Try to Solve
Boxes A, B and C contain 480 oranges altogether. 1/3 of the oranges in Box A were put into Box B. Then 1/5 of the Oranges in Box B were put into Box C. Finally, 1/6 of the oranges in Box C were put into Box A. There was an equal number of oranges in box at the end. How many oranges did each box contain at first.
Very interesting question
yeah its really an interesting one. I think I can solve.
At the end each box has 160 oranges so before we move 1/6 oranges from box C to box A so there must be (128,160,192) oranges in each box. So when we moved 1/5 of the oranges from box B to box C we must have had (128,200,152) oranges in the boxes. and when we moved 1/3 from A to B we must have had (192,136,152) oranges in each box.
Am I right?
Here is the work in order:
Configuration after third move: A=160, B=160, C=160
A+C = 320 and A + 1/6C = 160 and solve for "A" we get 128.
C = 320-128 = 192
Configuration after second move: A=128, B=160, C=192
B+C = 352 and C + 1/5B = 192 and solve for "B" we get 200.
C = 352 - 200 = 152
Configuration after first move: A=128, B=200, C=152
A+B = 328 and 1/3A + B = 200 and solve for "B" we get 136.
A = 328 - 136 = 192
At the beginning we have: A=192, B=136, C=152
Thanks!
Great thinking and you also showed your work.
Nice post
@larryphang pls help to resteem my post...its not getting the attention it deserves...thanks
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