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Be sure to check last week's solution. It will help you for Thursday's forthcoming Unit test. Further, this week's POTW is for Algebra practice as well! Good luck.
- oldest child: 1000 and 1/10 of what remains - second oldest: 2000 and a 1/10 of what remains - thrid oldest: 3000 and a 1/10 of what remains - rest was distrubuted and each child receives the same amount. - how many children were there?
I'm guessing we have to solve this algebraicly because it's called "practice for algebra unit test"..... So, guess I'm going to have to do that first.
Let... - x represent the original amount of money - y represent how much each child receives - meaning that y/x would be the number of children
First child got y= 1000 + (x-1000)/10 Second child got y= 2000+ (x-2000)/10 Third child y- 3000 + (x-2000)/10 and so on.
We can use the first and second children to solve this question. x= 1000 + (x-1000)/10 Multiply by each side by 10 to get rid of the "/10" 10x= 10000 + y-1000 Reverse to find y y= 10x - 9000
Second child: Gets 2000 and 1/10 of the remainder after the first child receives their fair share fo the money. Remove 2000. Do the same as the first child. x= 2000 + 1/10 (y-x-2000) 10x= 20000 + y - x - 2000 y= 11x - 18000 Put both into equation since they are equal (question said they received the same amount) 10x-9000 = 11x - 18000 If 10x= 9000 and 11x is 18000, then x must be 9000 since 11x-10x=x and 18000-9000=9000 If I replace in the euqation again, i can find y. y= (10 x 9000) - 9000 y = 90000 - 9000 y = 81000
Number of children is y/x = 81000/9000 = 9. There were 9 children and each of them recieved $9000. :)
My tab got deleted earlier because of WiFi issues, so I'll start where I left off. Let x be the amount of money at the start, let y be the amount of money each child receives.
Use the first and second children to answer the main question. X = 1000 + (X-1000)/10 10x = 10000 + y-1000 y= 10x - 9000
Repeat the equation above but using 2000 instead of 1000. X = 2000 + 1/10 (Y-X-2000) 10x= 20000 + Y-X-2000 y= 11x - 18000
Then compare the two simplified equations to find the amount of money received in total as both children received the same amount of money 10X - 9000 = 11X - 18000 After simplifying, 11x - 10x = x, and 18000-9000 = 9000. X is equal to 9000. Then I can substitute X in the equation. Y = (10 x 9000) - 9000 Y = 90000 - 9000 Y = 81000 The number of children can be found by dividing 81000 by 9000 because each received $9000. 81000/9000 = 9. 9 children received money.
let x = the amount of money that each child recieves let y = the total amount of money
First child: x = 1000 + 1/10(y-1000) x - 1000 = 1/10(y-1000) 10x - 10000 = y - 1000 10 x - 9000 = y y = 10x - 9000
Now I know the value of y and I can substitute it in later to find the value of x.
Second child: x = 2000 + 1/10(y-x-2000) 10x = 20000 + y - x - 2000 11x = 20000 + y - 2000 11 x = 18000 + y 11x - 18000 = y
Now that the y is isolated, I can substitute it. y = 11x - 18000 10x - 9000 = 11x - 18000 10x + 9000 = 11x 9000 = x This means that each child receives $9000. Now I know the value of x and I can plug it into the equation to find the value of y.
y = 10x - 9000 y = 10(9000) - 9000 y = 90000- 9000 y = 81000 This means that there the father gave a total of $81000 to his kids.
To find out how many kids there are, I have to divide the total amount of money by the amount of money that each kids receives.
Grade 7/8 POTW I did the question earlier, but I think it didn't send so I will briefly talk about it. I got 3 answers for this question. For the first one, the one considered right by the actual POTW answer in the next question, I got 9 using pretty much the same algebraic method as some of the other students and the answer itself. I don't want to talk about that descriptively since it was what literally everyone pretty much did. My next answers are 0 and 1. This is because technically, the question did say that he gave it to his children, but if he had no children, $0 to no child and that is kind of it. I'm not sure if this counts though. For 1, if there is 1 child, he might have $1000 and he gives $1000 and then 1/10 of $0. I don't know if that counts either. Anyways, there were 0 or 1 or 9 children.
