Math POTW #1 2018/2019 - Math Enrichment and Extensions
Welcome back all returning Cranny Intermediates, and hello to all of you new Intermediates! (i.e. Grade 7s and newcomers to our school). I know Alan L. was one of our great POTW contributors last year and the year before (he's off to the Gr. 9 MaCS program now, wow!) He passed the torch along to the likes of Seayrohn, Fiona, Esther, Jessica, among many others, who did an amazing job each week in Grade 7. Keep it up for Grade 8! I hope other Grade 8s can also contribute answers each and every week, as well as help support their Grade 7s peers. I wonder who in Grade 7 will be our outstanding Math POTW contributor this year?
As Fiona, Maxwell, and Seayrohn can attest, I can also use questions that YOU have created if you'd like. Just come see me (Mr. Milette) with your GAPPS file (question AND full solution provided).
As most of you know, this website is a space where Mr. Milette seeks to challenge you with the Math Problem of the Week! (A.K.A. POTW). Usually, two questions will be posted each week. One question is geared toward the Grade 7s and a second question is more for the Grade 8s. You are challenged to answer both questions each week, but only your specific-grade question is mandatory. Both of the Grade 7 & 8 questions can be considered enrichment-type math problems, so they’re a great way to be challenged!
The most important requirement of this website is YOUR participation. This may come in the form of:
- sharing your POTW answer
- posting your comments, questions, and replies to others
- seeking out the answers to POTWs (answers are posted a week after the question has been posted)
- reading and following new posts
With that being said, it is absolutely pivotal that you remind yourself about online netiquette. You must always consider the feelings of others when posting comments. Before you press 'Submit' or 'Send' or 'Publish' please ask yourself the following two questions: "Would I say this in-person to someone, face-to-face?" and "Is my comment helpful in any way?" If the answer to either of those questions is "No" then YOU MUST revise your comment.
You will find below the first POTW for the 2018-2019 school year! Please submit any and all responses regardless if you think someone else has the same answer. Both Grade 7 and Grade 8 students can submit answers in the same reply section, please just state which question you’ll be answering at the beginning of your post. Remember in Mathematics you can always show your work in a multitude of different ways!
Be sure to also comment on other students' work, provide feedback to others, and ask clarifying questions.
Always remember to be signed-in to your gapps account before submitting a response. Enjoy the POTW! (Please ask a one of the Grade 8 POTW pros if you have issues with signing-in)
Grade 7 POTW #1 Question:
Grade 8 POTW #1 Question:
Grade 8 POTW:
ReplyDelete(I didn't expect this to up this fast)
Info:
- 9 students split into 2 groups
- Heights are 151, 153, 157, 161, 157, 156, 159, 154
- No Nano is taller than any Techno, but one of the Nanos is the same height as one of the Technos.
- Two of the Nanos are the same height
Find:
- The difference between the mean height of a member of the Technos and the mean height of a member of the Nanos (find whos in which group and calculate the mean. Subtract)
Since no Nano is taller than any Techno, that means that the heights of the Nanos are the shortest of the 9. The heights are 151, 153, 153, 154, 156, 157. Since one of the Technos is the same height as one of the Nanos, two Nanos are the same height (153) and no Nano is taller than a Techno, the same Nano-Techno height cannot be 153. The mean of the Nanos is 154 and the mean height of the Technos is 159. The two subtracted would be 5.
Yes Fiona! You were an amazing contributor last year, I see that you'll be at it again this year. Great work.
ReplyDeleteGrade 8 POTW
ReplyDeleteFirst I will order the heights of the 9 students from least to greatest:
151, 153, 153, 154, 156, 157, 157, 159, 161
Okay, so we know that no Nanos is taller than a Technos, but one Nanos IS the same height as a Technos. We also know that 2 Nanos are the same height. So far, we don't know the number of people that could be in a specific group (Nanos or Technos). However we could use the fact that one Nanos is the same height as one Technos as well as two of the Nanos are the same height to help us figure out the groupings. The smallest repeated value would guaranteed be part of the Nanos as well as any value less than that. Therefore, 151, 153 and 153 are apart of the Nanos already. Using the fact that a Nanos and Technos are the same height proves that the next repeated values could be that same height or other repeated values past that. However, the only repeated numbers after 153 is 157 meaning that that is where all values over 157 including 157 is apart of the Technos grouping and 157 and below is apart of the Nanos. We can now make the groups.
Nanos:
151, 153, 153, 154, 156, 157
Technos:
157, 159, 161
The mean of any # of values is the sum of the values divided by the # of values. Knowing this, we can now calculate the mean of both groups.
Nanos (6 terms/values):
151 + 153 + 153 + 154 + 156 + 157 = 924
924/6 = 154
Nanos Mean = 154
Technos (3 Terms/Values):
157 + 159 + 161 = 477
477/3 = 159
Technos Mean = 159
The difference between the mean height of the Technos to that of the Nanos would be 159 - 154 = 5.
