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When processing iron ore into stainless steel, high-purity oxygen
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06 Nov 2020, 00:29
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When processing iron ore into stainless steel, high-purity oxygen is blown through, removing 1.5% of all impurities. Then only 30% of the remaining metal alloy can be used to create rust-free stainless steel. If 5 kg of stainless steel is desired, how much starting iron ore is needed?
a. 19.61 kg b. 16.9 kg c. 6.73 kg d. 7.64 kg e. 16.5 kg
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Re: When processing iron ore into stainless steel, high-purity oxygen
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06 Nov 2020, 04:57
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rajatchopra1994 wrote:
When processing iron ore into stainless steel, high-purity oxygen is blown through, removing 1.5% of all impurities. Then only 30% of the remaining metal alloy can be used to create rust-free stainless steel. If 5 kg of stainless steel is desired, how much starting iron ore is needed?
a. 19.61 kg b. 16.9 kg c. 6.73 kg d. 7.64 kg e. 16.5 kg
Let x kg of iron ore be required. After blowing (100 - 1.5) = 98.5% remains and then 30% of this 98.5% is used for manufacturing rust free stainless steel.
Solving for x, we get x = 16.92 which is approximately 16.9 kg.
because of the numbers, the calculations might take some time. We can use an approximate method. Let us assume that there was no loss due to blowing, then we would need \(\frac{5 * 100}{30} = 16.65 \) kg. Now we have 2 options, 16.5 kgs and 16.9 kg. Now if we factor the loss due to impurities, we would need a little bit more than 16.65 and therefore 16.9 kg is the closest possibility.
Re: When processing iron ore into stainless steel, high-purity oxygen
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06 Nov 2020, 10:26
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rajatchopra1994 wrote:
When processing iron ore into stainless steel, high-purity oxygen is blown through, removing 1.5% of all impurities. Then only 30% of the remaining metal alloy can be used to create rust-free stainless steel. If 5 kg of stainless steel is desired, how much starting iron ore is needed?
What is the source? The question doesn't make sense. It says that the oxygen process "removes 1.5% of all impurities". That does not mean anything remotely like "reduces the amount of ore by 1.5%". Presumably the ore is only partly "impurities" (it would be strange to extract iron ore that was entirely "impurities", because then there's no iron in it, but I'm guessing that's actually what the question is trying to say). As the question is worded, if we don't know how much of the ore is impurities, and how much is actual iron, we have no way to even use the "1.5%" figure.
And then the question starts introducing new terms, and I don't know what they mean. It talks about "30% of the remaining metal alloy". Is that the same thing as "30% of the remaining iron ore"? Or is the alloy only part of the ore? I'm not a professional smelter, so I have no idea. Then it talks specifically about making "rust-free stainless steel", but only asks about how much "stainless steel" we're making. Can some of that stainless steel be non-rust-free? Or is "rust-free stainless steel" the same thing as plain "stainless steel". I imagine that's the inference we're meant to draw, but if so, what reason can there be for including the description "rust-free" sometimes but not always? The right answer here is also only an approximation, and any time a real GMAT question is not asking for an exact value, the question will always say so.
Real GMAT questions are far more careful with language than this one is, so I'd suggest practicing from those, or from other sources that word their questions more carefully. Mathematically, if I take my best guess what this question means, we start with some amount of ore, remove a negligible (1.5%) portion of it that we can basically ignore since the answers are far apart. Then 30% of what we have is 5 kilograms. So (3/10)x = 5, and x = 50/3 which is roughly 16.7. Since we removed a tiny amount to begin with, the answer is slightly larger than that, and B is approximately correct.
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