For these weeks, please try all of the questions, including the optional ones. But again, some are optional, so enjoy your Winter Break first and foremost! And...some of the questions this week are great practice for your Spatial Sense Assessment OF Learning this Tuesday.
Grade 7/8 Math POTW Optional Winter Break Question #2 (#1 is down below):
Grade 8 Math POTW #15 Question:
Gr. 7 Math POTW #15 Question (Spatial Sense Practice):
Fiona POTW #2 (Question from last week, Week #14) Solution:
This question deals with the Primitive Pythagorean Triples, also known
as the Pythagorean Triples. All you need to know are the following triples
and you can solve this problem:
3-4-5
5-12-13
7-24-25
8-15-17
20-21-29
First, we need to look at triangle ABD because we need to know side BD in
order to find side DC. We could use the pythagorean theorem to solve this,
but if you notice, both side lengths are divisible by 4. Dividing everything by 4,
we get:
52/4 = 13
48/4 = 12
This looks familiar because this is ⅔ of one of the triples (5-12-13). All we
need to do now, is take 5 and multiply it by the common factor, in this case it is 4.
5 x 4 = 20
Now that we know side BD, we can use that to find that triangle BCD is
also very familiar. It is part of the 20-21-29 triple. Because we already
have the 20 and the 21, side length CD has to be 29.
Therefore side length CD has to be 29.
POTW #14 Gr. 7 Solution:
POTW #14 Gr. 8 Solution:
Math POTW Optional Winter Break Grade 8 Question #1:
Grade 7 Math POTW Optional Winter Break Question #1:
Gr. 7 Math POTW #15 Question (Spatial Sense Practice):
ReplyDeleteDone on grid paper 60 km traveled the shape is the border of one square in the middle with 5 by 5 sides and 1 square of the same size on each of the 4 sides.
A cross or a dodecagon since it has 12 sides.
Area is 125km²
Grade 7 Math POTW Optional Winter Break Question #1:
Looks something like this: (Letn=no water and w=yes water)
w
n w
w w n
w n w n
w w n
n w
w
Grade 7 Math POTW
ReplyDeleteI've mapped out the boy's route on grid paper.
They traveled 60 kilometres starting from S and ending at S using the pattern in the question.
The shape is shaped like a plus sign +. Since the shape as 12 sides, it is a Dodecagon.
In order to find the area of the shape, I will divide it into 5 squares of equal area and perimeter. The area of one square is the product of its length and width. Length=5 . Width=5 . 5x5=25km squared.
The area of one square is 25km squared. Now I will multiply 25km by the number of squares(5) to get 125km squared as the area of the shape.
Grade 8 POTW:
ReplyDeleteWe can narrow the problem down by considering the four possible roads to take to travel to the top section of the… graph. The following roads are:
- Road A - the road to the left of King’s Lake
- Road B - the road in between King’s Lake and King’s Park.
- Road C - the road to the right of King’s Park
- Road D - the road to the far right.
If we look at Road A, we can zigzag starting from going north until we reach the summer house. This will be 15 km
If we look at Road B, we can zigzag starting from going right. This will be 13 km.
If we look at Road C, we can zigzag staring from going right. This will be 17 km.
If we look at Road D, we can zigzag staring from going north. This will be 15 km.
We can conclude that Road B will be the shortest distance because it is only 13km while the others are longer than that.
I did this on paper. The shape enclosed looks like a plus sign. I counted the sides and got 12. Therefore it is a dodecagon. To find the area i will separate the shape inot 5 equal squares. Then i will calculate the area of each.
ReplyDeleteSince the length and width is both5 of each square and 5x5=25 each sqaure has an area of 25. 25 times 5 is 125 there fore saying that the area of the shape is 125km2
Grade 7 POTW
ReplyDeleteI mapped out the route on paper, however I know that they traveled a total of 60 kilometers, before they returned to the start. The shape they traveled resembled a + sign. To find the area of the shape, I broke it down into 5 squares, each with a side length of 5. I know that 5*5=25, and 25*5=125, therefore they traveled 125 km squared.
Hi, it's me again! Today we're going to try to calculate the area of this shape without looking at it!