Therefore, the difference between the mean heights of the 2 groups' mean is 5.
Grade 7 POTW
ReplyDeleteFirstly, the average of any values is the sum of the values divided by the number of values. To make this easier and in more math related terms, we can consider her marks in variables. Her first mark can be considered a, second, b and so on to sixth being f.
a) The overall average of Bea's mark can simply be put as the average of (a and b), (c and d), and (e and f) and the average of those 3 averages. It is easier this way since the values were already given. The average of a and b is 71, the average of c and d was 66 and the average of e and f were 82.
71 + 66 + 82 = 219, 219/3 = 73.
Therefore, the average of all of Bea's marks is 73.
b) In order to solve for Bea's average of her first and sixth mark, you have to make an algebraic statement, and simplify it until you find out the sum of a and f.
(a + b) + (b + c) + (c + d) + (d + e) + (e + f) = (71 x 2) + (75 x 2) + (66 x 2) + (68 x 2) + (82 x 2) = 142 + 150 + 132 + 136 + 164 = 724 = a + 2b + 2c + 2d + 2e + f
From here, we can't do much without getting rid of or cancelling out the 2b, 2c, 2d and 2e. We can substitute 2b + 2c + 2d + 2e by making it into 2(b + c) + 2(d + e) which we do have values to:
724 = a + 2b + 2c + 2d + 2e + f = a + 2(b + c) + 2(d + e) + f = a + 2(150) + 2(136) + f = a + 300 + 272 + f = 724
That'll simplify to:
724 - 300 - 272 = 152 = a + f
Since the average is the sum of the terms divided by the number of terms, (a + f)/2 = 152/2 = 76
The average of Bea's first mark and sixth mark is 76.
Great to have you back in the POTW Seayrohn!
DeleteGrade 7 POTW:
ReplyDelete(because why not :)
(all the letters pertain to the marks. a is the first one, b is the second and so on)
Info:
- 6 marks
- Average of a and b=71
- Average of b and c=75
- Average of c and d= 66
- Average of d and e= 68
- Average of e and f= 82
Find:
- Overall average
- Determine the average of Bea's first and sixth marks.
First we need to find the sums of each average writing above. To do this, we need to multiply each average by 2. That gives us.
a+b=142
b+c=150
c+d=132
d+e=136
e+f=164
To find the average, we need to find the sum of a+b+c+d+e+f then divide by 6.
a+b+c+d+e+f=142+132+164
a+b+c+d+e+f=438
(a+b+c+d+e+f)/6=73.
Bea's overall average is 73.
To find the average of Bea's first and sixth marks (a+f) and divide by 2.
(a + b) + (b + c) + (c + d) + (d + e) + (e + f) = 142 + 150 + 132 + 136 + 164
a+2b+2c+2d+2e+f=724
a+2(b+c)+2(d+e)+f=724
a+2(150) +2(136) + f= 724
a+300+272+f=724
a+f=152
(a+f)/2=76
The average of Bea's first and sixth marks is 76%
Gr. 8 POTW (Attempt)
ReplyDelete9 students
Order heights from least to greatest to see which Nano is the same height as a Techno and which two Nanos are the same height.
151, 153, 153, 154, 156, 157, 157, 159. 161
As we know no Nano is taller than a Techno, it's safe to assume the two with heights of 153 cm should be the two Nanos of the same height, and it cannot be 157 cm as Nanos are shorter than Technos, so the two 157 cm heights should be the Nano as tall as one of the Technos. So now we know 3 Nano heights and 1 Techno's height. Knowing now that the tallest Nano is 157 cm tall, that means the heights more than 157 cm will be the heights of each Techno.
So now, we know the height of all 3 technos are 157 cm, 159 cm, and 161 cm.
Onto Nanos, by deduction, every height under 157 cm is a Nano, so there are 6 Nanos, measuring 151 cm, 153 cm, 153 cm, 154 cm, 156 cm, and 157 cm.
Now to calculate the average (mean) of all 6 Nanos and 3 Technos.
To calculate mean, simply add all data and divide the sum by the amount of numbers in the data set.
So for Nano, it's (151 + 153 + 153 + 154 + 156 + 157) /6. This is simplified to 924/6, which is 154 cm. Which means the mean of all the Nano's heights are 154 cm.
As for Technos, the procedure is the same, (157 + 159 + 161) / 3 = 477/3 = 159 cm is the mean.
Now to find the difference, just subtract Nano mean from Techno mean, which is 159 - 154 = 5 cm.
Therefore the difference of both means from Nano and Techno is 5 cm.
grade 8 potw:
ReplyDeletefirst I notice the two doubles, 157 and 153. since group nano has two people with the same height, I can assume it would be 153 since nanos are shorter except for one guy that is the same height. the other double would be because there are two people from each group that are the same height. any height under 157 would be in nano and above would be in techno. group nano's heights are 151, 153, 153, 157, 154, 156 and techno 157, 159, 161. calculate the avg of each group leaves me at 159(techno avg) - 154(nano avg) which gives me 5.
therefore the difference of the averages of the two groups is 5.
