and 29 chickens in his farm.
ReplyDeleteTo solve this potw, I used algebra. My solution was that there are 19 dogs and 29 chickens.
To begin, let x represent the number of dogs, and y represent the number of chickens.
From the given information, the number of animals in total is 48, meaning that the number of dogs + the number of chickens = 48. Using x and y,
x+y=48
Let's call this the heads equation
I can also use common sense and see that each dog has 4 legs, and each chicken has 2. So, the number of dogs times 4 + the number of chickens times 2 = 134 ( the total number of legs). Using x and y,
4x + 2y = 134
Let's call this the legs equation.
So, with those two equations, let's start.
First, I solved for x. To do so, I had to isolate y in one of the equations in order to substitute y into the other equation, making an equation in x.
x + y = 48
y = 48 - x
Then, I substituted this y into the legs equation, giving me an equation in which I could easily find x.
4x + 2 (48 - x) = 134
So, I solved for x
4x + 96 - 2x= 134
2x = 134 - 96
2x = 38
x = 19
Therefore, the number of dogs ( or x) is 19.
But I also need the number of chickens. So, I return to the heads equation and substitute x for 19.
x + y = 48
19 + y = 48
Then, I could easily solve for y.
y = 48 - 19
y = 29
Therefore the number of chickens ( or y) is 29.
Verification:
Heads equation:
x + y = 48
19 + 29 = 48
48 = 48
Legs equation:
4x + 2y = 134
4(19) + 2(29) = 134
76 + 58 = 134
134 = 134
So that's my solution and how I found it. I'm pretty sure it's correct but I might be wrong.
Even though I hadn't done this question before, I decided to solve the problem using algebra.
ReplyDeleteFirst, I knew d + c equals 48, which represents the heads.
Second, I know that 4d + 2c = 134, which represents the legs, as dogs have 4 legs and chickens have 2.
Now, I had the information to start solving for d:
I did c = 48 - d, using the knowledge from the heads.
I substituted the c into the following equation:
4d + 2 (48-d) = 134.
With this information, I solved for x.
48 x 2 = 96
4d + 96 - 2d = 134
2d = 134 - 96
134 - 96 = 38
2d = 38
38/2 = 1d
38/2 = 19
Now I knew that d = 19 and I could easily find c.
48 - 19 = 29.
c = 29.
So Mr. Milette owns 19 dogs and 29 chickens.
Is parent help allowed? I kind of forgot some math over the summer.
ReplyDeleteNot really Renee. The POTWs are for a challenge. I want you to try them on your own. If you are really stuck, it is ok to try trial and error (guess and check). Look at Eddie's post (he is one of the "unknown" posters) for a great way to try trial and error.
DeletePOTW #2:
ReplyDeleteMr. Milette owns a chicken farm. In order to keep his free-range chickens behaving properly he also has many dogs to act as security. So, the only animals on his farm are dogs and chickens. One day in March he wanted to count his animals for tax-return purposes. He needed to know exactly how many chickens and how many dogs he had. He got a bit scrambled in his counting (number sense tripped him up a bit). He ended up counting a total of 134 legs and 48 heads (again, total). Figure out how many chickens and how many dogs Mr. Milette has. Show your work. See if you can solve it in a variety of ways, or in ways that differ from your classmates. One sentence responses ARE NOT ENOUGH! For those of you that may have already done a problem like this in the past, I challenge you to use algebra in your solution.
48 heads= 48 animals.
2 legs= 1 animal (chicken)
4 legs = 1 animal (dog)
if they are all chicken, then i multiply all of the head count by 2 legs. 48 x 2 = 96. clearly this is too small and the information provided said that mr.milette had dogs to guard the chickens. therefore, this is an unreasonable extreme.
The next extreme i tried ws dogs. to do this, i multiplied each head by 4 legs, there for giving me a number for the amount of legs in total if it all dogs. i established before that if they were all chickens, then it would have 96 legs in total. knowing that dogs have twice the amount of legs per animal than to chickens, i took a short cut and multiplied the amount of chicken legs in total from my previous equation by 2. 96 x 2 = 192. this is also clearly way above the leg total counted, therefore it is unreasonable to think that it would be all dogs. also, it said that it was a chicken farm, so if it were to be all dogs, then it wouldn’t make sense for it to be a chicken farm, because there is no chickens.
