Question 1196875
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12 of the 150 marbles were removed from B at the end, leaving 138 marbles in the two boxes.<br>
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.<br>
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.<br>
(Algebraically, that is {{{(2/3)x=46}}} --> {{{x=46(3/2)=69}}})<br>
And that means the number originally in B was 150-69 = 81.<br>
ANSWER: 81 marbles in B originally<br>
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)<br>