ReplyDeleteFirst, I'll dilate the instructions by 1/5. Why? This makes it so that it is easier to determine the shape. We know that we have made a U, since we made 2 turns "right", or rather, clockwise. Then, we make a turn in the opposite direction. This means we need to line up 4 us, because we need to turn counterclockwise 4 times. (the reason for this being that we make a u, then we turn in the opposite direction, making us repeat this turn until the condition is satisfied). What shapes can you make with 4 us, each side unit length, pointing in different directions? A cross, and practically only a cross!
Next, to calculate the area of this cross, we can dilate the instructions back. We know that we have 5 squares in this cross, because each regular cross can be divided into 5 unit squares (each "u" is a square, and there is one in the middle". We also know that each of these squares have a side length of 5, because in the dilation by 1/5, each square had a side length of 1. Therefore, we only need to calculate the area of one square, and multiply it by 5.
Square = 5x5 = 25km^2
25km^2 x 5 = 125km^5
The same can be applied to the u's to find perimeter.
U = 5 + 5 + 5 = 15km
15 x 4 = 60km
Therefore, the area of the cross is 125 km^2, and it's perimeter is 60km.
Hi. This is Tin. Here is Math. (The reindeer question)
ReplyDeleteSo the total possibility is equal to 8!, since there are 8 spots for 8 reindeer to take, which equals 40320.
So for the total possibilities where the arrangement is correct:
We can start but placing 4 circles (representing reindeer without the letter "r"), and placing 5 boxes around them like this: []O[]O[]O[]O[]. Basically, the 4 reindeer can go in any of the boxes. Why 5? This is because while on first Glance, it may seem like the arrangement has to look like this (r represents reindeer with the letter "r") rOrOrOrO or vice versa, it actually doesn't state that reindeer without the letter "r" can't be beside each other, so there is also arrangements other than the pattern above like this: rOrOrOOr. In this case, the box between the third and fourth O is empty, and in fact, any single box could be empty. So, this leaves us with 5! correct arrangements instead of 4! Also, using simple math, we can find that the total possibilities of the reindeer without "r" is 4!, as in 4 reindeer in 4 different spots. However, the question also states that two reindeer, one with "r" and one without, can't be beside each other. This subtracts 8 different possibilities, which leaves us with a fraction of 5!*4!-8/8!
which equals 120*24-8/40320
which equals 2880-8/40320
which equals 2872/40320
which equals 718/10080
which equals 359/5040, which converts to around 7.123%, meaning on average, it would take them 14 attempts to get the arrangement right,
Grade 7/8 Math POTW Optional Winter Break Question #2:
ReplyDeleteWe know that there are 4 reindeer with r's in their name. Because there are only 8 reindeer we can only put the 4 starting on the first or the second. Then it comes down to how many ways can we rearrange the other 4 reindeer's. 4! = total amount of ways to rearrange. 4! = 4*3*2*1 = 24
But we also have to take into account that you can rearrange the reindeer's that have a r in their name. That's also is 24. 2*24*24= 1152
Now we have to subtract the order's with Donner and Blitzen are adjacent. There are 8 spots where Blitzen and Donner can be. Each one leaving the other 6 reindeers to be rearranged. That gives 3!*3!. Then because the order can be Blitzen first OR Donner first you multiply by 2. (2*3!*3!) Then add the number of ways to rearrange Blitzen and Donner's order without counting order because we already did that with the *2. 8*2*3!*3!. This is actually wrong though, because if Blitzen or Donner are placed in the 8th slot then the other one can only be placed behind them. That mean you have to subtract from the ways that Blitzen and Donner can be rearranged. In other words it turns the 8 into a 7. 7*2*3!*3!=7*2*3*2*1*3*2*1=504
1152-504 = 648
648 ways to rearrange correctly. Total ways is 8!=40320
probability that they will rearrange correctly is 648/40320
Simplified into 9/560
The possibility of the randomly ordered reindeer line to be correct is 9/560
Grade 8 POTW
ReplyDeleteOkay, so first, let's indicate the paths we MUST take to even get to the summer house. We can either go to the left side of King's Lake where there is only one road we can take (to get to the other side), in between King's Lake and Park where there is only one road we can take, and to the left of King's Park where there are two roads we can take. For all the options, to take the least distance, we can't move away from the road when we don't have to (for example, continue to go left and then go back right) as that would take up distance that doesn't need to be taken. Also, to explain easier, I will explain in the view of the map and not the view of the driving.
