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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 ?

Solution

Correct Option: 3

Bag1 = 180250 g, Bag2 = 270050 g. Total = 450300 g (preserved). Need Bag1 = 23\frac{2}{3} Bag2, so Bag1 : Bag2 = 2 : 3, total 5 parts. Bag1 = 25×450300=180120\frac{2}{5}\times 450300=180120 g. Transfer = 180250180120=130180250-180120=130 g.

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