POTW #4 Grade 7 Solution:
POTW#4 Grade 8 Solution:
POTW #5 Grade 7 Question:
POTW #5 Grade 8 Question:
Thursday, October 6, 2016
POTW #5 - New Friday Due Date
Hello again, although a new due date of each Friday has been announced, please don't leave your work until the last minute. Many students who struggled with the past week's POTW were asking questions of their peers in class and getting some tutorial support from those who had already finished. It was great to see! The answers to the previous POTW #4 and the newest POTW #5 are posted below:
POTW #4 Grade 7 Solution:
POTW#4 Grade 8 Solution:
POTW #5 Grade 7 Question:
POTW #5 Grade 8 Question:
POTW #4 Grade 7 Solution:
POTW#4 Grade 8 Solution:
POTW #5 Grade 7 Question:
POTW #5 Grade 8 Question:
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There would be 216 animals inside the whole farm. How I got this answer is because I just did simple trial and error. So first, since pigs share the trough with 8 pigs, we would do the pigs first. So first I did 24 pigs, which would make the amount of troughs to 3. Then since cows share it 3 to one, then it would be 8 troughs in order to make it the same amount. And same for the Horses as they would be at 12 troughs. But this isnt enough troughs as there should be 69. so we continue on to 48 as it is a sure chance that it will be able to have a divisible number. But since 69 is an odd number, and there will be no odd numbers which means it is impossible to get 69. Now we move on to 72. This would work because many the amount of pigs would need 9 troughs, cows would need 24 troughs, and horses would need 36 troughs. This would add up to 69.
ReplyDeleteTherefore, the answer is 216 animals.
First I found the LCM of the farm animals, and it was 24. I needed to calculate the LCM because there must be the same amount of animals on in that farm. That means if there are 24 pigs, then there would be 3 troughs. If there are 24 cows, there would be 8 troughs and if there are 24 horses, there would be 12 troughs. If you add all the troughs up, then that would equal 23 troughs, but that isn't enough, and the question is asking for 69 troughs. So now lets make the amount of each animal to 48. That would mean if there are 48 pigs, there are 6 troughs, if there are 24 cows, then there would be 16 troughs, and if there 48 horses, you will need 24 troughs. The added troughs would equal to 46, but that still isn't enough. Keeping the same amount of animals, we move up the amount of animals to 72, which is the next common factor. That means if there are 72 pigs, there are 9 troughs. If there are 72 cows, there would be 24 troughs and 72 horses would need 36 troughs. If you add all the troughs up, then that would equal to 69 troughs. Now we know the amount of animals, we have to multiply 72 by 3 because there are 3 types of animals. That gives us 216.
ReplyDeleteThere are 216 animals in his farm.
To solve this question, I would need to find a common multiple of 3, 8, and 2 that will result in the answer.
ReplyDeleteMultiples of 3, 8 and 2
2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72
3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 54, 57, 60, 63, 66, 69, 72
8: 8, 16, 24, 32, 40, 48, 56, 64, 72,
The first the common divisors of 2,3 and 8 are 24, 48 and 72.
24: There would be 12 horse troughs, 8 cow troughs, and 3 pig troughs. This adds up to 23 troughs which is way under the 69 troughs Old MacDonald has. He needs 3 times more troughs so I try 72 (common multiple of 2,3 and 8 and 3 times greater than 24).
72: There would be 36 horse troughs, 24 cow troughs and 9 pig troughs. This adds up to 69 troughs which is exactly the amount of troughs Old MacDonald has. There for there are 72 of each animal.
72 + 72 + 72 or 72*3 = 216 animals.
Old MacDonald has 216 animals in total.
Here is how I did the Grade 8 POTW:
ReplyDeleteBecause the father will be four times as old as the son, the total amount of parts there will be is 4+1, since the son takes up 1 part and the father would take up 4 to be 4 times as old as the son.
This means that after the years pass the sum of the son and father's years would end up to be a multiple of five, with a ones digit of 5 or 0.
But, I know that 0 is impossible since the sum is an odd number. Each time a year passes, the sum doesn't increase by 1, but 2 because both the father and son get older by a year. Since I started off with an odd number, I would end with an odd number.
Since the sum is 43 right now, there can be 45, 55, 65, 75, and any other number that ends with 5.
So, there are many different answers to this question, and I will show 45 and 55.
45/5 = 9, so the age of the son would be 9.
9*4 = 36, so the age of the father would be 36.
To double check, I subtract 1 from both ages and add them together, seeing if I will get 43. 9-1+36-1=43. The answer is possible.
For 55, the age of the son would be 11, and the father 44. From double checking, I subtract 6 from each year since 55-43 is 12, and divided by 2 is 6. I divide by 2 because each year would have 2 added to the sum. 11-6+44-6=43, so the answer is correct as well.
