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We hope the winter weather is finally behind us and we can start to see that warm sun again! Keep up the POTW! Can you solve these week's questions algebraically?
Info: - 69 troughs - 2 horses per trough - 3 cows per trough - 8 pigs per trough - Same number of each animal
The first thing we need to do is find the LCM of 2,3 and 8 because the number of animals is the same. 2 - 2 3 - 3 8 - 2^3 ——————- 2^3 x 3 = 24
This means that there are 24 animals at the very least, t = 24 Neighers = t/2 = 24/2 = 12 Moo-ers = t/3 = 24/3 = 8 Oinkers = t/8 = 24/8 = 3 12 + 8 + 3 = 23
69 total troughs/23 minimum troughs = 3 times more animals that we need
24 x 3 = 72
72 x 3 types of animals = 216 total animals.
There are 216 total animals and now Old MacDonald is going broke because he has so many mouthes to feed.
Sorry, I forgot to add the Let statement (originally t was going to be troughs, but I realized that was wrong and that I had mistaken something). Let t be the minimum number of each animal.
Grade 7 POTW One way to solve this problem is guess and check. We know that 8 pigs=one trough, 3 cows=one trough and 2 horses=one trough. In order to ensure that the amount of animals we guess is divisible by 2, 3, and 8, we have to find their LCM. The LCM of 2,3, and 8 is 24. We can try 24. 24 pigs means 3 troughs, 24 cows means 8 troughs, and 24 horses means 12 troughs. All of these troughs equal to 23, which is 3 times smaller than 69, so we have to multiply each of the amounts of animals by 3 and add them up to find our answer. 24x3=72x3=216
There are 216 total animals on the farm. (I tried to solve algebraically using substitution and was not successful)
I missed the last one ;-; So first I tried to use guess and check, which didn't work for some reason. Then, I tried to solve it algebraically. So First I wrote out the amount for each group: Gr11: x(x-1)2/2 = x^2-x Gr12: y(y-1)4/2 = 2(y^2-y) = 2y^2-2y Both: 3xy Then, I use some algebra magic to find the answer; x^2-x+2y^2-2y+3xy=1392 x^2+y^2+2xy+xy+y^2-x-2y (x+y)^2+xy-(x+y)-y+y^2 30^2-30+(30-y)y-y+y^2 30^2-30+30y-y^2-y+y^2 900-30+30y-y-y^2+y^2 870+29y=1392 29y=1392-870 29y=522 y=18. So since y was the number of Gr 12 students, the number of Grade 11 students in the club was 12.
Grade 8 POTW There are 30 students. You must make every combination of pairs possible. A grade 11 pair is 2 hours, 12 pair is 4 hours, 11 and 12 is 3 hours. The total hours is 1392. To do this, we first need to give values to the number of grade 11/12 students. Grade 11 Students: x Grade 12 Students: 30 - x We can now make expressions for each possible pair. Grade 11 & 11 Pair: x(x - 1)/2 Grade 11 & 12 Pair: x(30 - x) Grade 12 & 12 Pair: (30 - x)(30 - x - 1)/2 = (30 - x)(29 - x)/2 We still aren’t done as we need to find the time in hours in total. Grade 11 & 11 Pair: (x(x - 1)/2)2 = x(x - 1) Grade 11 & 12 Pair: (x(30 - x))3 = 3x(30 - x) Grade 12 & 12 Pair: ((30 - x)(29 - x)/2)4 = 2(30 - x)(29 - x) Now we just have to put this into an equation using 1392 hours to find the value of x. x(x - 1) + 3x(30 - x) + 2(30 - x)(29 - x) = 1392 x^2 - x + 90x - 3x^2 + 2(870 - 30x - 29x + x^2) = 1392 x^2 - x + 90x - 3x^2 + 2(870 - 59x + x^2) = 1392 x^2 - x + 90x - 3x^2 + 1740 - 118x + 2x^2 = 1392 -29x + 1740 = 1392 -29x + 1740 - 1740 = 1392 - 1740 -29x = -348 -29x/-29 = -348/-29 x = 12 30 - 12 = 18 Therefore, there are 12 grade 11 students, and 18 grade 12 students.