Sorry for not being on top of the POTW. 1000 for first child+1/10 of remaining 2000 for second child+1/10 of remaining 3000 for third child+1/10 of remaining.
Let x = Amount of money child recieves Let y = Total amount of money
First child: x=1000+(y-1000)/10 10x=10000+y-1000 y=10x-9000
Second Child: x=2000+(y-x-2000)/10 10x=20000+y-x-2000 10x+x=20000+y-2000 11x=18000+y y=11x-18000
Since both equations now equal y, we can make them both equal each other 10x-9000=11x-18000 -9000+18000=11x-10x 9000=x
Now we plug 9000 into our first equation y= 10(9000)-9000 y=90000-9000 y=81000
POTW: Let y = total amount of money let x = amount of money each child gets
expressions for amount of money each child gets: oldest child: 1000 + 1/10(y-1000) second oldest: 2000 + 1/10(y-x-2000) third oldest: 3000 + 1/10(y-2x-3000)
Since each child gets the same amount of money, the expression for the oldest child should be equal to the expression for the second oldest child.
In the end, I found out that there were 9 children in total. Each one got $9000.
- oldest child: 1000 and 1/10 of what remains
ReplyDelete- second oldest: 2000 and a 1/10 of what remains
- thrid oldest: 3000 and a 1/10 of what remains
- rest was distrubuted and each child receives the same amount.
- how many children were there?
I'm guessing we have to solve this algebraicly because it's called "practice for algebra unit test"..... So, guess I'm going to have to do that first.
Let...
- x represent the original amount of money
- y represent how much each child receives
- meaning that y/x would be the number of children
First child got y= 1000 + (x-1000)/10
Second child got y= 2000+ (x-2000)/10
Third child y- 3000 + (x-2000)/10
and so on.
We can use the first and second children to solve this question.
x= 1000 + (x-1000)/10
Multiply by each side by 10 to get rid of the "/10"
10x= 10000 + y-1000
Reverse to find y
y= 10x - 9000
Second child:
Gets 2000 and 1/10 of the remainder after the first child receives their fair share fo the money. Remove 2000. Do the same as the first child.
x= 2000 + 1/10 (y-x-2000)
10x= 20000 + y - x - 2000
y= 11x - 18000
Put both into equation since they are equal (question said they received the same amount)
10x-9000 = 11x - 18000
If 10x= 9000 and 11x is 18000, then x must be 9000 since 11x-10x=x and 18000-9000=9000
If I replace in the euqation again, i can find y.
y= (10 x 9000) - 9000
y = 90000 - 9000
y = 81000
Number of children is y/x
= 81000/9000
= 9.
There were 9 children and each of them recieved $9000.
:)
Shouldn't it be x/y?
DeleteI think you switched up your variables.
There were a total of nine children receiving $9000 each in the will.
ReplyDeleteGrade 8 POTW
ReplyDeleteThere was a total of $81,000 divided evenly among 9 children, with each child recieving a $9000 inheritance. I did my work on paper.
My tab got deleted earlier because of WiFi issues, so I'll start where I left off.
ReplyDeleteLet x be the amount of money at the start, let y be the amount of money each child receives.
Use the first and second children to answer the main question.
X = 1000 + (X-1000)/10
10x = 10000 + y-1000
y= 10x - 9000
Repeat the equation above but using 2000 instead of 1000.
X = 2000 + 1/10 (Y-X-2000)
10x= 20000 + Y-X-2000
y= 11x - 18000
Then compare the two simplified equations to find the amount of money received in total as both children received the same amount of money
10X - 9000 = 11X - 18000
After simplifying, 11x - 10x = x, and 18000-9000 = 9000. X is equal to 9000.
Then I can substitute X in the equation.
Y = (10 x 9000) - 9000
Y = 90000 - 9000
Y = 81000
The number of children can be found by dividing 81000 by 9000 because each received $9000.
81000/9000 = 9.
9 children received money.
Grade 8 POTW:
ReplyDeleteMy work is done here:
https://docs.google.com/document/d/1Q6eTTob9ozzsX21xtl0wAgE6EMXTuuAXu2U9diol4KA/edit
There were a total of 9 children.
let x = the amount of money that each child recieves
ReplyDeletelet y = the total amount of money
First child:
x = 1000 + 1/10(y-1000)
x - 1000 = 1/10(y-1000)
10x - 10000 = y - 1000
10 x - 9000 = y
y = 10x - 9000
Now I know the value of y and I can substitute it in later to find the value of x.