Hi.I'm Tin.This is my first POTW of the year. *yay*
ReplyDeleteSo first I saw that there were supposed to be two heights shared by two differnt people, the one between Nanos and the one between a Nano and a Techno. I saw that no number was repeated three times, but there were two numbers repeated twice (153 and 157). Also, since i know that Nanos were either equal or shorter than Technos, I know that 153 cm is the height shared by two Nanos, meaning that 157 cm is the height shared between a Nano and a Techno. That meant that all heights above 157 was a Techno, and all heights below 157 was a Nano. Then, I took the Nano heights (151, 153, 153 again, 154, 156, and 156) and found the mean (924/6 = 154) and I took the Technos (157, 159, 161) and to find the mean, I saw that the shorter Techno was 2 cm shorter than the middle, and the tallest was 2 cm taller than the middle, so they equal out to 159 (The height of the dude/dudette in the middle). So the mean/average of the nanos was 154, and the mean/average of the technos was 159, meaning there was a 5 cm difference between the nanos and technos
So my answer is 5cm difference.
:)
Hi Tin
DeleteGrade 8 POTW
ReplyDeleteSo, from the clues we can figure out:
- that one of the Nanos is the same height as the Technos (and probably the tallest Nano seeing as they are all supposed to be shorter than the Technos)
- that two of the Nanos are the same height
Now, seeing as there are only two pairs in the set of heights (the 153s and the 157s)and that one of the pairs has a Techno, which should probably be the tallest, we can easily tell which pair goes with which height:
1 Nano and 1 Techno are 157, and the other 2 Nanos are 153.
To figure out the rest of the Nanos, we can just take every number under 157, since that is the height of the tallest Nano(which is the tallest because they are the same height as a Techno who should be taller than the group of Nanos).
This means that the Nanos' heights are 157, 156, 154, 153, 153, and 151, and the Technos' heights are 157, 159, and 161.
Now we can find out the mean of each team, starting with the Nanos.
151+153+153+154+156+157= 924
924 divided by 6 (because there are 6 members on team Nano) equals 154.
The mean height of the Nanos is 154.
Time to figure out the mean of the Technos:
157+159+161= 477
477 divided by 3 equals 159, so the mean height of the Technos is 159.
To find out the difference, we subtract.
159-154= 5
We can now conclude that the difference between the mean height of a Techno member and the mean height of a Nano member is 5.
Grade 8 POTW
ReplyDeleteSo, from the clues we can figure out:
- that one of the Nanos is the same height as the Technos (and probably the tallest Nano seeing as they are all supposed to be shorter than the Technos)
- that two of the Nanos are the same height
Now, seeing as there are only two pairs in the set of heights (the 153s and the 157s)and that one of the pairs has a Techno, which should probably be the tallest, we can easily tell which pair goes with which height:
1 Nano and 1 Techno are 157, and the other 2 Nanos are 153.
To figure out the rest of the Nanos, we can just take every number under 157, since that is the height of the tallest Nano(which is the tallest because they are the same height as a Techno who should be taller than the group of Nanos).
This means that the Nanos' heights are 157, 156, 154, 153, 153, and 151, and the Technos' heights are 157, 159, and 161.
Now we can find out the mean of each team, starting with the Nanos.
151+153+153+154+156+157= 924
924 divided by 6 (because there are 6 members on team Nano) equals 154.
The mean height of the Nanos is 154.
Time to figure out the mean of the Technos:
157+159+161= 477
477 divided by 3 equals 159, so the mean height of the Technos is 159.
To find out the difference, we subtract.
159-154= 5
We can now conclude that the difference between the mean height of a Techno member and the mean height of a Nano member is 5.
Oh, it posted twice...
DeleteSorry, the page wasn’t loading and I didn’t see the wait for approval notice, so I copy-pasted it into a new comment to do again.
No problem Hailey!
DeleteFaustina's POTW:
ReplyDeleteThe heighs in order are: 151, 153, 153, 154, 156, 157, 157, 159, and 161
Now as Nanos are the shorter group, but one is the same height as the Technos, we can assume that one Nano is either 153 or 157.
The other piece of information is that there are two Nanos of the same height, so there are two Nanos that are 153 cm tall, meaning that Technos must be over 157 cm.
Nanos: 151, 153, 153, 154, 156, 157
Technos: 157, 159, 161
Mean is average, which means that the mean of Nanos is 154 cm, while the mean of Technos is 159.
The two subtracted is equal to 5 (cm).
Esther's Grade 8 POTW:
ReplyDeleteClues:
- No Nano is taller than a Techno
- One Nano is the same height as one Techno
- Two Nanos are the same height
To organize the numbers I first put them in order from least to greatest:
151, 153, 153, 154, 156, 157, 157, 159, and 161
If a Nano is the same height as another Nano, the only 2 possibilities is 153 cm and 157 cm because there are doubles of each height. However, no Nano is taller than a Techno, meaning it must be 153 cm. This means that the remaining double height; 157 cm must be the height in which a Techno and Nano share the same height.
