It's interesting how many of the 6 different Work Habits and Learning Skills come-out in the Weekly POTW...Responsibility, Organization, Initiative, Self-Regulation, etc.
POTW #7 Solution:
POTW #8 Solution:
POTW #9 Question Grade 8
POTW #9 Question Grade 7
Grade 7 POTW
ReplyDeleteFirst, I'm going to list the facts.
- 3 trees
- 4 blue jays, 2 orioles
- 15 blue-jays, 14 orioles
- Largest number of birds per tree
First, I'm going to calculate the mandatory birds.
4 x 3 = 12 (blue-jays)
2 x 3 = 6 (orioles)
2 + 4 = 6. 6 Birds/tree so far.
Now I'm going to find the leftovers.
15 - 12 = 3
14 - 6 = 8
All three blue jays can go on one tree.
3 + 6 = 9
There are 7 available spots for orioles.
7 + 2 = 9
The last oriole can go on other trees. Now for the total.
9 + 9 = 18
There can be a maximum of 18 trees in a tree.
Re-Do
DeleteI was looking at other comments and realized no one had the same answer as me. I re-looked it and when I was adding the blue-jays to one tree, I added 3 to 6?
I probably used 6 due to all the 6s.
3 + 4 = 7.
Now, there can be + 5 orioles to the tree.
5 + 2 = 7.
Can the orioles fit into the other trees?
Let's see.
8 - 5 = 3.
2 + 2 and 2 + 1. It can work for other trees.
7 + 7 = 14.
A maximum of 14 birds can go on one tree.
Also, I saw 13 birds a lot in the answers. In fact, it was the mode of answers. Can someone explain (as in how they got 13, not the mode part)?
This comment has been removed by the author.
DeleteGrade 8 Potw
ReplyDeleteFirst, I’ll list the facts.
- 4 teams
- 4 teams play against other 3 teams 4 times
- Each team earned 3 points for a win
- Each team earned 1 point for a tie
- Each team earned 0 points for a loss
- Lions earned 22 points
- Tigers earned 19 points
- Mounties earned 14 points
- Royals earned 12 points
I’m just going to do the math now through trial and error. By the looks of the question, there is only one answer so I can just do one by one. However, this isn’t completely trial and error because I know that I have to make enough points for 12 games per team (4 x 3) and make it equal to the total points. Also, the number of wins in total must equal the number of losses in total. Ties must be an even number. Even with all of this, I still need to double check.
Trial and Error (ish):
Lion: 3, 3, 3, 3, 3, 3, 3, 1, 0, 0, 0, 0
Tigers: 3, 3, 3, 3, 3, 1, 1, 1, 1, 0, 0, 0
Mounties: 3, 3, 3, 3, 1, 1, 0, 0, 0, 0, 0, 0
Royals: 3, 3, 3, 1, 1, 1, 0, 0, 0, 0, 0, 0
The truth is. This was the right answer. I checked and it’s right (unless I messed up). However, my trial and error was trial and error, but I changed the numbers every once and awhile to try to make it more right. My work isn’t really shown too much so sorry, but the final answer is there and I explained what I was doing. Anyways, I still need to answer the question.
Wins: 19
Ties: 5 (games that are tied come twice)
Grade 8 POTW:
ReplyDeleteThis question is easy when you know some specific facts.
First of all, I will figure out how many games that they played in total. Since each team played against another team 4 times, first of all, I have 3+2+1 = 6 combinations of 4 games played, since team 1 plays against team 2, 3, 4, team 2 plays against team 3, 4, and team 3 plays against 4. Then, I can multiply this combination amount by 4, which would give me 24. This means that 24 games were played in total between the 4 teams.
Now comes the hard part. I will first need to find the total amount of points that could have been earned. Since I have 24 games, I can multiply it by 3 since I know that one side must have won. This gives me 24*3 = 72 points in total.