Second child:
x = 2000 + 1/10(y-x-2000)
10x = 20000 + y - x - 2000
11x = 20000 + y - 2000
11 x = 18000 + y
11x - 18000 = y
Now that the y is isolated, I can substitute it.
y = 11x - 18000
10x - 9000 = 11x - 18000
10x + 9000 = 11x
9000 = x
This means that each child receives $9000.
Now I know the value of x and I can plug it into the equation to find the value of y.
y = 10x - 9000
y = 10(9000) - 9000
y = 90000- 9000
y = 81000
This means that there the father gave a total of $81000 to his kids.
To find out how many kids there are, I have to divide the total amount of money by the amount of money that each kids receives.
81000 / 9000 = 9
There are a total of 9 children.
(Check was done on paper)
Grade 7/8 POTW
ReplyDeleteI did the question earlier, but I think it didn't send so I will briefly talk about it.
I got 3 answers for this question. For the first one, the one considered right by the actual POTW answer in the next question, I got 9 using pretty much the same algebraic method as some of the other students and the answer itself. I don't want to talk about that descriptively since it was what literally everyone pretty much did.
My next answers are 0 and 1. This is because technically, the question did say that he gave it to his children, but if he had no children, $0 to no child and that is kind of it. I'm not sure if this counts though.
For 1, if there is 1 child, he might have $1000 and he gives $1000 and then 1/10 of $0. I don't know if that counts either.
Anyways, there were 0 or 1 or 9 children.
Sorry for not being on top of the POTW.
ReplyDelete1000 for first child+1/10 of remaining
2000 for second child+1/10 of remaining
3000 for third child+1/10 of remaining.
Let x = Amount of money child recieves
Let y = Total amount of money
First child:
x=1000+(y-1000)/10
10x=10000+y-1000
y=10x-9000
Second Child:
x=2000+(y-x-2000)/10
10x=20000+y-x-2000
10x+x=20000+y-2000
11x=18000+y
y=11x-18000
Since both equations now equal y, we can make them both equal each other
10x-9000=11x-18000
-9000+18000=11x-10x
9000=x
Now we plug 9000 into our first equation
y= 10(9000)-9000
y=90000-9000
y=81000
81000/9000=9
So there were a total of 9 children.
Grade 7/8 POTW:
ReplyDelete- oldest child: $1000 + 1/10 of what remains
- second eldest: $2000 + 1/10 of what remains
- third eldest: $3000 + 1/10 of what remains
Let x = amount of money each child receives
Let y = the total amount of money
First Child:
x = 1000+(y-2000)/10
Second Child:
x = 2000+(y-x-2000)/10
Let's figure out the first and second child's uh... okay, I don't know what to call it. But let's figure out what they received.
First Child:
10x = 10000+y-1000
y = 10x-9000
Second Child:
10x = 20000+y-x-2000
10x+x = 20000+y-2000
11x+x=20000+y-2000
y=11x-18000
So, both equations equal y. As both equal the same thing (says so in the equation) we can put them both into the equation.
10x-9000 = 11x-18000
-9000+18000 = 11x-10x
9000 = x
If the answer is indeed 9000, we can use this to find y by plugging it into the first equation:
y = 10(9000)-9000
y = 90000-9000
y = 81000
Now we divide y by x.
81000/9000 = 9
There were 9 children in total.
POTW:
ReplyDeleteLet y = total amount of money
let x = amount of money each child gets
expressions for amount of money each child gets:
oldest child: 1000 + 1/10(y-1000)
second oldest: 2000 + 1/10(y-x-2000)
third oldest: 3000 + 1/10(y-2x-3000)
Since each child gets the same amount of money, the expression for the oldest child should be equal to the expression for the second oldest child.
In the end, I found out that there were 9 children in total. Each one got $9000.
I'm not sure if my previous answer sent, so I'll just send another one to be sure.
ReplyDeleteThere were 9 kids in total and each got $9000 out of the $81000.