Since no Nano can be taller than a Techno, the remaining heights below 157 cm are Nanos and the remaining heights greater than 157 cm are Technos:
Nanos: 151, 153, 153, 154, 156, 157
Technos: 157, 159, 161
To find the means of each team, I need to add up all of the heights and divide the sum by the number of team members.
For the Nanos I would add up 151, 153, 153, 154, 156, 157 and divide by 6, which is 154 cm.
For the Technos, I would add up 157, 159, 161 and divide by 3, which is 159.
The difference is found by:
159 - 154 = 5
Therefore the difference between the mean height of the Technos and the mean height of the Nanos is 5 cm.
Grade 7 POTW (Leah):
ReplyDeleteTo average her overall scores I should find the mean of her averages but some are repeated. For example, first and second and second and third (second is repeated). So, I will just use some of the scores. I am using her first and second, her third and fourth, and her fifth and sixth. That is 71%, 66%, and 82%. To find the average or mean, you add all your numbers and then divide them by the amount of numbers. The numbers added together are 219. That divided by three (because I am using three scores) is 73.
This means her overall average score is 73%
To make it easier when putting it into an algebraic expression, I am going to convert each mark into a letter. With that, I know the averages of:
A and B= 71
B and C= 75
C and D= 66
D and E= 68
E and F= 82
To find my answer, I am just going to use an algebraic expression and keep simplifying
(A+B)+(B+C)+C+D)+(D+E)+(E+F)= 142+150+132+136+164
A+2B+2C+2D+2E+F=724
A+2(B+C)+2(D+E)+F=724
A+2(150)+2(136)+F=724
A+300+272+F=724
724-300-272=152
152÷2=76
The average of her first and sixth scores is 76%.
Great job Leah!
DeleteGrade 7 POTW (Maryam)
ReplyDeleteTo find Bea’s overall average of her marks, I will use the averages already given to me. These are the averages of her 1st and 2nd marks (71%), 3rd and 4th marks (66%), and the 5th and 6th marks (82%). After I do so, I will divide the sum by the amount of integers I added.
71+66+82=219
219÷3=73
The overall average of Bea’s marks is 73%.
b) To find the average of Bea’s 1st and 6th marks, I am going to make an algebraic equation. But before that is done, I will convert each mark into a variable to make this equation simpler.
1st mark = a
2nd mark = b
3rd mark = c
4th mark = d
5th mark = e
6th mark = f
Now, I need to find the total marks for each average.
a+b = 71x2=142
b+c = 75x2=150
c+d = 66x2=132
d+e = 68x2=136
e+f = 82x2=164
Finally, I will start the equation.
(a+b)+(b+c)+(c+d)+(d+e)+(e+f) = 142+150+132+136+164
a+2b+2c+2d+2e+f = 724
a+2(b+c)+2(d+e)+f = 724
a+2(150)+2(136)+f = 724
a+300+272+f = 724
a+572+f = 724
a+f+572-572 = 724-572
a+f = 152
From here, since we have the total marks of a and f (marks 1 and 6), we can simply divide 152 by 2 to get to the average.
a+f = 152 ÷ 2 = 76
The average of Bea’s 1st and 6th marks is 76%.
Grade 7 POTW (Hilary)
ReplyDeletea)
To simplify it, I'm going to make all the averages variables. (a, b, c, etc) To find the average of all of it, I'm going to find out the total of the 2 marks instead of the average. Since you get an average by a+b÷2, you can use the same equation to solve 6 of them. You get the total by multiplying the average by 2 to get the total again. a+b+c+d+e+f÷6=average of all 6.
a+b - 71×2=142
b+c - 75×2=150
c+d - 66×2=132
d+e - 68×2=136
e+f - 82×2=164
so if you use the equation from before you can get the average of all her marks. You can simplify it by using (a+b)+(c+d)+(e+f)=437. Now since there were 6 marks, you divide it by 6 and get 73. So the average of all her scores are 73%
a)
To get the average of the 1st and 6th mark, you have to add them and then divide them by 2. To do that I just start with all of the marks added together and then keep simplifying
(a+b)+(b+c)+(c+d)+(d+e)+(e+f) = 142+150+132+136+164
a+2b+2c+2d+2e+f= 724
a+2(b+c)+2(d+e)+f = 724
a+2(150)+2(136)+f=724
a+300+272+f= 724
a+f= 152
a+f÷2=76
the average of her 1st and 6th score is 76%
Nice first post Hilary
DeleteGrade 8 POTW (Not Mandatory but Why Not)
ReplyDeleteFirst, I will arrange the students' heights from least to greatest.