So, there are a number of answers, but two of them are 9 and 36, and 11, and 44.
-Alan
I got 216 animals. I got this by first finding the multiples of 2,3 and 8 the amount of animals in 1 trough (up to 30)
ReplyDelete2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
8: 8, 16, 24
The only number overlapping all 3 is 24. So if we have 24 of each animal we can calculate the amount of troughs in total. 24/2=12 24/3=8 24/8=3 12+8+3=23. If we do 69/23 we get 3. From there we know 24 of each animal is 1/3 of the total. So if you do 24+24+24*3 you would get the total amount which is 216.
Grade 7 POTW:
ReplyDeleteFirst we need to compare the multiples of numbers to get a common one.
2:2,4,6,8,10,12,14,16.18.20.22.24.26.28.30.32...
3:3,9,12,15,18,21,24,27,30,33,36...
8:8,16,24,32,40...
In this we can see the only number that's a multiple of all three is 24. The total amount of troughs if there are 24 animals of each is 23 troughs. Obviously that isn't enough to reach 69 troughs but if you multiply that by 3 you will reach the number needed (69). Now let's calculate amount of animals. 24+24+24x3=216 animals on his farm.
Grade 7 POTW:
ReplyDeleteFirst I had to figure out the LCM of all the numbers. The LCM would be 24. The total of troughs needed would only be 23, which would not be enough. Now, we keep going with the multiples of 24, and I reach the value of 72, which shows that 69 troughs are needed. Then, I multiply it by 3 to get 216 animals.
Alan
Grade 7 POTW:
ReplyDeleteFirst we start off with figuring out the LCM (Least Common Multiple):
2: 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40...
3:3,6,9,12,15,18,21,24,27,30,33,36,39,42...
8:8,16,24,32,40,48...
As you can see, the LCM is 24, which mean that there could be 24 of each animal, but that would mean they would only have 23 troughs. So we keep on finding multiples of 24, which leads to 72. That would mean that there are 36 horse troughs, 24 cow troughs, and 9 pig troughs. So we multiply 72 by 3, giving me 216.
Therefore, Old McDonald has 216 animals on his farm.
To solve this question I used algebra and plugging in numbers. Here is my work:
ReplyDeleteX=Son's age
(43-x)+n=4(x+n)
43-x+n=4x+4n
43+n=5x+4n
5X=43-3n
After the last equation I began plugging numbers in that could be n. After plugging in different numbers I got the final ages of:
8,5,2 for the kid and 35,38,41 for the father.
I did the POTW for grade 8 in m math notebook and got the three answers of 41_2, 38_5 and 35_8
ReplyDeleteI got 41 and 2, 38 and 5, and 35 and 8. I solved it before but it didn't save, so i just wrote my answers down.
ReplyDeleteI did POTW hard copy, and I got the Kid ages of 2, 5, and 8 and the dad's age 41, 38, and 35 respectively.
ReplyDeleteGreat use of the word "respectively"
DeleteTo solve the grade seven potw, I first had to find a combination of animals and troughs that had an equal amount of animals. To do this, I had to find the LCM (Lowest Common Multiple).
ReplyDelete2: 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
3: 3 6 9 12 15 18 21 24 27 30 33
8: 8 16 24 32
The LCM is 24. So, 24 pigs, 24 horses, and 24 cows, and 12 horse troughs, 9 cow troughs, and 3 pig troughs, and 23 in total.
23*3 is 69, meaning that there would be 216 farm animals on Old Mcdonald's farm because 24*3 is 216.
Edit: I made a typo on my answer, instead of 24*3=216, it was supposed to be 72*3.
ReplyDeleteHere is how I went about solving the Grade 7 POTW:
ReplyDeleteSo we start off by finding the LCM (lowest common multiple) of the three animals. We know what numbers to use because it states in the question that there can be two horses to a trough, three cows and eight pigs. After doing some calculations, you find that the LCM of 2, 3 and 8 is 24. That means if there are 24 of each kind of animal, there would be 12 horse troughs (24/2), 8 cow troughs (24/3) and 3 pig troughs (24/8). This would bring us to a total of 23 troughs (12 + 8 + 3). But since there are 69 troughs in total, we would need three times more of each animal. This would mean that we have to multiply the total of each animal (24) by 3. This would give us a product of 72, and since there are three types of animals, we do, 72 + 72 + 72 and that equals 216. Therefore, Old MacDonald had 216 animals on his farm.
To solve this, we would need to find a common multiple of 2, 3 and 8.
ReplyDelete2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72
3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72
8: 8, 16, 24, 32, 40, 48, 56, 64, 72
24 is the first common multiple from the three groups but if use that, there would only be 23 throughs. If we use the other common multiple 48, it would still not be enough as there would only be 46 throughs in total so we use the next common multiple 72. Now for 72, it would actually equal to 69 throughs so if we multiply 72 by 3 (3 different animals), we would get a final answer of 216. Old MacDonald has 216 total animals on his farm.
