POTW #20 Gr. 7
POTW #20 Gr. 8
POTW #19 Answer Gr. 7: (if you want the full solution please see Mr. Milette
POTW #19 Solution Gr. 8 (if you want the full solution please see Mr. Milette
Thursday, February 15, 2018
POTW #20 - Xin Nian Kuai Le!
Happy Lunar New Year/Chinese New Year to those who celebrate! I know I'll be eating a ton.
POTW #20 Gr. 7
POTW #20 Gr. 8
POTW #19 Answer Gr. 7: (if you want the full solution please see Mr. Milette
POTW #19 Solution Gr. 8 (if you want the full solution please see Mr. Milette
POTW #20 Gr. 7
POTW #20 Gr. 8
POTW #19 Answer Gr. 7: (if you want the full solution please see Mr. Milette
POTW #19 Solution Gr. 8 (if you want the full solution please see Mr. Milette
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There will be 11 different amount of stacks that can be created
ReplyDeleteThe way I got this is first creating some rules. Since the blue coins are have a $5 value and since each stack must equal $750 there must be an even amount of blue coins. Then since each stack must contain 1 coin of each color using only blue coins and only red coins won't work. I then just listed out all the options starting from the maximum amount of blue coins used and replacing 2 blue coins for 5 red coins after each stack
28 Blue 5 Red
26 Blue 10 Red
24 Blue 15 Red
22 Blue 20 Red
20 Blue 25 Red
18 Blue 30 Red
16 Blue 35 Red
14 Blue 40 Red
12 Blue 45 Red
10 Blue 50 Red
8 Blue 55 Red
6 Blue 60 Red
4 Blue 65 Red
2 Blue 70 Red
That counts 14 different stacks, not 11.
DeleteGrade 7 POTW:
ReplyDeleteInfo:
- 1x1
- 2x2
- 3x3
- Table is 84x112
- Minimum???
The example says that if you use only 1x1, you will use 9408 tiles in total. The best way to cover space is to use the largest tile size. First, I need to find out if 84 or 112 are divisible by 3. 8+4 =12, 84 is divisible by 3. 1+1+2=4, 112 is not divisible by 3. Instead of using 2x2 tiles, I can use 2x2 tiles until I end with a number that is divisble by 3. I need to divide 84 by 3 to find out how many tiles I need to use for the one side. 84/3=26. Now I need to find out how many 2x2 tiles are needed before I reach the closest number to 112 (without going over) that is divisible by 3. The closest number is 108 (1+8=9). That uses 2 2x2 tiles. The total number of tiles used so far is 28 (26+2). Now, I need to divide 108 by 3. 108/3=36. To find the total number of tiles used, I can multiply 36 by 28= 1008. That's the minimum number of tiles used. :D
And so on, not just 1, 2 and 3. Also, you can only use 1 tile type.
DeleteGrade 7 POTW
ReplyDeleteWhat you do is you take the greatest common factor of 112 and 84. The factors of 84 are 1,2,3,4,6,7,12,14,21,28,42,84. The factors of 112 are 1,2,4,7,8,14,16,28,56,112. The greatest common factor of 112 and 84 are 28. That means the least number of tiles would be the number of 28 by 28 tiles. On the 112 side, 112/28 = 4. On the 84 side, 84/28 = 3. 3 x 4 = 12. The least number of square tiles are 12 28 by 28 tiles.
GCF of 84 and 112
ReplyDelete= 28
There can be 3 28x28 squares by 4 28x28 squares. 3x4 = 12. Sara will need 12 squares of 28x28 to cover her table.
9408/9 = 1044.444444444444…
ReplyDelete9*1044 = 9396
9408 - 9396 = 12
12/4 = 3
9396/9=1044
Use 1044 3cm * 3cm and 3 2cm*2cm tiles to cover the table top.
the minimum amount of tiles is 1047.
ReplyDeleteI first found a number that is less than 9408 but is the biggest number that would be a factor of 9 since 3 x 3 is 9, and that is the biggest tile. I got a number of 9396. I also needed the number to end in an even number so it could be divided by 4. divide 9396 by 9 to give you 1044, then add the quotient of the remaining amount divide by 4. the final answer is 1047.
Grade 8 POTW
ReplyDeleteOkay, so to do this question, I don’t know how to explain really. All I can explain is that there has to be an even number of blue coins at all times.
I will now somewhat list the coins.
