POTW #21 Solution:
POTW #22 Question: (there are different ways to solve it, can you use your algebra unit knowledge to do so?)
Friday, March 2, 2018
POTW #22 - Keep Up the GREAT Work!
It was great to talk with students this week about the latest POTW. Keep asking those critical questions about the POTW and don't forget to submit your answers, even if your solutions were done on hard copy. For this week's POTW, Mr. Milette challenges you to try using algebra in your solution.
POTW #21 Solution:
POTW #22 Question: (there are different ways to solve it, can you use your algebra unit knowledge to do so?)
POTW #21 Solution:
POTW #22 Question: (there are different ways to solve it, can you use your algebra unit knowledge to do so?)
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Grade 7/8 POTW
ReplyDeleteFirst, I'll list the facts.
- a is less than b
- a is less than c
- b is less than c
- the sum of 2 variables is 2986
- the sum of 2 variables is 3464
- the sum of 2 variables is 3550
Since there are only 3 values with no repeats of one, there can only be 3 sums. In that case, there can only be a larger value, with larger variables. Therefore:
a + b = 2986
a + c = 3464
b + c = 3550
I can now determine the numbers by finding the difference.
a + b = 2986
a + c = 3464
That means that:
3464 - 2986 = c - b = 478
a + c = 3464
b + c = 3550
That means that:
3550 - 3464 = b - a = 86
a + b = 2986
b + c = 3550
That means that:
3550 - 2986 = c - a = 564
Let's use these values.
c is 478 more than b.
c is 564 more than a.
b is 86 more than a.
Let's now use algebra even more to solve this.
2986 = a + (a + 86)
3464 = c + (c - 564)
3550 = b + (b + 478)
Note: I have to include each variable twice in each statement to solve.
This is equivalent to
2986 = 2a + 86
3464 = 2c - 564
3550 = 2b + 478
We can now use algebra to solve.
a:
2986 - 86 = 2a + 86 - 86
2900 = 2a
2900/2 = 2a/2
1450 = a
c:
3464 = 2c - 564
3464 + 564 = 2c - 564 + 564
4028 = 2c
4028/2 = 2c/2
2014 = c
b:
3550 = 2b + 478
3550 - 478 = 2b + 478 - 478
3072 = 2b
3072/2 = 2b/2
1536 = b
a = 1450
b = 1536
c = 2014
Time to check:
1450 + 1536 = 2986
1450 + 2014 = 3464
1536 + 2014 = 3550
So yes, they are correct (and I used algebra like the question states).
a = 1450
b = 1536
c = 2014
The value of the largest number is 2014.
POTW:
ReplyDeleteThere is a simpler way without algebra, but with algebra I reach the same solution.
a+b = 2986
a+c = 3464
b+c = 3550
a+b+a+c+b+c = 2986+3464+3550
2a+2b+2c = 10000
a+b+c = 5000
a = 5000 - b - c
a = 5000 - 3550 = 1450
b = 5000 - a - c
b = 5000 - 3464 = 1536
c = 5000 - a - b
c = 5000 - 2986 = 2014
Overall: a = 1450, b = 1536, c = 2014.
Largest number would be 2014.
-Alan
1: a+b=2986
ReplyDelete2: a+c=3464
3: b+c=3550
Equation 2 -Equation 1= c-b=478 = Equation 4
Equation 4 + Equation 3 = 2c =4028
c=2014
Grade 7/8 POTW:
ReplyDeleteInfo:
- a is less than b
- a is less than b
- b is less than c
- sums: 2986, 3464, 3550
- determine the largest value.
Since the sums have a "least", a "middle" and a "greatest", we can determine that:
a+b=2986
a+c= 3464
b+c= 3550
We can determine a by finding the difference.
a+b=2986
a+c=3464
3464-2986=c-b=478
(a+b)-(a+c). Cross out the a on both sides. (c)-(b)
Determine c:
a+c=3464
b+c=3550
3550-3464=b-a=86
a+b=2986
b+c=3550
3550-2986=c-a=564.
c is 478 more than b.
c is 564 more than a.
b is 86 more than a.
So:
2986 = a + (a + 86)= 2a+86
3550 = b + (b + 478) =2b+478
3464 = c + (c - 564)=2c+564
1. 2986-86=2900/2= 1450=a
2. 3464+564=4028/2=2014=c
3. 3550-478=3072/2=1536=b
Largest number is 2014
Grade 8 POTW
ReplyDeleteA is 1450, B is 1536 and C is 2014. I did my work on paper.
That's not what the question is asking for though...
DeleteI mean, basically it is because you can see what the largest value is...
DeleteIt isn't. If you have an answer on a test and don't say the therefore statement or anything like that, you lose marks, even if it is right.
DeleteThe value of c is 2014
ReplyDeleteI solved this by finding the difference in each variable first, Since A+B is 2986, A+C is 3464 and B+C is 3550. Since 3464-2986=478 and that equation is the same as (A+C)-(A+B) or C-B we know that C is 478 greater than B. Then 3550-3464=86 we know that B is 86 more than A since that is the same as (B+C)-(A+C) or B-A. Now taking those numbers we can solve for their individual number. Since B is 86 more than A we can make this equation 2986=A+(A+86)or 2A+86. We then simplify the equation by subtracting 86 from each side to get 2900=2A. We then divide each side by 2 to get A=1450. We then add 86 to A to get B since A+86=B or 1450+86=B or B=1536 and 1536+1450 does equal 2986. We then add 478 to be to get C since B+478=C or 1536+478=C or C=2014. Adding 2014 and 1536 does get us 3550 so C=2014.