Grade 7 POTW There are 69 troughs. 2 horses per trough, 3 cows per trough, 8 pigs per trough. There are the same number of each animal. First, we already know that there are an equal number of each animal. This can only happen if the number of each animal is a factor of 2, 3 or 8. The least common multiple (lcm) is 24 (8 x 3). Therefore, there are at least 24 animals of each type, or a multiple of that. If there are 24 animals each, we need to see how many troughs that would take: 24/2 + 24/3 + 24/8 = 12 + 8 + 3 = 23 troughs 69/23 = 3 Therefore, we need 3 times as many troughs or 3 times as many animals. 24 current animals x 3 animal types x 3 times more needed = 24 x 9 = 216 animals. Therefore, Old MacDonald has 216 animals on his farm.
So I began with finding the LCM of 2, 3, and 8. (Two horses per each trough, 3 cows per each trough, and 8 pigs per each trough). LCM: 24. 24 / 2 = 12 24 / 3 = 8 24 / 8 = 3 12 + 8 + 3 = 23 69 total troughs. 23 x 3 = 69. 23 is three times 69, so we multiply number of animals by 3, and add them all up. 24 x 3 = 72 72 x 3 = 216. Total number of animals on Old MacDonald's farm: 216.
Grade 8 POTW (Assuming that every grade 11 must play a duet with another grade 11 and a grade 12):
ReplyDeleteLet x be the number of grade 11 students and y be the number of grade 12 students.
Since there are a total of 30 students…
x = x
y = (30-x)
11-11 duets: x(x-1)/2
12-12 duets: (30-x)(30-x-1)/2 or (30-x)(29-x)/2
11-12 duets: x(30-x)
11-11 duets time: (x(x-1)/2)2
12-12 duets time: ((30-x)(30-x-1)/2 or (30-x)(29-x)/2)4
11-12 duets time: (x(30-x))3
Total time = 11-11 pairs + 12-11 pairs + 12-12 pairs
1392 = 2((x(x-1)/2) + 3(x(30-x)) + 4((30-x)(30-x-1)/2 or (30-x)(29-x)/2)
1392 = x^2 - x + 3(30x-x^2)+ 2(870 - 59x + x^2)
1392 = x^2 - x + 90x - 3x^2 +1740 - 118x + 2x^2
1392 = -29x +1740
-348 = -29x
29x = 348
x = 12
Therefore, there are 12 students in Grade 11, and 18 in Grade 12.
(I didn't use square brackets, which may have confused some things)
Grade 7 POTW:
ReplyDeleteInfo:
- 69 troughs
- 2 horses per trough
- 3 cows per trough
- 8 pigs per trough
- Same number of each animal
The first thing we need to do is find the LCM of 2,3 and 8 because the number of animals is the same.
2 - 2
3 - 3
8 - 2^3
——————-
2^3 x 3 = 24
This means that there are 24 animals at the very least, t = 24
Neighers = t/2 = 24/2 = 12
Moo-ers = t/3 = 24/3 = 8
Oinkers = t/8 = 24/8 = 3
12 + 8 + 3 = 23
69 total troughs/23 minimum troughs = 3 times more animals that we need
24 x 3 = 72
72 x 3 types of animals = 216 total animals.
There are 216 total animals and now Old MacDonald is going broke because he has so many mouthes to feed.
Sorry, I forgot to add the Let statement (originally t was going to be troughs, but I realized that was wrong and that I had mistaken something). Let t be the minimum number of each animal.
DeleteGrade 7 POTW
ReplyDeleteOne way to solve this problem is guess and check. We know that 8 pigs=one trough, 3 cows=one trough and 2 horses=one trough. In order to ensure that the amount of animals we guess is divisible by 2, 3, and 8, we have to find their LCM. The LCM of 2,3, and 8 is 24. We can try 24. 24 pigs means 3 troughs, 24 cows means 8 troughs, and 24 horses means 12 troughs. All of these troughs equal to 23, which is 3 times smaller than 69, so we have to multiply each of the amounts of animals by 3 and add them up to find our answer. 24x3=72x3=216
There are 216 total animals on the farm.
