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# Solving an Algebra I (8th Grade) Word Problem

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Joined: 15 Jan 2013
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15 Jan 2013, 00:21
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4 kids have a total of 92 marbles

Bill has 8 more than Allen

Charles has 2 times more than Bill

Daryl has 3 times more than Charles

How many marbles does Charles have?

My daughter has gotten this far:

Bill = 8 + A

Allen = A

Charles = 2B

Daryl = 3C

8+A + A + 2B + 3C = 92

8 + 2A + 2B + 3C = 92

Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her?

Thanks
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15 Jan 2013, 01:54
Consider Bill has x marbles.
Bill has 8 more than Allen, i.e. Allen has (x-8) marbles.
Charles has 2 times more than Bill, i.e. Charles has 2x marbles.
Daryl has 3 times more than Charles, i.e. Daryl has 3*2x = 6x marbles.

All 4 kids have total 92 marbles.
Hence $$x + (x-8) + 2x + 6x = 92$$
i.e. $$10x = 100$$
i.e. $$x = 10$$

Therefore, Charles has 2x=20 marbles.

Calculating the rest, Allen=2, Bill=10, Charles=20, Daryl=60, Total=92
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15 Jan 2013, 04:19
tara15225 wrote:
4 kids have a total of 92 marbles

Bill has 8 more than Allen

Charles has 2 times more than Bill

Daryl has 3 times more than Charles

How many marbles does Charles have?

My daughter has gotten this far:

Bill = 8 + A

Allen = A

Charles = 2B

Daryl = 3C

8+A + A + 2B + 3C = 92

8 + 2A + 2B + 3C = 92

Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her?

Thanks

2B = 2*(8 + A) = 16 + 2A

3C = 3*2B = 6B = 6*(8 + A) = 48 + 6A

Picking up from where you left

8 + 2A + 2B + 3C = 92

8 + 2A + 16 + 2A + 48 + 6A = 92

10A + 72 = 92

10A = 92 - 72 = 20

A = 2
B = 2 + 8 = 10
C = 2*10 = 20
D = 3*20 = 60
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15 Jan 2013, 21:48
tara15225 wrote:
4 kids have a total of 92 marbles

Bill has 8 more than Allen

Charles has 2 times more than Bill

Daryl has 3 times more than Charles

How many marbles does Charles have?

My daughter has gotten this far:

Bill = 8 + A

Allen = A

Charles = 2B

Daryl = 3C

8+A + A + 2B + 3C = 92

8 + 2A + 2B + 3C = 92

Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her?

Thanks

Allen =x
Bill=x+8
Charles=2(x+8)
Daryl=6(x+8)

Hence, x+x+8+2x+16 +6x+48 =92
10x + 72=92
x=20/10=2
Hence, Charles has 2(2+8) =20
Re: Solving an Algebra I (8th Grade) Word Problem   [#permalink] 15 Jan 2013, 21:48
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