Okay, so for the first option, the left of the lake, going right would mean we HAVE to go up, then we can turn left meaning we have to go up, but then we can't go up again to get to the other side. So instead of going right in the beginning, we can go up, then right, then up, then left, then up, then right, then up, then right, then down, then right, then up, then right, then down, then right, then up. This would be 15 km.
For the second option, if we go up, it means we reach a dead-end, so we'll go right (to not make this too long I'll skip this part). The path for this would be right, up, right, down, right, up, right, up, left, up, right, up, right. This would be 13 km so the first option is out.
For the third option and first road, the shortest path would be up, right, down, right, up, right, down, right, up, right, up, right, up, left, up. This would be 15 km so the option is out.
The last option would be right, up, right, down, right, up, right, down, right, up, right, up, left, up, left, up, right. This would be 17 km so the option is out.
Therefore, the shortest route is 13 km and is right, up, right, down, right, up, right, up, left, up, right, up, right. This would be the shortest route as the road it uses takes 13 km as the fastest route using that road, while the other fastest routes for the other roads is 15 km and 17 km.
Grade 7 POTW
ReplyDeleteOkay, so instead of doing this on paper as this is online, we can just visualize the steps and answer the question like that, but first I'll list the exact questions.
a) the perimeter of the shape (that is what the question asks in other words)
b) the name of the shape
c) the area of the shape
Okay, so the car first faces up. It drives 5 km, turns right, drives 5 km, turns right, drives five kilometers, turns left. It then repeats this. If we think about this, the car would've made a five by five square without the bottom (scale is 1km per unit). The car would also be facing towards the right now and would've traveled 15 km. Now after repeating this process, the same bottomless square would be made, except rotated 90 degrees cw and the old left side would become the top, the top would become the right, the right would become the bottom, and the left would now be opened up. At the end of this, another 15 km would've been made and the shape would appear to be 2 squares, one without a bottom, another without a left side, and be attached by the corner and be diagonal from one another. Repeating this another time would be a flipped version of the original, and the same relationship of the original and the second version. 45 km would be covered. Doing it one last time would complete the shape making it 60 km as the distance traveled or perimeter. The process was performed 4 times with each time consisting of 3 sides or a total of 4 x 3 = 12 sides. It appears to be an addition or plus sign. Since it has 12 sides and isn't regular, it is an irregular dodecagon. For the last question, the area, we know that it is a plus sign. Instead of counting every unit square on the grid, we can just use what we know. We know that each square without a side is 5 by 5 meaning it has an area of 25 square units (assuming it does have a side). The part where the squares are opened forms another 5 by 5 making it 25 more square units. In other words, the area is 25 x 4 + 25 or 25 x 5 = 125 km^2.
Therefore:
a) the perimeter of the shape: 60 km
b) the name of the shape: irregular dodecagon
c) the area of the shape: 125 km^2
Grade 8 Optional POTW
ReplyDeleteThe question is basically just asking us how many pairs of 4-digit palindromes have a sum of a five digit palindrome.
Okay, so instead of just going trial and error with a bunch of palindromes (would take so long), we can take advantage of the fact that they are palindromes. Palindromes must be read the same backwards, so digit 1 must be digit 4 and digit 2 must be digit 3. We can establish variables:
abba + cddc = efgfe
In this situation any variable can be the same as another one.
*Note Palindromes can't begin with 0 (in the case of numbers).
Anyways, since it is going to be a 4 digit number added to a 4 digit number, we can establish the value of e. No matter what the case, any 2 4 digit numbers added to each other would result in either a 4 digit number (not what we want) or a 5 digit number (what we want) that begins with 1. Therefore, e = 1.
abba + cddc = 1fgf1
In this situation, to make it a 5 digit number, a + c = 11, as if it is 1, then the number couldn't be a 5 digit number as there would be no carrying. The possible values for a and c are as shown.
a = 2, 3, 4, 5
c = 9, 8, 7, 6
In this case, I'm not giving a and c the other values that the other variable has because it would just be repeating numbers. Continuing from before:
abba + cddc = 1fgf1
We know that a + c = 11, so that means that f can either be 1 or 2. This is because a + c = 11 carrying the 1 to the ten thousands column and leaving the 1 in the thousands column, or the hundreds column carries a 1 to the thousands column making a + c + 1 = 12 leaving a 2 in the thousands and 1 in the ten thousands.