I am not done with these extremes yet. if i take away the leg count by the chicken leg count and then divide it by 2,it should give how many dogs are in the mix. i can do the same with dogs only that i take away the amount of dog legs by the total counted and then divide the difference by 2 to get how many dogs are in the mix. i will use the dogs to check if the chicken calculation was correct. 134 - 96 = 32. 32 / 2= 16, 48 - 16 = 29 chickens, 19 dogs, 42 heads. sadly, this only added up to 128 legs. so i went and changed 3 chickens to dogs because for every chicken that is turned to a dog, there are 2 extra legs. now to do it with the dogs. 192-134 = 58. 58/2= 28 chickens. 48 - 28 = 20 dogs. 20 x 4 = 80. 28 x 2 = 56, 56 + 80= 136. since i am two legs above what i am supposed to have, i swapped out one dog for a chicken. which is again 19 dogs and 29 chickens. this confirms my previous answer. therefore, mr.milette has 19 dogs and 29 chickens in his chicken farm.
Great explanation Ramzi. I want to mention some whole-class feedback about your post tomorrow.
DeleteOk so i got 19 dogs and 29 chickens.
ReplyDeleteI used algebra
statement: let x be number of chickens and y be the number of dogs
x+y=48
2x+4y=134
y=48−x
4(48−x)+2x=134
192−4x+2x=134
192−134=4x−2x
58=2x
29=x
so there are 29 chickens.
48-29=19
so there are 19 dogs.
there was probably a faster way to solve this. but nonetheless it's solved. hopefully the equation solving part made sense.
The answer that I got was 29 chickens and 19 dogs. To solve this problem I used algebra:
ReplyDeleteDogs=4 legs each, Chickens=2 legs each
4x + 2y = 134; x + y = 48
x = 48 - y
4 x (48 - y) + 2y = 134
192 - 4y + 2y = 134; 192 - 2y = 134
-2y = 134 - 192; -2y = -58; y = 29
x + 29 = 48; x = 48 - 29; x = 19
4 x 19 + 2 x 29 = 134 = 19 dogs & 29 chickens.
I got 19 dogs and 29 chickens. How i got this answer was that i first that i made x the amount of dogs and y the amount of chickens. so i know that dogs have 4 legs and chickens have 2 legs so i did 4x + 2y = 134 legs and they each have 1 head so i did x + y = 48 heads. then i did y = 48 - x to find x first. so then i did 4x +2(48 - x) so i replaced c with 48-x and then i multiplied 48 by 2 and x by 2 to get 96 and 2x. Then the equation turns into 4x + 96 -2x. then since 96 - 2d is basically 2y imi going to add 2d and 96 which has to get 134 to try and find how many dog there are. so that means 2x= 134 - 96 which gets you 38. But that gives you the amount for 2 dogs and you only want 1 dog. so x= 1/2 38 which is 19. so then i did 48, the total amount of animals, minus 19 to get the rest of the chickens which gave me 29. Therefore, there are 19 dogs and 29 chickens
ReplyDeleteMr. Milette has a total of 19 dogs and 29 chickens. To solve this problem I decided to use the method of guess and check. I first divided the number of heads up equally and then multiplied them by 2 for chicken legs and 4 for dog legs. I got the sums of 96 and 48 I then added those two sums together and got a total of 144 which was 10 more legs then I wanted. I decided to add four more chickens and took away 4 dogs. I then did the same thing and added it up to a total of 136 which was only off by two! Finally I added one more chicken and one less dog to finally get a sum of 134 legs. Therefore Mr. Milette has 19 dogs and 29 chickens.
ReplyDeleteFirst off, I recorded the important information...
ReplyDeleteInformation:
- 48 heads
- 134 legs
Then, I represented chickens and dogs with x and y...
x= Chickens y= Dogs
After that, I created 2 equations...
1. 2x+4y =134
2. x+y =48
Next, I found the value of x through y in order to eliminate the second unknown variable...
x=48-y
Then, I substituted `x` with 48-y as it is equivilent...
2(48-y)+ 4y=134
Afterwards, simply simplified the equation...