Based on process of elimination, C=2014. So, 2014 is the largest number.
ReplyDeleteSince A to C is smallest to largest, we can assume that the sums are using smallest to largest variables. A + B = 2986, A+C= 3464, and b+c= 3550. Knowing this, you can just find the differences based on each equation. b is 478 less than c, a is 86 less than b, and a is 564 less than c. Using this, 2986 - 86 = 2900. 2900/2 = 1450. a = 1450, b is 1536. 3550 - 1536 = c. c is 2014. The largest number is 2014.
ReplyDeleteI did my work on a sheet of paper, but essentially;
ReplyDeletea+b=2986
a+c= 3464
b+c= 3550
And if we work through, we can get;
3464-2986=478 (c+478=b)
3550-3464=86 (b+86=a)
3550-2986=564 (c+564=a)
So now, if we know b+86=a and a+b=2986, we just subtract 86 from 2986 then divide it by 2 to get 1450.
So now we can just go through since we know the differences between each one. And we end up as 2014=c. The largest number is 2014.
Grade 8 POTW:
ReplyDeletea is equaled to 1450, b is 1536 and c is 2014. Therefore, the largest value is 2014. I did my work in my math notebook.
I turned the problem into this:
ReplyDeletea + b = 2986
a + c = 3464
b + c = 3550
I was able to find the differences in the numbers using this:
a + 86 = b
a + 564 = c
b + 478 = c
Instead of a + b = 2986, I plugged in a + 86 for b
a + a + 86 = 2986
2a = 2900
a = 1450
I plugged a into a + c = 3464
1450 + c = 3464
c = 2014
I plugged c into b + c = 3550
b + 2014 = 3550
b = 1536
a = 1450
b = 1536
c = 2014
Check:
1450 + 1536 = 2986
2986 = 2986
1450 + 2014 = 3464
3464 = 3464
1536 + 2014 = 3550
3550 = 3550
It all works.
a = 1450
b = 1536
c = 2014
I didn't answer the question.
DeleteThe largest value is 2014.
Good stuff Avi
DeleteIm kind of reaaaally late, but here's my answer + explanation:
ReplyDeleteSo the question tells us that a < b < c and that when the numbers added in pairs, the sums are 2986, 3464, and 3550.
Judging by the sums and the given values of the variables, I can tell that 2986 is the sum of a + b. 2986 is the smallest sum, and a as well as b are the two smallest numbers added together.
I can assume that 3464 is the sum of a + c, since a + c would be the next biggest number and 3464 is so.
I can also assume that the last sum, 3550, is the sum of b + c because those are the two biggest numbers. So:
a + b = 2986
a + c = 3464
b + c = 3550
Now, I need an equation. Since I know that 2986 - b = a and 3464 - c = a, I can put those in an equation. 2986 - b = 3464 - c
Now, we need the same variable in order to be able to solve for c. I can substitute b for 3550 - c. So: 2986 - (3550 - c) = 3464 - c
Now I can solve for c.
The final answer is that c is equal to 2014.
Better late than never (great algebra practice too!)
DeleteI'm a bit late, ehuehuehue. (sorry)
ReplyDeleteHere's the information we have:
- There are three different numbers
- They are written in order, smallest to largest
- a < b < c
- When they numbers are added in pairs, the sums 2986, 3464 and 3550 are obtained.
We have to determine the value of the largest number.
Ookkkaaaayy. So, judging by the sums and what we know about the variables, we know that the smaller sum would be the result of the two smallest variables, which are a and b. So, a + b is equal to 2986.
We can also assume that b + c (being the two LARGEST variables) equal the largest sum, which is 3550, and that a + c is the next biggest number, which is 3464. So, know we know this:
a + b = 2986
a + c = 2464
b + c = 3550
Now, we need an equation to be able to find out what the values are. So, let's do it this way: we know that these numbers are interlinked (which is my way of saying that all of them are added with all of them to create different sums) which means that two numbers subtracted by different numbers would inevitably equal the same number (if there are 3).
...I read that over and realized that makes sense, but it'll be understood through my next step.
Let's use the smallest number, a. That means that the sums have to have b and c taken away from them, as that will equal a, which means that the end result will equal each other.
So it'll look like this:
2986 - b = 3464 - c
Next, we need to find some way to be able to solve for c. To do this, we need to incorporate c in such a way that the only variable we need to find is c. To do this, we need to find the variable in which c is added to, which in this case, is b. Now, we need to reverse what we did before - which is to say, subtract c from the sum for b.
So now it looks like this:
2986 - (3550 - c) = 3464 - c
= 2986 = 7014 - 2c
= c = 2014
CHECK:
2014 x 2 = 4028
= 7014 - 4028
= 2986
The largest variable (c) is 2014.
méi guānxi Faustina! Better late than never
Delete