First of all, horses have more troughs than cows which have more troughs than pigs. I will do some trial and error. First of all, the middle, 23 troughs for each animal. 23 x 2 = 46, 23 x 3 = 69 and 23 x 8 = 184. So there are way to many pigs and i removed 10 pig troughs and put 7 into the horse troughs and 3 into the cow troughs. 13 x 8 = 104 for pigs, 30 x 2 = 60 for horses and 26 x 3 = 78 for cows. There were still to many pigs so then i removed 3 more pig troughs and put 2 into the horse troughs and 1 into the cow troughs. That gave me 80 pigs, 64 horses and 81 cows. Now i have to many cows and pigs so i took one from each and put it into the horses. That left me with 72 pigs, 68 horses and 78 cows. So then I took 2 cows and put them into the horses which gave me 72 pigs, 70 horses and 72 cows ad then i added to more to the horses because i forgot to add an extra two in my calculations somewhere which gave me 72 of each animal. 72 + 72 + 72 = 216 animals total.
ReplyDeleteGrade 7 POTW #5:
ReplyDeleteTo figure out the POTW, I listed some of the multiples for each number or animals for one trough:
2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74
3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80
8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80
Three numbers are the same, 24, 48, and 72.
24: For this number there would be 12 horse troughs, 8 cow troughs and 3 pig troughs, but that only adds up to 23. Which is not enough to add up to 69.
48: There will be 24 horse troughs, 16 cow troughs and 6 pig troughs. That adds up to a total of 46. Still not 69.
72: Total of 36 horse troughs, 24 cow troughs and 9 pig troughs. That is a total of 69.
There are 72 of each animal. 72 x 3 = 216
Therefore, there are 216 animals in Old Macdonald's farm.
Lowest common multiple = 24
ReplyDelete24 of each = 23 troughs
24 x 3 = 72
72 = 69 troughs
72 x 3 = 216
The answer is 216
To solve this question, you have to find the LCM that fits the criteria.
ReplyDeleteMultiples of 3, 8 and 2
2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72
3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 54, 57, 60, 63, 66, 69, 72
8: 8, 16, 24, 32, 40, 48, 56, 64, 72
For the number 24(each animal), there would be 8 cow troughs, 3 pig troughs and 12 horse troughs, but if you add them all together that only equals 23 which is way under 69, instead of looking at the next LCM, I concluded that if there were 24 cow troughs, 9 pig troughs and 36 horse troughs, that would equal to 69 which meets the criteria! And since 9 is 3 times more than 3, that would mean all I have to do is 24*3 which is 72. 72 of each animal so 72*3=216.
So there are 216 animals in total.
First i had to find out the LCM (least common multiple), which is 24. That means 24 pigs, horses and cows. Since its 8 pigs for 1 trough, and there are 24 pigs, all of the pigs would take up only 3 troughs. There are 24 horses and 2 horses go per trough. That means the horses would take up 12 troughs. Now we move onto cows. 3 cows per trough, and we have 24 cows which means that the cows would take up 8 troughs. Now if we add all of our troughs together we would end up with 23 troughs which isn't enough since MacDonald has 69. Now we can move onto the second multiple which is 48, but this one also doesn't work because all the troughs add up to 46 (We need 69). Now we move onto the 3rd multiple which is 72. This means the all 72 of the pigs would take up 9 troughs. The horses would take up 36 troughs, and the cows would take up 24 troughs. If we add up all the troughs we'd get exactly 69 which is what we need. But we need to find out how many animals are on the farm. Since there are the same amount of animals, we'd have to do 72(number of animals)x3(Different types of animals)=216. That means that there are 216 animals on MacDonald's farm.
ReplyDeleteSince Old MacDonald has the same amount of cows, horses and pigs, we have to figure out how many of each he has to answer the question. Two horses drink from one horse trough, three cows drink from one cow trough, and eight pigs drink from one pig trough. Since the number of each animal must be a number divisible by the number of said animal drinking from that animal's trough, we must find the LCM of 8,2 and 3, which is 24. That means that there would be 12 horse troughs, 8 cow troughs and 3 pig troughs, which is equal to 23 troughs.
ReplyDeleteHowever, since there are 69 troughs, there needs to be 3x the amount of troughs, and that means there needs to be 3x the amount of each animal as well. This means there would be 72 of each animal on the farm. Since there are 3 different types of animals on the farm, there are 216 animals on Old Macdonald's Farm.
I did my work in Hard copy and the answers I got were 8,5 and 2 for the son and 35,38,41 for the Father Respectively.
ReplyDelete