Question

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?

• A. 0.13
• B. 13
• C. 130
• D. 500

Answer 1

The Correct Option is C

Given two bags of iron ore, one weighing 180 kg 250 gm and the other 270 kg 50 gm. We need to transfer some iron ore from the first bag to the second such that the first bag's weight becomes two-thirds of the second bag's weight.

Convert the weights of both bags to grams for easier calculations:

• First bag: \(180 \text{ kg } 250 \text{ gm} = 180,250 \text{ gm}\)
• Second bag: \(270 \text{ kg } 50 \text{ gm} = 270,050 \text{ gm}\)

Let \( x \) be the amount of iron ore in grams transferred from the first bag to the second bag. After transferring, the new weight of the first bag will be \(180,250 - x\) and for the second bag, it will be \(270,050 + x\).

According to the problem, the weight of the first bag should be two-thirds of the second bag after transferring. Therefore, we have the equation:

\[\frac{2}{3}(270,050 + x) = 180,250 - x\]

Solving for \( x \):

• \(2(270,050 + x) = 3(180,250 - x)\)
• \(540,100 + 2x = 540,750 - 3x\)
• \(2x + 3x = 540,750 - 540,100\)
• \(5x = 650\)
• \(x = \frac{650}{5} = 130\)

Thus, 130 grams of iron ore should be taken from the first bag and added to the second to satisfy the condition.