The number of points won was 22+19+14+12 = 67 points. The difference between this is 72-67 = 5. Since both sides are given 1 point in a tie, and in a win one side is given 3, the number of ties is 5/(3-2) = 5/1 = 5 tied games. This means that the number of games won is 24-5 = 19 games won.
If we double check this, it would be 19*3+5*2 = 57+10 = 67 points in total. This is what the question gave us, so my answer is correct.
-Alan
Question for the Grade 7 POTW:
ReplyDeleteWhat is the maximum birds that can be on one branch? Because if there is no limit then it means that I am able to put all birds on one branch.
But there has to be a least 4 Blue jays and 2 orioles in each tree
Deleteeach tree has to have at least 4 blue jays and 2 orioles, you putting all the birds on one tree makes no birds on the other two trees which doesn't follow the question
DeleteYeah. I agree. There would most likely not be a trick question for POTW.
DeleteGrade 7 POTW:
ReplyDeleteInfo:
- 15 Blue Jays
- 14 Orioles
- At least 4 Blue Jays on each tree and 2 Orioles on each tree
- 3 trees
- Always Blue Jays>Orioles.
Since there have to be at least 4 Blue Jays and 2 Orioles, I have to first put the 6 birds on each tree.
Tree 1= 4 Blue Jays, 2 Orioles
Tree 2= 4 Blue Jays, 2 Orioles
Tree 3= 4 Blue Jays, 2 Orioles
There are 3 Blue Jays and 8 Orioles still not on a tree. So I can add a couple of the Orioles to each tree so it is a 1:1 ratio.
Tree 1= 4 Blue Jays, 4 Orioles
Tree 1= 4 Blue Jays, 4 Orioles
Tree 1= 4 Blue Jays, 4 Orioles
Now, there are 3 Blue Jays and 2 Orioles still not on a tree. I can add all five of them to Tree 3 because there are more Blue Jays then Orioles left and I can add them all onto one tree. Instead of spreading them out within the three trees, I'm clumping them together to find the max number of birds per tree.
Tree 1= 4 Blue Jays, 4 Orioles
Tree 1= 4 Blue Jays, 4 Orioles
Tree 1= 7 Blue Jays, 6 Orioles
The max number of birds is 13 birds.
:)
Oh woops. When it says:
DeleteTree 1= 4 Blue Jays, 4 Orioles
Tree 1= 4 Blue Jays, 4 Orioles
Tree 1= 4 Blue Jays, 4 Orioles
I meant to say
Tree 1= 4 Blue Jays, 4 Orioles
Tree 2= 4 Blue Jays, 4 Orioles
Tree 3= 4 Blue Jays, 4 Orioles
I was copying and pasting from the first set and I forgot to change it. Oops.
First of all, I know that each tree must have at least 4 blue jays and 2 orioles. That means that I will give each tree 4 jays and 2 orioles.
ReplyDelete#1: 4J, 2o
#2: 4J, 2o
#3: 4J, 2o
Now, there are 3 blue jays left, because 4 x 3 = 12. 15 - 12 = 3. There are also 8 orioles left, because 2 x 3 = 6, and 14 - 6 = 8.
Now, since I want the largest possible numbers of birds in one tree, I can give all three blue jays onto one tree.
#1: 7J, 2o
#2: 4J, 2o
#3: 4J, 2o
Since there has to be more blue jays than orioles, I can also add 5 orioles to that tree (#1).
#1: 7J, 7o
#2: 4J, 2o
#3: 4J, 2o
There are now still 3 orioles left. I can add 1 to tree #2 and 2 to tree #3.
#1: 7J, 6o
#2: 4J, 3o
#3: 4J, 4o
The largest number of birds in a tree is 13. 7 of them are blue jays, and 6 of them are orioles.
grade 8 potw:
ReplyDeleteIMPORTANT/RELEVANT INFORMATION
-4 teams (L, T, M, R)
-3 points per win, 1 point per tie, 0 for a loss
-L = 22
T = 19
M = 14
R = 12
DEDUCTIONS:
first we can assume that there is 24 games played in total. Each team plays 12 games. the total amount of points each team can receive is 36. based off that I just used guess and check to figure out which combination of wins/loses/ties would make sense.