151, 153, 153, 154, 156, 157, 157, 159, 161
Since I know that no Nano is taller than any Techno, I can assume that the shortest height (151) is a Nano. After that, I also know that two of the Nanos are the same height. This means that the two heights tat are the same yet one of the shortest are those of the to Nanos. The last piece of info given is that a Nano and a Techno have the same height. The only pair of same heights left are two '157'. Now, I can easily tell which of the remaining heights belongs where (the heights shorter than 157 are all those of Nanos and vice versa for the Technos).
Nanos: 151, 153, 153, 154, 156, 157
Technos: 157, 159, 161
Now that I know the members of each group, all I need to do now is add the Nanos heights and Technos (separately) and divide the sum by the number of memebers in each group. Finally, I will need to subtract the heights to find the difference.
Nanos: 151+153+153+154+156+157 = 924/6=154
Technos: i57+159+161 = 477/3=159
159-154=5
The difference between the mean height of a memeber of the Technos and Nanos is 5.
Keep up doing the grade 8 version too, awesome!
DeleteA)
ReplyDeleteScore 1=69 71% Average(With score 2)
Score 2=73 75% Average(With score 3)
Score 3=77 66% Average(With score 4)
Score 4=55 68% Average(With score 5)
Score 5=81 82% Average(With score 6)
Score 6=83 Average=???(With score 1)
Now that I know all 6 scores, by using the average to find her scores. For example, score 1 is 69, and since the average with score 2 is 71%, then score 2 has to be 73. I use this tactic to find all the scores.
69+73+77+55+81+83=438
438/6=73
Her Average is 73.
B)
69+83=152
152/2=76
Her average of her first score and last score is 76%
Please try to sign in next time, and/or add your first name to the post. The grade 8s can help any 7s troubleshoot this.
DeleteOh sorry. I thought i was already signed in. This was Allen's
DeleteGrade 7 POTW (Edward)
ReplyDeletea) This question is a question that has to do with average which means that I have to find the mean. In order to find the mean, I have to add all of the percents and divide them by the number of percents. I am trying to add all of the test scores but some of them are the same, the first and second average is 71% and the second and third is 75%.I cannot add 71 and 75 because the second test is in both of them. In this way, I will have to add 1 and 2, 3 and 4, and 5 and 6.
71+66+82=219 divided by 3 is 73.
The average test score is 73%.
b) In order to find the first and sixth mark average, I will take away marks 2 3 4 and 5 to get left with test 1 and 6.
1+2=142
2+3=150
3+4=132
4+5=136
5+6=164
1+2-2-3+3+4-4-5+5+6= 142-150+132-136+164
=
152
Take out 2,3,4,5 and get left
with 1+6. 1+6=152 so 152 divided by 2= 76
What I did was I took out 2,3,4,5 by subtracting and adding each of the tests to take out each of the spare marks. The spare marks are the first and second tests and second and third which means I took out one of the 2's. I did that to 3,4 and 5 and then took out the 2,3,4,5 to be left with 1 and 6. I then converted the marks into their percents and subtracted and added those according to the tests. I was left with 152 which was 1 and 6 combined. I then divided 152 by 2 and got 76.
76 is the average between the first and sixth marks.
Grade 7 POTW (Edward)
ReplyDeletea) This question is a question that has to do with average which means that I have to find the mean. In order to find the mean, I have to add all of the percents and divide them by the number of percents. I am trying to add all of the test scores but some of them are the same, the first and second average is 71% and the second and third is 75%.I cannot add 71 and 75 because the second test is in both of them. In this way, I will have to add 1 and 2, 3 and 4, and 5 and 6.
71+66+82=219 divided by 3 is 73.
The average test score is 73%.
b) In order to find the first and sixth mark average, I will take away marks 2 3 4 and 5 to get left with test 1 and 6.
1+2=142
2+3=150
3+4=132
4+5=136
5+6=164
1+2-2-3+3+4-4-5+5+6= 142-150+132-136+164
=
152
Take out 2,3,4,5 and get left
with 1+6. 1+6=152 so 152 divided by 2= 76
What I did was I took out 2,3,4,5 by subtracting and adding each of the tests to take out each of the spare marks. The spare marks are the first and second tests and second and third which means I took out one of the 2's. I did that to 3,4 and 5 and then took out the 2,3,4,5 to be left with 1 and 6. I then converted the marks into their percents and subtracted and added those according to the tests. I was left with 152 which was 1 and 6 combined. I then divided 152 by 2 and got 76.
76 is the average between the first and sixth marks.
A)
ReplyDeleteThere are 6 total marks. 72, 70, 80, 52, 84, 80. I know this because all you have to do is add the 1st and 2nd mark together which is 142 then separate them based on the percentage of the other mark. So, if I separate 142 based of the two averages (1 and 2) and (2 and 3) the 1st mark would likely be 70 and the second 72. Then you continue on. 2 and 3 is 75% multiplied by 2 which makes the combined percentage 150 and then you split up the marks based on the next 2 averages. Then it will go on until you get get to the last average. So, after doing all this you should be able to find that all the six marks are 70, 72, 80, 52, 84 and 80. Then you add them all up and divide the sum by 6 to get an overall average of 73
B)
For this, all you have to do is add the first mark and the sixth mark so you would just do 72 and 80 and then divide it by 2 to get an average of 76 for 1 and 6's average
This is Jordan BTW
DeleteGrade 7 POTW (Danielle Merisanu)
ReplyDeletePOTW #1
First we can take the six averages and line them up.
