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?
A. 250 kg
B. 275 kg
C. 300 kg
D. 266.66 kg
Answer: Option D
Solution(By Examveda Team)
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
This Question Belongs to Arithmetic Ability >> Percentage
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Let's solve this problem step by step:
1. We know that the ore contains 25% of an alloy that has 90% iron. This means that in every kilogram of the ore, there is 0.25 kg of the alloy.
2. In the 0.25 kg of alloy, 90% is iron. So, the amount of iron in every kilogram of the ore is 0.25 kg * 0.90 = 0.225 kg.
3. We want to obtain 60 kg of pure iron. So, we need to find out how many kilograms of the ore are required to obtain this amount.
4. Let's assume that x kg of the ore is needed to obtain 60 kg of pure iron.
5. From step 2, we know that in every kilogram of the ore, there is 0.225 kg of iron. Therefore, in x kg of the ore, there will be a total of 0.225 kg * x kg of iron.
6. Since we want to obtain 60 kg of pure iron, we can set up the equation: 0.225 kg * x kg = 60 kg.
7. Solving for x, we have: x = 60 kg / 0.225 kg = 266.6667 kg.
Rounding to the nearest whole number, we get x ≈ 267 kg.
Therefore, the answer is D. 266.66 kg.
how 22.5 kg of iron contains 100kg of ore