Question 1196875: Hi
Boxes A and B contained 150 marbles altogether. When 1/3 of the marbles in box A were transferred to box B and 12 marbles were taken out from B there were twice as many marbles in B than in A.
How many were in B in the beginning.
Thanks
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
x = amount in box A in the beginning
150-x = amount in box B in the beginning
This is because x and 150-x add to 150 marbles total
x is some positive whole number.
Now take 1/3 of the marbles in box A.
So we'll take 1/3 of x to get (1/3)x or x/3
This amount gets subtracted from the amount in box A
x - (1/3)x = (2/3)x
So if we dump 1/3 of the contents of A, then we're left with 2/3 of the original contents as the remainder.
The amount x/3 is then transferred to box B
So we'll add it to the (150-x) mentioned earlier.
(150-x)+x/3
450/3 - 3x/3+x/3
(450 - 3x+x)/3
(450 - 2x)/3
This represents how many marbles are in box B after adding on the x/3 amount. This is before the 12 marbles are taken away.
Then we take away 12 marbles from box B. I'm assuming these are just tossed aside and not placed in box A (since the instructions don't mention these 12 going to box A)
We'll subtract 12 from the previous expression
(450 - 2x)/3 - 12
(450 - 2x)/3 - 36/3
(450 - 2x - 36)/3
(414 - 2x)/3
This is the final marble count for box B. This is twice as much compared to what's in box A, which was (2/3)x
Amount in box B = 2*(amount in box A)
(414 - 2x)/3 = 2*( (2/3)x )
(414 - 2x)/3 = (4x)/3
414 - 2x = 4x
414 = 4x+2x
6x = 414
x = 414/6
x = 69
At the start, there are 69 marbles in box A and 150-x = 150-69 = 81 marbles in box B
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Check:
69 marbles in box A at first
81 marbles in box B at first
1/3 of 69 = (1/3)*69 = 23 marbles taken out of box A and moved to box B
69-23 = 46 marbles in box A now (i.e. (2/3)*69 = 46)
81+23 = 104 marbles in box B now
Now take 12 away from box B
104 - 12 = 92
As mentioned before, I'm assuming these 12 are tossed aside and not placed in box A.
We have this
46 marbles in box A
92 marbles in box B
The ratio of which is 92/46 = 2
Which shows 92 = 46*2, i.e. the amount in B is twice of that compared to A.
The answer has been fully verified.
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Answer: 81 marbles in box B in the beginning
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
12 of the 150 marbles were removed from B at the end, leaving 138 marbles in the two boxes.
At that point, there were twice as many in B than in A -- i.e., 2/3 of the 138 marbles were in B and 1/3 of them in A. That makes 46 in A and 92 in B.
Prior to that, 1/3 of the marbles in A had been moved to B, so the number of marbles remaining in A was 2/3 of the number originally in A. We know that number of marbles was 46; so the number of marbles originally in A was 46*(3/2) = 69.
(Algebraically, that is  -->  )
And that means the number originally in B was 150-69 = 81.
ANSWER: 81 marbles in B originally
CHECK:
start: A = 69; B = 81
move 1/3 of the marbles in A (23) to B: A = 69-23 = 46; B = 81+23 = 104
remove 12 from B: A = 46; B = 104-12 = 92.
92 = 2(46)
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