71% - Mark 1 = 69%
Mark 2 = 73%
To figure this out all you need to do is find out what two numbers' average is 71%. For example the average of 70% and 73% is 71%. So you can take the 71 and add a certain amount to it to get the first mark and then subtract the same amount to get the second mark. You can use any combination of numbers. I used 69 and 73.
75% - Mark 2 = 73% (which is 75 - 2)
Mark 3 = (So 75 + 2 =) 77%
Because I knew that the second mark Bea got was 73% I can then just add two to 75% to get my third mark, 77%. I know to do this because I had to subtracted two to get my second mark so to get my third mark I have to add two.
66% - Mark 3 = 77%
Mark 4 = 55%
77% - 66% = 11. This means I need to subtract 11 from 66% to get my fourth mark. 66% - 11 = 55%.
68% - Mark 4 = 55%
Mark 5 = 81%
68% - 55% = 13. This means I need to add 13 to 68% to get my fifth mark. 68% + 13 = 81%.
82% - Mark 5 = 81%
Mark 6 = 83%
82% - 81% = 1. This means I need to add 1 to 82% to get my sixth mark. 82% + 1 = 83%
A) 69+73+77+55+81+83 = 438. 438 divided by 6 = 73%. Bea's over all average is 73%.
B) To do this a add 69 and 83 which is 152. Now we take 152 and divide it by 2 which is equal to 76%. The average of Bea's 1st and 6th marks is 76%.
Grade 8 POTW (Edward)
ReplyDeleteI am going to go with the facts to find out which heights are Nanos and technos. Theere that no Nano is taller than a Techno but one Nano is the same height as a Techno.Lastly, 2 of the Nanos are the same height. According to this knowledge, I found out which heights are Nanos and Technos.
Nanos: Technos: - 61 goes in Techno since it's tallest
157 159 - 59 goes in techno since there are no other heights
153 161 and it is the second tallest
153 157 - 57 goes in Techno since there are 2 57's and one has
151 to go in Nanos and one in Technos
154 - all other heights would go in Nanos
156
I will now add all the heights of Nanos together and then divide it by the number of heights to obtain the mean of heights in Nanos.
157+156+154+153+153+151=924 divided by 6= 154
I will now do the same of Technos.
161+159+157=477 divided by 3=159
To find the difference between the 2 mean heights, I will subtract them to get the difference.
159-154=5
%
5 is the difference of the means of Nanos and Technos.
151, 153, 153, 154, 156, 157, 157, 159, 161
ReplyDeleteIf this is the data set, no nano can be taller than a techno yet one of each are the same size and two nanos are the same size, what is the average height of each?
Well, if you look above there are two incidents where two people are the same size (One case, 2Nano other case, Nano, Techno). And there are two incidents where two peices of data are the same (153/153, 157,157). Yet if no nanos are taller than technos, they can't be the pair of 157s as they would be taller than the Techno at 153. This means that two nanos are 153, one nano is 157 and one techno is 157. Yet if a Techno and Nano are 157, no Technos can be shorter than that and no Nanos can be teller or a Nano would be taller than a techno, breaking the rules. Therefore all the data higher than 157 belongs to the Technos while everything below belongs to the Nanos.
So out of 151, 153, 153, 154, 156, 157, 157, 159, 161
151, 153, 153, 154, 156 and 157 are Nanos while
157, 159 and 161 are Technos
Now to find the Nano mean height:
151+153+153+154+156+157=924
924/6=154
And for Technos
157+159+161=477
477/3=159
159(Techno average)-154(Nano average)=5
Therefore the difference between the mean height for Technos and Nanos are 5
Also, I'm so sorry if this was late
DeleteFirst, I tried trial and error, so I started with her first mark being 60%, and multiplied 71% by 2 to get 142%. Since 142% - 60% is 82%, we know the second mark. To find the third mark, we multiply 75% by 2 to get 150% and then subtract 82% from 150% to gt 68%. We continue to do this for all the marks and averages, and the first mark ends up as 60%, the second 82%, the third 68%, the fourth 64%, the fifth 72%, and the sixth ends as 92%. All the numbers fit properly. I then tried to see if you do it with the first mark being 61%, and the second is 81%, the third is 69%, the fourth is 63%, the fifth is 73%, and the sixth is 91%. So it doesn’t matter what number you start with as it always ends up as being 438/600 if you add all the scores divided by the total amount of points you could have gotten. 438 divided by 600 is 0.73 which turns into 73% when you convert it. The first and sixth marks add up to 152/200 which is 76%.
ReplyDeleteThe overall average is 73%.
The first and sixth average is 76%.