1. 75 red
2. 2 blue, 70 red
3. 4 blue, 65 red
4. 6 blue, 60 red
5. 8 blue, 55 red
6. 10 blue, 50 red
7. 12 blue, 45 red
8. 14 blue, 40 red
9. 16 blue, 35 red
10. 18 blue, 30 red
11. 20 blue, 25 red
12. 22 blue, 20 red
13. 24 blue, 15 red
14. 26 blue, 10 red
15. 28 blue, 5 red
16. 30 blue
There are 16 possible different stacks. I could’ve just multiplied to quickly get the answer, but that’ll be harder to show how that works.
NM. I messed up. I didn't see that we needed at least one blue and one red coin. So that rids 2 possibilities. There are actually 14 different possibilities.
DeleteThere would be 14 combinations:
ReplyDelete70 * 10 dollar coins + 2 * 25 dollar coins
65, 4
60,6
55,8
50,10
45,12
40,14
35,16
30,18
25,20
20,22
15,24
10,26
5,28
These solutions are all possible for the grade 8 POTW.
Because you can do it with 112 and 84, you can just take the greatest common factor of it, which is 28. So there can be 3 28x28 tiles on one side, and 4 28x28 tiles on the other side. This means that 3 x 4 is 12, which is the number of tiles necessary, and then that means you need 12 28x28 tiles and all that.
ReplyDeleteGrade 8 POTW:
ReplyDeletePossible combinations:
b = blue
r = red
28b, 5r
26b, 10r
24b, 15r
22b, 20r
20b, 25r
18b, 30r
16b, 35r
14b, 40r
12b, 45r
10b, 50r
8b, 55r
6b, 60r
4b, 65r
2b, 70r
This adds up to 14 different possible combinations.
There are 14 different stacks that can be made.
Grade 8 POTW:
ReplyDeleteBy finding the pattern of the different combinations, we can figure out this question an easier way. I know that the pattern in which you can get to $750, the $25 coins in each combination would decrease by 5, starting at 70 red coins. By dividing our pattern rule with the red coins by the amount we start with, we can get our answer.
70/5 = 14
There, there are a total of 14 different possibilities.
GRADE 8 POTW
ReplyDeleteFirst, I started with 70 red coins and 2 blue coins, the combination with the max. number of red coins. Then, I found two other combinations, 65 red and 4 blue coins and 60 red and 6 blue coins. I realized that as the number of red coins goes down by 5, the number of blue goes up by 2. Using that information, I made a list of possibilities.
R B
70 2
65 4
60 6
55 8
50 10
45 12
40 14
35 16
30 18
25 20
20 22
15 24
10 26
5 28
Therefore, there are 14 different stacks of coins that can be made.
POTW Grade 8 #20
ReplyDeleteLike many others, I chose to manually find the combinations by starting with the easiest combination to think of.
1. 28 Blue, 5 Red -> 25(28) + 10(5) = 700 + 50 = 750
2. 26 Blue, 10 Red -> 25(26) + 10(10) = 650 + 100 = 750
3. 24 Blue, 15 Red -> 25(24) + 10(15) = 600 + 150 = 750
4. 22 Blue, 20 Red -> 25(22) + 10(20) = 550 + 200 = 750
5. 20 Blue, 25 Red -> 25(20) + 10(25) = 500 + 250 = 750
6. 18 Blue, 30 Red -> 25(18) + 10(30) = 450 + 300 = 750
7. 16 Blue, 35 Red -> 25(16) + 10(35) = 400 + 350 = 750
8. 14 Blue, 40 Red -> 25(14) + 10(40) = 350 + 400 = 750
9. 12 Blue, 45 Red -> 25(12) + 10(45) = 300 + 450 = 750
10. 10 Blue, 50 Red -> 25(10) + 10(50) = 250 + 500 = 750
11. 8 Blue, 55 Red -> 25(8) + 10(55) = 200 + 550 = 750
12. 6 Blue, 60 Red -> 25(6) + 10(60) = 150 + 600 = 750
13. 4 Blue, 65 Red -> 25(4) + 10(65) = 100 + 650 = 750
14. 2 Blue, 70 Red -> 25(2) + 10(70) = 50 + 700 = 750
Since the question stated that there must be at least one red coin and one blue coin in each stack, the combinations:
15. 30 Blue, 0 Red
16. 0 Blue, 75 Red
would not work.
To make that(^) easier to understand,
I found a pattern rule for the blue coins and red coins.
Blue Coins: Start at 28, decrease by 2 each term.
Red Coins: Start at 5, increase by 5 each term.
Therefore, you can make 14 different stacks of coins with different combinations.
Grade 8: There are 14 different stacks of coins that fit the criteria.
ReplyDelete