2 x 48 -2y +4y =134
96-2y+4y=134
-2y+4y=134-96
2y= 38
y= 19
Now I know there are 19 dogs
48-19 = 29
So... that means there are 29 cats
Finally I checked my response by plugging in the values into the equations...
(2 x 29) +( 4 x 19)= 134
58+76= 134 I know that is correct
And just to double check...
29+ 19= 48 I know that is also correct
In conclusion, there are 19 dogs and 29 cats on the farm.
Very clear.
Deletea is total chickens and b is total dogs.
ReplyDeletesince there are 48 heads and neither chickens nor dogs have 2 heads, there are 48 animals.
a+b=48
the number of legs is 134: 2 legs per chicken and 4 per dog
2a+4b=134
now theres 2 algebraic equations:
a+b = 48
2a+4b=134
now to remove either 2a or 4b i need to double the smaller one and get 2a+2b=96
2a+4b=134
2a+2b=96
now that I have two, i can subtract one from another
2a+4b=134
-2a+2b=96
_____________
2b=38
This tells me that 2x the number of dogs is 38, so the number of
dogs is 19. and the number of chickens is 48-19=29.
Checking: 19+29=48
Final Answer: There are 19 dogs and 29 chickens.
C=number of chickens D=number of dogs
ReplyDelete1. C+D=48heads
2. 2C+4D=134legs
We want to get rid of the dogs inorder to find how much chickens there are from subtracting it from the total of 48 animal. This is how......
3. (1 x 4)(they all need to be multiplied by 4) 1C+1Dx4=4C+4D=192legs
4. (3-2) 4C+4D-2C-4D=192-134. 2C+0=58. C=29. D=48-29=19
Numbers of chickens= 29 number of dogs=19
I've done a question like this before and since Mr. Milette would like to know how to solve this question with algebra here it is:
ReplyDeletea=number of chickens
b=number of dogs
2a=number of chicken legs x number of chickens
4b= number of dog legs x number of dogs
(a+b=48)
(2a+4b=134)
a=(48-b)
2(48-b)+4b=134
(2x48-2xb)+4b=134
(96-2xb)+4b=134
(96-2b)+4b=134
96+(4b-2b)=134
96+2b=134
2b=134-96
134-96=38
b=38 divided by 2
b=19
(a+b=48)
(a=19=48)
a=48-19
a=29
check your work:
29+19=48 ( total number of animals)
2x29=58 ( total amount of chicken legs)
4x19=76 (total amount of dog legs)
58+76=134 (total number of legs)
There is a total of 29 chickens and 19 dogs on Mr. Milette's farm because after isolating b I found out that b=19 therefore a would equal 48(the number of animals in total)-19 which is 29. I even checked my work by adding 29 to 19 to get the number of animals in total which was 48 and got 48. I multiply the number of legs on a chicken to the amount of chickens to get 58. I multiplied the number of dog legs to the number of dogs to get 76. I finally added them together to get 134 the total amount of legs. Sorry that I can't explain how I isolated b, I just didn't really know how to explain it but my work is there if you want to see. I solve the question with algebra! It isn't that hard Mr. Milette, you said it was a hard algebra question!(I don't really like algebra by the way)
Wonderful explanation!
DeleteI did guess and check, which I know isn’t the fastest way, but I did it anyway. Also, I have never done this before but you said it was a challenge to use algebra, so I used algebra.
ReplyDeleteX= Dogs Y=Chickens
First I wrote out these 2 questions.
Legs: X×4+Y×2=134
Heads: X+Y=48
I did heads first. I thought, what would be a not so obvious way to make 8? (You said it was challenging) So I did 9+9. That equals 18. I’m still short 30. I decided it to split it into 10 and 20. Add 9 to both. Now I have 29 and 19. I added them together to check, and I got 48. But which was chickens, and which was dogs? Dogs have more legs, I thought. So there should be less. So I decided that 19 would be dogs, and 29 would be chickens.
Now to test it on the legs question.
I replaced the X with 19, and the Y with 29. 19×4+29×2= 134. That means I did it right! The think the answer is:
There are 19 dogs and 29 chickens on Mr. Milettes farm. Or X=19 and Y=29.