First, I tried trial and error, so I started with her first mark being 60%, and multiplied 71% by 2 to get 142%. Since 142% - 60% is 82%, we know the second mark. To find the third mark, we multiply 75% by 2 to get 150% and then subtract 82% from 150% to gt 68%. We continue to do this for all the marks and averages, and the first mark ends up as 60%, the second 82%, the third 68%, the fourth 64%, the fifth 72%, and the sixth ends as 92%. All the numbers fit properly. I then tried to see if you do it with the first mark being 61%, and the second is 81%, the third is 69%, the fourth is 63%, the fifth is 73%, and the sixth is 91%. So it doesn’t matter what number you start with as it always ends up as being 438/600 if you add all the scores divided by the total amount of points you could have gotten. 438 divided by 600 is 0.73 which turns into 73% when you convert it. The first and sixth marks add up to 152/200 which is 76%.
ReplyDeleteThe overall average is 73%.
The first and sixth average is 76%.
This the link to a picture of my POTW done by hand
ReplyDeletehttps://gm1.ggpht.com/31RiIY1LQlxBLSK-qnPXTReQ3pHhgOEXu67qqzvK0co6B6ktlSwJCycLxwU-MnRgxyGOCJ0kBsQy08unORQJGj26YRxa2pB_5Z0fDWriEjOMvXH-nfQTkKM8my2COpxbsuMtnB7sO1kdXlPSJDr8y1QbSqB5Or4_I8iaxtdi0jo_jEg9sU25EXWiJykSvRhth276yK6AUi6lDUsvg4UVV8Eeuuuj2o2JP4YxlFeB2M6S4YslqO-ctv2FCbi2jTEupqsMqLo-9YsL4qIiVknpbDeRe7bMGklUTBSU5dKNTS7yxjQxHHRDBQdAMfh52NrQretIJZ3OjGDELB73D8G3BXUSlIfBICqu530k5tWboJBcbV5BqpF7r9rs9Ug2MBK9S9UDQU1TLk0EqJNOWb5CGsXrlZITJO4kJZC1wpOhohPXPxphsb6jKnOeOtDORIv_KeZdjJOz2wqKMcKYEzeTW-co6x3McAC2vBAfcZLKdQpfSYliUy2So2yGfaqoj4xehM5teKNWec96NJAqh1u4mCdsTyzETwaUkv2s9DCsc7VkDNtMCpwR7OkmBOLzpQzjLj92d3OAg4vqiHNmQA-NHiWJCTmoMzye_1Iz6z2UlSqEwFOH2TBGDSROzKwkHNOE4Fv96H-pffv9wbPASePqflJcvbCW7ixV_Z_VwSjEZ-ThlC9tIl2wMRlqF633AeeGfs9PoT0Z=s0-l75-ft-l75-ft
Maksim Anntsipovitch
ReplyDeleteTo solve the grade 7 question you needed to choose any number to be the first mark and then it would balance out the rest of the marks(The number had to be more than 42 and a least 100). Ti solve the questions you will first need to figure out all of the marks. To do so you will need to use the percentages that were given to us in the problem (marks 1-2=71%, 2-3=75%, 3-4=66%, 4-5=68% and 5-6=82%). I used mean to find the marks so we know that to find the mean of 2 numbers you will need to add them and then divide by the number of numbers that we added. As I said, we can chose any number to be the first mark;I chose 62. Then I thought (62 plus what, divided by 2 equals 71.)It was 80. That would be the second mark. Then I did that to all the marks. The question was what is Bea's overall average. Average= mean (marks 1-6) 62+80+70+62+74+90=438 and divided by 2= 73. The second question was:Determine the average of marks 1 and 6. Mark 1= 62 Mark 6= 90. (90+62)/2= 76
Therefore. a)Bea's overall average was 73%
b) The average of mark 1 and 6 is 76%
the average of the first mark and the second mark= 71
ReplyDeletethe average of the third mark and fourth mark= 66
the average of the fifth and sixth mark=82
the total average is 219/3= 73
b)
a+b+b+c+c+d+d+e+e+f= 724
a+2b+2c+2d+2e+f= 724
a+2x150+2x136+f=724
a+300+272+f=724
a+f=724-572
=152/2=76
a) To find the average, you have to add up the marks and divide. But some marks repeat so I got rid of the marks that repeated which left me with 1 and 2, 3 and 4, and 5 and 6. So then you add them all together:
ReplyDelete71+66+82=219 divided by 3 is 73.
The average test score is 73%.
b) First I added up the sums of each mark:
a+b=142
b+c=150
c+d=132
d+e=136
e+f=164
To find the average of Bea's first and sixth marks (a+f) and divide by 2.
