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Bunuel
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Fr12milgap
I don’t quite agree with the solution. The First Statement does not mean that there is only one, we simply know that one weighs 70 or more.
Had it been written " only ", we would have understood it like it is presented in the explanation.
Pls fix.
There’s nothing to fix.

Statement (1) says one student weighs more than 70. The question stem says the average of any two students is less than 70. So if anyone else also weighed ≥70, pairing them would push the average to 70 or more — violating the stem.

So yes, even though the word "only" isn’t used, the logic forces it. No other student can weigh 70 or more. That makes (1) sufficient. The explanation is spot on.

Also, the question asks, "How many students in the class weigh 70 kg or more?" — if Statement (1) had said "only one student weighs more than 70 kg" then it would have directly answered the question. What kind of GMAT problem would that even be?

Please take more time to carefully review both the question and the solution
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I think statement 2 is also sufficient in answering the question. If the median weight is 68 kg then it states that there could be only 1 student with weight 70 kg or more. The only other possibility that comes would be 0 students with 70 kg or more weight.
Bunuel
Please correct me if I am wrong.
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Abhishake03
I think statement 2 is also sufficient in answering the question. If the median weight is 68 kg then it states that there could be only 1 student with weight 70 kg or more. The only other possibility that comes would be 0 students with 70 kg or more weight.
Bunuel
Please correct me if I am wrong.
Thank you

You’re contradicting yourself. You gave two possible cases: one with 0 students ≥ 70 and one with 1 student ≥ 70. That alone proves that Statement (2) doesn’t lead to a unique answer, which means it’s not sufficient. The solution already explains this clearly. Please read it more carefully.
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I don’t quite agree with the solution. i feel the 2nd part is poor quality
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BinodBhai
I don’t quite agree with the solution. i feel the 2nd part is poor quality

The second statement is absolutely correct. A median of 68 only fixes the middle value when the weights are ordered but does not restrict the others. All five could weigh 68 (giving zero students at or above 70), or the set could be something like {66, 68, 68, 68, 70} (giving one student at or above 70). Since both are possible, the statement is not sufficient.
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I like the solution - it’s helpful. Most of us are confused and don't agree with the solution because when we read the stem, we assume that one would be 70 kg. We should not. The stem says the average is less than 70 kg, so it could be anything..

The next question is, how many are 70kg? Statement (1) tells you that while with statement (2) we have a yes and a no situation, as explained by Bunuel.
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I like the solution - it’s helpful.
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I might have a logical flaw here, but doesn't the question itself already infer how many students can weigh 70kg or more?

If we randomly take any two students from the group of five and their arithmetic mean is below 70, means as soon as there is more than one student weighing >70kg, it pushes the mean above 70?

E.g. Student A: 70kg, Student B: 70kg -> (70+70)/2 = 70, which is not less than 70, but exactly equal to it.
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repellatearum
I might have a logical flaw here, but doesn't the question itself already infer how many students can weigh 70kg or more?

If we randomly take any two students from the group of five and their arithmetic mean is below 70, means as soon as there is more than one student weighing >70kg, it pushes the mean above 70?

E.g. Student A: 70kg, Student B: 70kg -> (70+70)/2 = 70, which is not less than 70, but exactly equal to it.
From the stem alone, the number of students who weigh 70 kg or more can only be 0 or 1,
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