WINS : 19 games
TIES : 5 games
so the obvious thing to do first is put two trees with the minimum requirement which makes the last tree have the most amount of birds left. the only problem is that there can't be more orioles than blue jays and making the first two trees with 4 blue jays and 2 orioles makes the last tree have 10 orioles and 7 blue jays which wouldn't work.
ReplyDeleteso instead I made the biggest possible number of blue jays and orioles on one tree while having the other two trees meet the requirement. I know that the maximum number of blue jays on one tree is 7 since you need at least 4 blue jays on the other trees(15 - 8 = 7). So that means the biggest possible number of birds for orioles is 7 since orioles isn't bigger and the other two trees have 4 blue jays and 4 orioles; it all works out.
now all I have to do is add the amounts of birds on the biggest tree( 7 + 7 ) which gives me my final answer of 14.
Grade 7 POTW
ReplyDeleteFirst, I'm going to review the information I already know from the text.
- There are 15 Blue Jays (BJ)
- There are 14 Orioles (O)
- There must be at least 4 Blue Jays and 2 Orioles on each tree
- There are 3 trees
- There must be more Blue Jays on each tree than Orioles
First, I'm going to start with expectations, which are at least 4 Blue Jays and 2 Orioles on each tree. I will name the trees; the first tree being named T1, the second tree being named T2, and the third tree being T3.
T1: 4 BJ, 2 O
T2: 4 BJ, 2 O
T3: 4 BJ, 2 O
Now, because each tree has 4 Blue Jays and 2 Orioles, I know how many are left. There are 3 Blue Jays left. (3 x 4 = 12, 15 - 12 = 3) There are also 8 Orioles left. (2 x 3 = 6, 14 - 6 = 8)
Now, since I want the largest possible number of birds in one tree, I'll just give all the Blue Jays to one tree, as this won't matter when it comes to the regulations set. (There are still more Blue Jays - obviously - than Orioles, so we're still OK.)
T1: 7 BJ, 2 O
T2: 4 BJ, 2 O
T3: 4 BJ, 2 O
Now, to continue, I have eight Orioles left. The thing is, I'd rather not spread them out within the trees, and instead I want to all add them to one tree, to find the maximum number of birds possible on one tree. As T1 already has 7 Blue Jays on it, it seems reasonable to add most of the Orioles to T1, but only 6, because there must be more Blue Jays than Orioles.
T1: 7 BJ, 6 O
Now that T1 is taken care of, I'm going to start spreading out the remaining birds (3 Orioles) to the other trees.
T2: 4 BJ, 3 O
T3: 4 BJ, 4 O
Since now all the birds are on trees, as requested, and all requirements are fulfilled, I can find the maximum possible number of birds, which I can find from T1, as that is the tree I crammed the most birds onto. On T1, there are 7 Blue Jays, and 6 Oriels, which results in 13 birds altogether.
Therefore, the highest possible number of birds on one tree is 13.
Grade 7 POTW
ReplyDeleteTree #1: 4 Blue Jays, 4 Orioles
Tree #2: 4 Blue Jays, 4 Orioles
Tree #3: 7 Blue Jays, 6 Orioles
The largest amount of birds that can be on one tree is 7 Blue Jays and 6 Orioles. When I tried to put the lowest amount of birds on a tree which is 4 Blue Jays and 2 Orioles, that left me with 7 Blue Jays and 8 Orioles on one tree. That couldn’t work because there were more Orioles than Blue Jays on one tree. Instead, I decided to put the same amount of Blue Jays and Orioles so that there are 7 Blue Jays and 6 Orioles.
Take an oriole from another tree and add it to 6 orioles. All rules are followed so 13 isn't the maximum. 14 is.