(a + b) + (b + c) + (c + d) + (d + e) + (e + f) = 142 + 150 + 132 + 136 + 164
a+2b+2c+2d+2e+f=724
a+2(b+c)+2(d+e)+f=724
a+2(150) +2(136) + f= 724
a+300+272+f=724
a+f=152
(a+f)/2=76
So the average of Bea's first and sixth marks is 76%
Nanos: (151, 153, 153, 154, 156, 157) Technos:(157, 159, 161)
ReplyDeleteThere are two pairs of the same values, since one pair are Nanos, then the other pair is the shared height of one Nano and one Techno, and is the split point for the two groups with all Technos being taller than Nanos.
Then I found the mean by adding all the values and dividing by the number of values and was left with the Nanos and Technos means, 154cm and 159cm respectively. The difference is 5 cm.
The difference between the mean height on the Nanos and Technos is 5cm.
Grade 7 POTW (Ananya)
ReplyDeletea) To find Bea’s overall average, we add the averages of her first and second marks, third and fourth marks, and fifth and sixth marks, and divide it by 3.
71 + 66 + 82 = 219
219 / 3 = 73
So, Bea’s overall average is 73%
b) To get the average of the first and sixth marks, first let’s list all of the average marks:
x + y = 71 * 2 = 142
y + z = 75 * 2 = 150
z + a = 66 * 2 = 132
a + b = 68 * 2 = 136
b + c = 82 * 2 = 164
To find the average of the first and sixth marks we must to use an algebraic expression
(x+y) + (y+z) + (z+a) + (a+b) + (b+c) = 142 + 150 + 132 + 136 + 164
x + 2y + 2z + 2a + 2b + c = 724
x + 2(y+z) + 2(a+b) + c = 724
x + 2 * 150 + 2 * 136 + c =724
x + 300 + 272 + c = 724
x + c = 724 - 572
x + c = 152 / 2
x + c = 76%
So, the average of Bea’s first and sixth marks are 76%
a) Basically, to find the average of Bea's marks, you would have to add all the marks and then afterwards divide. Since marks tend to repeat, I got rid of those and i had 1,2,3,4,5, and 6 left. Then you add. 71+66+82 divided by 3 leaves you with 73, meaning the average is 73%.
ReplyDeleteb) The first thing that i did was add all of the sums of her marks.
a+b=142
b+c=150
c+d=132
d+e=136
e+f=164
In order to get the average of her first and sixth marks, you add a+f and divide it by 2.
a+b+b+c+c+d+d+e+e+f= 724
a+2b+2c+2d+2e+f= 724
a+2x150+2x136+f= 724
a+300+272+f= 724
a+f=152
152(a+f)/2= 76
So that means the average of her first and her sixth marks is 76%
a) First, in order to find Bea's overall average, I need to add all of the marks up. However, since some of the numbers are repeated, this only leaves me with 71, 66, and 82.
ReplyDelete71 + 66 + 82= 219. Now, I need to divide 219 by 3 so that I can find the average of 219. 219 ÷ 3= 73. Since the average must be in percentages, this would be 73%.
b) Note: First mark=a, second mark=b, third mark=c, fourth mark=d, fifth mark=e, sixth mark=f.
a+b/2= 71 ... a+b= 142
b+c/2= 75 ... b+c= 150
c+d/2= 66 ... c+d= 132
d+e/2= 68 ... d+e= 136
e+f/2= 82 ... e+f= 164
Now, I need to add all of these sums and to divide the final product by 2 to get the average of Bea's first and sixth marks.
a+2(b+c+d+e)+f= 724
a+2(150+136)+f= 724
a+(2x286)+f= 724
a+f= 724-572= 152
Now, I need to divide.
152/2= 76
This means that the average of Bea's first and sixth marks is 76%.
mine is on paper
ReplyDeleteaverage = 73
1 and 6 average = 76
(Brian)
ReplyDeleteScore Average
71 71
71 75
79 66
53 68
83 82
81
A) (71+71+79+53+83+81)/6 = 73
B) 71+81/2 = 76
I love POTW
ReplyDeleteSo we know that no Nano is taller than any Techno. That is our first giveaway. The second is that 2 Nanos are the same height. Since no Nano is taller than a Techno, the same heights that one Nano and one Techno share has to be 157 instead of 153, or else two Nanos will be higher than one or more Technos.
ReplyDeleteNanos:
157
Technos:
157
From the second giveaway, we can assume the two Nanos that share the same height are 153, because 153 repeats twice too.
Nanos:
157
153
153
Technos:
157
Now, we know that 157 is the base line for Technos and Nanos, Higher than 157 is a Techno, Lower is a Nano. This is because if a Nano is higher, than a it'll be higher than a Techno. If a Techno is lower than 157, than a Techno will also be lower than a Nano. So we can't allow that.
Nanos:
157
153
153
151
156
154
Techno:
157
161
159
So now we'll calculate the averages:
157 + 153 + 153 + 151 + 156+ 154 = 924
924/6 = 154
154 is the Nano's mean
157 + 161 + 159 = 477
477/3 = 159
159 is the Techno's mean
(I submitted an answer before, but it isn't here):
ReplyDeleteNaumana's POTW:
My POTW is on paper:
a) 73%
b) 76%