I think that there are 19 dogs and 29 chickens. I think that this is the answer because, first of all, there are usually less "security" (in this case the dogs) than the "civilians" (the chickens) and in my answer there is less dogs than chickens. In addition I decided to use guess and check to find the answer so I tried many ways such as 20 dogs and 28 chickens but there was 136 legs when there was supposed to be 134 so i knew that there should be less dogs and more chickens so i tried 19 dogs and 29 chickens which also equals to 48 heads ( because 19 + 29= 48) and 19 x 4 = 76 and 29 x 2 = 58 and 76 + 58 = 134, which means there are 134 legs and 48 heads which is correct
ReplyDelete-RACHEL
To find the answer to this POTW I knew that since there were 48 heads there had to be 48 animals. To answer this question I used guess and check. I started with 24 dogs and 24 chickens and I found that there would be 144 legs which is too much so I used more chickens since they have less legs. So I tried 20 dogs and 28 chickens and I got 136 which is 2 legs off of what I need to find so I used 19 dogs and 29 chickens and I got 134 legs which is what I needed to find. So the answer is Mr.Milette has 19 dogs and 209 chickens on his farm.
ReplyDeleteSince there are 48 heads in total I know that means that there are 48 animals as a dog and a chicken both only have 1 head. I also know that there are 144 legs meaning that there will be 48 animals with 144 legs. Since I know chickens have 2 legs and dogs have 4 I decided to use guess and check. I started out with 30 chickens and 18 dogs. I did 30 x 2 cause there are 2 legs per chicken to get 60 and 18 by 4 to get 72. I added that up to 132. Since that number is 2 off I know that I have to add one more dog to increase the amount of legs. So then I did 29 chickens and 19 dogs and got 134 total which is the answer.
ReplyDelete(Ps all the times I wrought 144 I meant 134 I'm typing on a iPad and I can't change it sorry)
That's OK. Thanks for the heads up William.
DeleteThe data which the question gave me was 134 legs and 48 heads. Since I know both of the animals in this problem have two heads, there has to be only 48 animals. To figure out how many chickens and dogs you can use algebra
ReplyDeleteY= chicken X= dogs
y+x=48 (how many animals)
2a+4b=134 (amount of legs and heads)
Now I need to get rid of either 2a or 4b so I need to multiply the number which is smaller
Now I have 2a+2b=96
Since now we have two equations:
2a+4b=134
2a+4b=96
If we subtract those equations you get 38, since we multiplied the equations we have divide it and we get 19. There are 19 dogs and to get the amount of chickens you subtract the number dogs to animals and you get 39 chickens.
On Mr.Milette's there are 19 dogs and 29 chickens
What I did was a simple guess and check method. I halved 48 to get 24 each and then multiplied by two or four to get 48 and 96 and then I added them to get a 144. Since it was 144 I knew I was close so I decided to add one head to the chickens and take away a head from the dogs. So it went on from 24:24 to 25:23 to 26:22 and so on to 29:19. By the time I got to 29:19 it was 58 chicken legs and 76 dog legs. Then I added them up to get 134. So in total Mr.Milette had 29 chickens and 19 dogs
ReplyDeleteTo answer the question I used guess and check. First I looked at how many heads they where which was 48 and I had to find out the number of each animal there was, so I guessed 20 dogs and 28 chickens to start with, and I had to multiply them by the number of legs they had so 20 by 4 and 28 by 2 and got 80 and 56 then I added them to get the total of legs which was 136 which was over by 2 which means I need to take away one dog and add one chicken which made my new numbers 19 and 29 and when I multiplied them by the number of legs they each had i got 134. So he had 19 Dogs and 29 Chickens on his farm.
ReplyDeleteOn Mr. Milette's farm, there are 48 animals in total: 29 chickens and 19 dogs. I found this (with some help from my mom) by using algebra (I'm pretty sure this is algebra) and variables.
ReplyDeleteChickens = x Dogs = y
x+y=48 animals in total
Each chicken has 1 head and 2 legs, and each dog has 1 head and 4 legs.
