An ore contains 25% of an alloy that has 90% iron. Other than this, in the remaining 75% of the ore, there is no iron. How many kilograms of the ore are needed to obtain 60 kg of pure iron?

Answer: Option D

Let there is 100 kg of ore.
25% ore contains 90% off Iron that means 25 kg contains;
$$\frac{25\times 90}{100}=22.5\text{kg}\text{iron}$$
22.5 kg Iron contains 100 kg of ore.
Then, 1 kg of iron contains = $$\frac{25}{100}\text{kg}\text{ore}$$
Hence, 60 kg iron contains
= $$\frac{100\times 60}{22.5}$$
= 266.66 kg ore