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Solving an Algebra I (8th Grade) Word Problem

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tara15225
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Solving an Algebra I (8th Grade) Word Problem [#permalink] New post   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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Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] New post   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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Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] New post   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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Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] New post   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




Answer:
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



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: Solving an Algebra I (8th Grade) Word Problem [#permalink]
15 Jan 2013, 21:48
Moderator:
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