1x (heads per chicken)+1y (heads per dog)=48 (number of heads)
Which means that 48 (number of heads) -1y=1x
And if 2x (legs per chicken)+4y (legs per dog) =134 legs total
Then 2 X (48-1y) + 4y = 134 legs total
2 X (48-1y) = 96 96+4y =134 legs total
96 - 2y +4y =134
134 - 96 =2y 134 - 96 =38 2y=38 2y÷ 2 = 1y 38÷ 2 = 19 1y = 19 dogs
48 animals - 19 dogs = y
48 animals - 19 dogs = 29 chickens y = 29
Double Check:
29 chickens have 29 heads. 19 dogs have 19 heads. 29 heads + 19 heads = 48 heads in total (which is correct).
29 chickens have 58 legs (2 each). 19 dogs have 76 legs (4 each). 58 legs + 76 legs = 134 legs in total (which is also correct).
Since, Mr.Milette counted 134 legs and 48 heads, to see how many dogs and chickens he has, I will start by estimating that he has 10 dogs which add up to 40 legs, considering the amount of legs he counted, 10 dogs is to little. My next estimate is 20 dogs which is 80 legs, I'll stick with 20 dogs because 80 is close enough to 134. There are 56 legs and 28 heads left.
ReplyDelete20+28=48 is the right amount of heads but 56+80 =136 is not the right amount of legs. Now I just take away animals to take away legs, if I take away a chicken I'll end up with 134 legs but 47 heads, so I'll try taking away a dog, I'll have 132 legs and 47 heads, now if I add a chicken it'll be perfect because it has the right amount of heads and legs.
Finally Mr.Milette has 19 dogs and 29 chickens adding up to a total of 134 legs and 48 heads which is exactly what he counted.
If there were 134 legs and 48 heads of chickens and dogs, you would get 19 dogs and 29 chickens.
ReplyDeleteI decided to just see how many legs there would be if there were only chickens. Since there were 48 heads and a chicken has two legs, I had to multiply 48 (amount of heads) by 2 (number of legs a chicken has). Getting me the product of 96 chicken legs. Obviously this wasn't enough because it doesn't include any dogs or equal to 134 legs. So I subtracted 96 from 134 to see how many more legs I was short of, and replace the amount with dogs I was left with 38 legs which meant I had to subtract that amount from the amount of chicken legs giving me the true number of chicken legs, 56 chicken legs. I still had 38 legs left so I had to divide it by 2 to see how many heads this would be, giving me 19 heads, 19 dog heads. Since 1 dog has 4 legs I had to multiple 19 by 4 to get the amount of legs for dogs getting me 76 dog legs.
Add 76 and 56 and you get 134 meaning I got the amount of legs per animal right. Add 19 and 29 and you get 48 heads, meaning I got the amount of heads per animal right.
This means that Mr. Milette has 29 chickens and 19 dogs
Great work everyone! I was thoroughly impressed by how many students used my feedback from the previous POTW (especially by explaining their work in more detail).
ReplyDeleteNow...my feedback is to remind you to try to solve problems in a variety of ways AND to always use COPs when posting responses (especially grammar, spelling, and punctuation revisions).
That being said, the math portions of your responses were great. And yes, the correct answer is that I do have a total of 29 chickens and 19 dogs. (In the word problem, not in real life).
d = dogs c = chickens
ReplyDeleteThere is 134 legs in total making : 4d+2c=134
There is 48 heads in total making: d+c=48
To get the chickens' legs are 48-d or the dogs' legs are 48-c, but I want to find the dogs first.
so, 4d+2 (2 is the chickens which is now replaced by 48-d)
=4d+96-2d (96 is 48 heads multiplied by 2 chicken feet)
=2d+96=134
2d=134-96=38
now we know that 38 is the amount dog feet
38 divided by 2 equals to 19 dogs leaving 29 chickens
I'm using algebra to solve the problem. I chose d as a variable for the amount of dogs in total and c as the amount of chickens in mr.milette's farm
ReplyDeleteI know that dogs and chickens are equivalent to 48 heads because they are the only animals talked about in all of the given information.
D + C = 48
C= 48 - D
I know the number of legs is 134 because it is also stated that Mr.Milette counted 134 legs.
4 x D + 2 x C = 134
4 x D + 2 (48 - D) = 134
4 x D + 96 - 2 D = 134
2d + 96 =134
2d = 34 -96
2s = 38
d = 19 dogs
D+C=48
C= 48- D
C= 48 -19
C = 29
I verified my results via 29 + 19 = 48 which is chicken + dog = 48, which I established first.