Bag1 = 180250 g, Bag2 = 270050 g. Total = 450300 g (preserved). Need Bag1 = 32 Bag2, so Bag1 : Bag2 = 2 : 3, total 5 parts. Bag1 = 52×450300=180120 g. Transfer = 180250−180120=130 g.
Two bags of iron ore weight 180 kg 250 gm and 270 kg 50 gm respectively. How much iron ore (in gm) must be taken out from the first bag and added to the second bag so that weight of the first bag may then be two-third of the second bag ?
Held on 7 Jun 2023 · Verified 13 Jul 2026.
0.13
13
130
500
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