Delete
ReplyDelete15 - 4 -4=7
14-3-4=7
The most birds that can fit on a tree is 14. 7 of each bird. I got this by first subtracting at least how many of each bird should be on the other two trees. I had to subtract at least 8 blue jays and 4 Orioles. Once I did that, I found out the most Blue Jays I can have on a tree is 7, as 15-8=7. When I subtracted 4 from 14 however, I had more Orioles than Blue Jays. That means that I have to subtract down to the amount where there are more or equal to the amount of Blue Jays. I then subtracted 3 more Orioles from 10 to get 7. That makes it so that there are less or equal to the amount of Blue Jays.
Grade 8 POTW
ReplyDelete19 Games ended in win, and 5 ended in a tie. I did my work in my math notebook
Maxwell's Grade 8 POTW:
ReplyDelete19 games were wins, and five games ended in tie at the end of the softball games. I did my work in my math notebook.
POTW #9
ReplyDeleteWhat I Know:
- 15 Blue Jays and 15 Orioles
- 3 trees
- Each tree has at least 4 Blue Jays and 2 Orioles
- Cannot have more Orioles than Blue Jays in one tree
What I Need to Find:
What is the largest amount of birds that can be in one tree.
Solving:
First, I know that 4 Blue Jays per tree times 3 trees is already 12 Blue Jays gone. 2 Orioles times 3 trees is 6 gone. There are now 3 Blue Jays and 8 Orioles left. To make the 4 Blue Jays and 2 Orioles equal in one of the trees, I added 2 Orioles since I can't add more.
Now there are 4 of each bird in a tree and 3 Blue Jays left with 6 Orioles left.
I added all of the Blue Jays to that tree. Now there are 7 Blue Jays and 4 Orioles. To make it equal, I can only add 3 more Orioles. In total there are 14 birds in one tree since 7 + 7 = 14.
The maximum number of birds that can be on one tree is 14.
There are 19 wins and 5 ties, based on looking at the team scores and using process of elimination.
ReplyDeleteGrade 8 POTW:
ReplyDeleteThis question had a lot of trial and error involved, which is what I used. I first listed the key info..
- Four teams
- 3 points for a win
- 1 point for a tie
- 0 points for a loss
Then, I looked at the scores that were given to the four teams...
Lions= 22
Tigers= 19
Mounties= 14
Royals= 12
After a LOT of trial and error, I came to this conclusion...
19 games were wins
5 games were ties
0 games were losses
Therefore, that is my answer!
POTW #9:
ReplyDeleteWe know that in total, 29 birds want to rest on three trees.
First we can put the minimum amount of birds on the trees to find out how much are left, so we can put the rest of the birds left on one of each trees.
Tree #1: 4 Blue, 3 Orio
Tree #2: 4 Blue, 3 Orio
Tree #3: 4 Blue, 3 Orio
There are:
3 Blue (4*3= 12, 15 - 12 = 3) and 8 Orio ( 2*3 = 6, 14 - 6 = 8)
Left not yet on the trees
We'll add all the blue to one of the trees because, we need more blue than orio and it won't break any rules whatsoever. But for the Orio, they need to be less than the Blue Jays.
Tree #1: 7 Blue, 3 Orio
Tree #2: 4 Blue, 3 Orio
Tree #3: 4 Blue, 3 Orio
Now, with the remaining Orio, we'll add as much orio as we can to Tree #1, enough to have more Blue but as much to make the total of the amount of birds on the tree higher.
Tree #1: 7 Blue, 6 Orio
Tree #2: 4 Blue, 2 Orio
Tree #3: 4 Blue, 2 Orio
Now, we have five remaining orios. We'll evenly place them on the other trees, not breaking the rules.
Tree #1: 7 Blue, 6 Orio
Tree #2: 4 Blue, 4 Orio
Tree #3: 4 Blue, 3 Orio
The most amount of birds on one tree is 13.
Grade 7 POTW:
ReplyDeleteThere are 15 Orioles and 15 Blue Jays distributed onto 3 different trees. Each of those 3 trees has 4 or more Blue Jays and 2 Orioles. But, you will not find more Orioles on a tree and Blue Jays.