(I tried to solve algebraically using substitution and was not successful)
I missed the last one ;-;
ReplyDeleteSo first I tried to use guess and check, which didn't work for some reason.
Then, I tried to solve it algebraically. So First I wrote out the amount for each group:
Gr11: x(x-1)2/2 = x^2-x
Gr12: y(y-1)4/2 = 2(y^2-y) = 2y^2-2y
Both: 3xy
Then, I use some algebra magic to find the answer;
x^2-x+2y^2-2y+3xy=1392
x^2+y^2+2xy+xy+y^2-x-2y
(x+y)^2+xy-(x+y)-y+y^2
30^2-30+(30-y)y-y+y^2
30^2-30+30y-y^2-y+y^2
900-30+30y-y-y^2+y^2
870+29y=1392
29y=1392-870
29y=522
y=18.
So since y was the number of Gr 12 students, the number of Grade 11 students in the club was 12.
Grade 8 POTW
ReplyDeleteThere are 30 students. You must make every combination of pairs possible. A grade 11 pair is 2 hours, 12 pair is 4 hours, 11 and 12 is 3 hours. The total hours is 1392. To do this, we first need to give values to the number of grade 11/12 students.
Grade 11 Students: x
Grade 12 Students: 30 - x
We can now make expressions for each possible pair.
Grade 11 & 11 Pair: x(x - 1)/2
Grade 11 & 12 Pair: x(30 - x)
Grade 12 & 12 Pair: (30 - x)(30 - x - 1)/2 = (30 - x)(29 - x)/2
We still aren’t done as we need to find the time in hours in total.
Grade 11 & 11 Pair: (x(x - 1)/2)2 = x(x - 1)
Grade 11 & 12 Pair: (x(30 - x))3 = 3x(30 - x)
Grade 12 & 12 Pair: ((30 - x)(29 - x)/2)4 = 2(30 - x)(29 - x)
Now we just have to put this into an equation using 1392 hours to find the value of x.
x(x - 1) + 3x(30 - x) + 2(30 - x)(29 - x) = 1392
x^2 - x + 90x - 3x^2 + 2(870 - 30x - 29x + x^2) = 1392
x^2 - x + 90x - 3x^2 + 2(870 - 59x + x^2) = 1392
x^2 - x + 90x - 3x^2 + 1740 - 118x + 2x^2 = 1392
-29x + 1740 = 1392
-29x + 1740 - 1740 = 1392 - 1740
-29x = -348
-29x/-29 = -348/-29
x = 12
30 - 12 = 18
Therefore, there are 12 grade 11 students, and 18 grade 12 students.
Grade 7 POTW
ReplyDeleteThere are 69 troughs. 2 horses per trough, 3 cows per trough, 8 pigs per trough. There are the same number of each animal. First, we already know that there are an equal number of each animal. This can only happen if the number of each animal is a factor of 2, 3 or 8. The least common multiple (lcm) is 24 (8 x 3). Therefore, there are at least 24 animals of each type, or a multiple of that. If there are 24 animals each, we need to see how many troughs that would take:
24/2 + 24/3 + 24/8 = 12 + 8 + 3 = 23 troughs
69/23 = 3
Therefore, we need 3 times as many troughs or 3 times as many animals.
24 current animals x 3 animal types x 3 times more needed = 24 x 9 = 216 animals.
Therefore, Old MacDonald has 216 animals on his farm.
POTW:
ReplyDeleteSo I began with finding the LCM of 2, 3, and 8. (Two horses per each trough, 3 cows per each trough, and 8 pigs per each trough).
LCM: 24.
24 / 2 = 12
24 / 3 = 8
24 / 8 = 3
12 + 8 + 3 = 23
69 total troughs.
23 x 3 = 69. 23 is three times 69, so we multiply number of animals by 3, and add them all up.
24 x 3 = 72
72 x 3 = 216.
Total number of animals on Old MacDonald's farm: 216.