For f = 1:
abba + cddc = 11g11
We know that a + c = 11 meaning that the tens column would have to have a 1 added to it. In other words, b + d + 1 = 1 or b + d = 0. This would only work if b AND d are 0, or there are negative numbers (not possible). Therefore there are the four possible a and c values x one possible b and d values = four combinations for f = 1.
For f = 2:
abba + cddc = 12g21
We know that a + c = 11 meaning that a 1 is carried into the tens column. This means b + d + 1 = 2 or 12, or b + d = 1 or 11. However, in the thousands column, b + d carries into the ten thousands column meaning it must equal 11. In other words, if f = 2, b + d = 11. The pairs of numbers for this is:
b = 2, 3, 4, 5, 6, 7, 8, 9
c = 9, 8, 7, 6, 5, 4, 3, 2
(This time, all 8 numbers are being included for both variables because the numbers won't be repeated because a and c are already different).
So there are four possible a and c values x eight possible b and d values = thirty-two combinations for f = 2.
Now we can add all the combinations:
4 + 32 = 36 combinations
Therefore, there are 36 pairs of four digit palindromes where their sum is equal to a five digit palindrome.
Grade 7 Optional POTW
ReplyDeleteI did this on paper, but I'll try to explain the question here. First, I'll label the islands so it'll be easier to explain each one. I will call each island #-# depending on the row and order it's in. For example, the one at the very top is in the first row, and is the first island from left to right so it is 1-1. The one to the far right would be 4-1, and so on.
So firstly, 4-1 has a 2 on it meaning 2 of its adjacent islands have water. It only has 2 adjacent islands, so 3-1 and 5-1 have water. Similarly 4-4 says 0 meaning none of its adjacent islands have water meaning 3-3 and 5-3 have no water.
Island 2-2 has the number 3 on it with 4 adjacent islands. We know 3-3 has no water meaning the other 3 islands, 1-1, 3-2, and 4-3 have water. Similarly, 6-2 has a 3 on it and with 4 adjacent islands and 5-3 has no water meaning 4-3 (like before), 5-2, and 7-1 has water.
Island 4-2 has the number 4 on it with 6 adjacent islands. We know 3-1, 3-2, 5-1, and 5-2 have water on them, meaning that 2-1 and 6-1 has no water on them.
Island 1-1 has a 2 on it with 3 adjacent islands. We know 2-1 has no water on it and 3-2 has water on it, meaning one more island needs water which has to be 2-2. Similarly, 7-1 has a 2 on it with 3 adjacent islands and 6-1 has no water while 5-2 does have water meaning 6-2 has water on it.
There are multiple islands we can use to help us now, but we don’t have to use them all, so I’ll only use as many as we need. Island 3-2 has a 4 on it with 6 adjacent islands. 1-1, 2-2, 4-3, and 5-2 have water on them while 2-1 has no water. We still need one more island to have no water meaning 4-2 has no water on it.
Island 3-1 has a 2 on it. Instead of listing the number of adjacent islands and all that stuff, we can just realise that adjacent island 5-1 is the only adjacent island that has water so far and there is only one available island meaning 4-1 has water on it.
Island 3-3 has a 2 on it with 2 adjacent islands already having water on them. This means the other adjacent island 4-4 has no water on it.
I’m just going to list the islands and their water statis once more to have it all together.
Therefore the islands
With water are: 1-1, 2-2, 3-1, 3-2, 4-1, 4-3, 5-1, 5-2, 6-2, and 7-1
Without water are: 2-1, 3-3, 4-2, 4-4, 5-3, and 6-1
This is a bit late, but here is my POTW # 15 (Spatial Sense):
ReplyDeletePOTW:
I drew out the route on paper, on a grid system. They traveled a total of 60 km, and the shape mapped out was a plus sign. I later then broke down this plus sign into 5 squares, all equal in size. L x W, 5 x 5 = 25, 25 x 5 = 125, making the area of this shape 125 km squared.