marbles in a box

eddy2017

eddy2017

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Hi, dear teacher and friends:

Two brothers have taken marbles out of a box. One took out 4/7 of the marbles, and the other took out 7/9 of the marbles left in the box. In the end only 10 marbles were left in the box, how many marbles were initially in the box?.

I'll appreciate your hints and clues!.
eddy.
 
eddy2017 said:
Two brothers have taken marbles out of a box. One took out 4/7 of the marbles, and the other took out 7/9 of the marbles left in the box. In the end only 10 marbles were left in the box, how many marbles were initially in the box?.
Click to expand...
What fraction of the original total was left after the first removal? What fraction of the original total was left after the second removal? That fraction of what number is 10?
 
Let x represent the number of marbles initially in the box.

One brother took 4/7 of the marbles. So how many marbles are left (in terms of x).

Then the other brother took 7/9 of the remaining marbles which means that he left what part? So what part is left? This part also equals 10.
 
Thanks a lot for replying.
I am going tovdo this:
I am gonnaxtry and answer your questions separately, for example, Dr Peterson asked me three questions. I will try to reason that out nad answer them
then I will read what Mr Jomo asked
and do that too.
that will help me get practice in reasoning things out.
 
eddy2017 said:
Thanks a lot for replying.
I am going tovdo this:
I am gonnaxtry and answer your questions separately, for example, Dr Peterson asked me three questions. I will try to reason that out nad answer them
then I will read what Mr Jomo asked
and do that too.
that will help me get practice in reasoning things out.
Click to expand...
Dr Peterson and I basically said the same thing. I added in that you should denote the total number of marbles as x.
My advice to you is to let x=the total number of marbles and then follow Dr Peterson's post.
 
Alright Mr J. I'll do that. Thanks.
Letting x be the total of marbles in the box,
the first brother took 4/7 of x = 4/7x
the second brother took 7/9 of x = 7/9x

Does that make sense?.
Thanks.

What fraction of the original total was left after the second removal?. (Need a hint here)
 
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We need to know not how much the brothers took out but rather how much remains. If the 1st brother took out 4/7 of x, then (3/7)x remains.

Now the 2nd brother doe NOT take out (7/9) of x. Why? Read the problem! He took out 7/9 of what was remaining after his brother took out some. So how much was left after the 2nd brother took out his marbles. There are two answers for this. One in terms of x and the other answer is 10.
Continue...
 
You are supposed to be learning algebra. So use it. There are three unknown numbers. Assign symbols to EACH of them.

[MATH]f = \text {number of marbles taken by first brother.}[/MATH]
[MATH]s = \text {number of marbles taken by second brother.} [/MATH]
[MATH]t = \text {total number of marbles.}[/MATH]
The idea of identifying and labeling the relevant numbers that you don’t know YET is far easier than trying to formulate them in your head in terms of a single unknown.

You identify and label what you do not know. That allows you to think in concrete terms and to communicate.

With three unknowns, we need three equations. This involves translating the information you have been given into algebraic terms.

[MATH]f = \dfrac{4}{7} * t = \dfrac{4t}{7}.[/MATH]
You translated that correctly. Well done.

[MATH]s = \dfrac{7}{9} * (t - f).[/MATH]
Why is that correct?

Now, written in algebra, the third thing we know is

[MATH]t - f - s = WHAT?[/MATH]
You may not know how to proceed from here. If so, please say so.
 
Right, right.
you asked at # 3 ;
Then the other brother took 7/9 of the remaining marbles which means that he left what part? So what part is left? This part also equals 10.

I think i got it.
The frist brother took 4/7 out of x amount
(4/7)x
the other brother took 7/9 out of the 3/7 left from the first removal, so i'll equal all that to 10 9the amount of marbels left in the box)
(4/7)x + (7/9)(3/7)x=10
 
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Thnaks Mr Jeff!. I will study your analisis. Is the way I presented above correct too?.
 
eddy2017 said:
Thnaks Mr Jeff!. I will study your analisis. Is the way I presented above correct too?.
Click to expand...
Sadly, your analysis is incorrect. 10/9 is greater than 1 so you are talking about more marbles than are in the box.

I’ll be honest: I am attacking this a different way from Jomo and Dr. Peterson. Theirs is the traditional way. If you find their way more intuitive than mine, by all means use their way. I have found, from students that I have tutored face to face, that some students find it easier to start by identifying what is unknown and then writing down algebraically what we do know.
 
No, no, i like your way. I just want to learn both ways.
Let me do it your way and see what i can come up with. Thanks a lot.

I think there is a mistake in the equation above. I should be:
(4/7)x + (7/9)(3/7)x + 10 = x
 
eddy2017 said:
Right, right.
you asked at # 3 ;
Then the other brother took 7/9 of the remaining marbles which means that he left what part? So what part is left? This part also equals 10.

I think i got it.
The frist brother took 4/7 out of x amount
(4/7)x
the other brother took 7/9 out of the 3/7 left from the first removal, so i'll equal all that to 10 9the amount of marbels left in the box)
(4/7)x + (7/9)(3/7)x=10
Click to expand...
It's important to label each quantity you write. In your equation, you are adding the amount taken by the first brother, plus the amount taken by the second brother, and equating that to the amount remaining! Does that make sense?

A small change will correct this. (There are a couple ways you can do it.)
 
I think i should have add 10 left in the box before equating it, and it should be equating to the unknown quantity, not to 10
(4/7)x + (7/9)(3/7)x + 10 = x
what do you think, now?
 
eddy2017 said:
I think i should have add 10 left in the box before equating it, and it should be equating to the unknown quantity, not to 10
(4/7)x + (7/9)(3/7)x + 10 = x
what do you think, now?
Click to expand...
eddy2017 said:
Solving for x,
x=105
Click to expand...

Good!

Another way would be to say that the amount left after the second turn is 2/9 of the 3/7 left by the first, so that 2/9*3/7 x = 10, which leads to the same result.
 
Now you should check your answer.

If you want to understand my way, let me know.
 
Here is how I would do it.
x = total number of marbles.
1st brother took took 4/7 of the marbles leaving 3/7 of the marbles behind. After 1st brother took his marbles there are (3/7)x marbles left.
The 2nd brother took 7/9 of the (3/7)x marbles leaving 2/9 of the (3/7)x marbles behind. Now there are (2/9)(3/7)x = (2/21)x marbles left.

So (2/21)x = 10 or x = 10(21/2) = 105. So originally there were 105 marbles.
 
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