I have to find the greatest number of birds that can fit into a tree.
There are at least 4 Blue Jays in each tree. So 4 Blue Jays in each of the 3 trees will sum up to 12 Blue Jays.
There are at least 2 Orioles in each tree. So 2 Orioles in each of the 3 trees will sum up to 6 Orioles.
The question stated that there are 15 of each species spread out though, so there are still 3 Blue Jays and 8 Orioles left.
I can give one tree another 2 Orioles, leaving that tree with 4 Blue Jays and 4 Orioles.
Tree1= 4 BJ and 2 OR
Tree2= 4 BJ and 2 OR
Tree3= 4 BJ and 4 OR
3 Blue Jays and 7 Orioles left.
I can give the rest of the Blue Jays to Tree 3 as well, as I am trying to find the largest amount of birds that can be on a tree.
Tree1= 4 BJ and 2 OR
Tree2= 4 BJ and 2 OR
Tree3= 7 BJ and 4 OR
7 Orioles left.
I can give Tree3 three more Orioles, as that is the maximum amount I can put on the tree without the rioles outweighing the Blue Jays.
Tree1= 4 BJ and 2 OR
Tree2= 4 BJ and 2 OR
Tree3= 7 BJ and 7 OR
4 Orioles left.
Then I can just distribute the rest of the Orioles on the other 2 trees to make them even.
Tree1= 4 BJ and 4 OR
Tree2= 4 BJ and 4 OR
Tree3= 7 BJ and 7 OR
Tree3, the tree with the most birds on it, has 7 Blue Jays and 7 Orioles on it. This totals up to 14 birds on the tree.
The greatest amount of birds on a tree can be 14.
First of all, you have to find the minimum amount of birds on 2 of the tress combined. If the minimum amount of blue jays on one tree is 4, and there has to be less or an equal amount of orioles and blue jays, there should be 4 blue jays and 2 orioles. If 4*2=8 and 2*2=4, than there are 7 blue jays and 10 orioles. if there has to be a lower or equal amount of orioles to blue jays, to get the highest amount you would have 7 of each bird in on tree and in one other tree you would add 2 orioles to make 4:4 and the final would be 4:3
ReplyDeleteThe final bird ratios per tree are 7:7, 4:4, and 4:3.
The most amount of trees you can have in a tree is 14 birds or 7 orioles and 7 blue jays
Grade 8 POTW -
ReplyDeleteThere were 24 games. 19 ended in wins/losses, 5 ended in ties.
Work in notebook
Grade 8 POTW:
ReplyDeleteThere are a total of 24 games (4!). If there are no ties and no losts, 72 points will be given. But 67 points were given (add all the points together). 72-67=5 5 ties.
24-5=19.
19 games ended in a win and 5 games ended in a tie.
19 wins and 5 games.
ReplyDeleteGrade 8 Potw:
ReplyDelete19 wins
5 ties
Work done by hand
Here is how I went about solving the Grade 8 POTW:
ReplyDeleteSince each team played every other team 4 times, I did "4!" and got 24. Now we know that 24 games took place in the league. Now the next part of the question states that 3 points will be given for a win. That means if there are no ties or losses, the team will be awarded 72 points. But in this case, the total amount of points received are 67 (22 + 19 + 14 + 12). Now we find the difference between the two (72-67=5) and divide by one, since that is the amount of points awarded if the game results in a tie (5/1 = 5). Now we subtract 5 from 24 to get 19.
Therefore, 19 games ended in a win and 5 games ended in a tie.
Grade 8 POTW:
ReplyDeleteSince:
- The total amount of points calculated is 67 (22+19+14+12)
- 24 games in total
- 24 * 3 = 72 which is the total possible amount of points if every team won every game
We can do:
72 - 67 = 5
to figure out how many ties there were.
Now that we know how many tie games there were, we can subtract that from the total number of games.
24 - 5 = 19
Therefore, 19 games were won and 5 games were tied.