M36-86

Question

In a certain class of 5 students, the average (arithmetic mean) weight of any two students in the group is less than 70 kg. How many students in the class weigh 70 kg or more?

(1) One of the students weighs more than 70 kg.
 
(2) The median weight of all 5 students is 68 kg.

Bunuel · Answer

Official Solution:

In a certain class of 5 students, the average (arithmetic mean) weight of any two students in the group is less than 70 kg. How many students in the class weigh 70 kg or more? 
 
(1) One of the students weighs more than 70 kg. 
 
There cannot be any other student who weighs more than or equal to 70 kg. because if there is, then the average (arithmetic mean) weight of this student and the student mentioned in the above statement would be more than 70 kg., which would contradict the stem. So, there is only one student in the class who weighs 70 kg or more. Sufficient.
 
(2) The median weight of all 5 students is 68 kg. 
 
Each of the 5 students can weigh 68 kg. and in this case the answer to the question would be NONE but it can be that the weights are {66, 68, 68, 68, 70} and in this case the answer to the question would be ONE. Not sufficient.

Answer: A

Fr12milgap · Answer

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.

Bunuel · Answer

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

Abhishake03 · Answer

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

Bunuel · Answer

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.

BinodBhai · Answer

I don’t quite agree with the solution. i feel the 2nd part is poor quality

Bunuel · Answer

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.

BinodBhai · Answer

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.

agrasan · Answer

I like the solution - it’s helpful.

repellatearum · Answer

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.

Bunuel · Answer

From the stem alone, the number of students who weigh 70 kg or more can only be  0  or 1,

Akshayjain806 · Answer

Hi All,

For A : {68,68,68,68,71}--it works as u can any two and avg will be less than 70.

Case 2:{68,68,68,70,71}--It will not work as avg of two 70 and 71 will be more than 70.

So, A as only answer which is 1.

For B: {{68,68,68,68,71}} use this case then ans will be 1 and if u use {68,68,68,68,70} then answer will be 0 and it is not impacting any info given like median is 68 and avg of any two is less than 70. So, it has two answers, so insufficent.

oshin29 · Answer

Wha if the guy above 70 weights 90 kg and he is paired with a guy weighing 50. still the avg of 2 would be more than 70

Bunuel · Answer

"The average (arithmetic mean) weight of any two students in the group is less than 70 kg" means  every possible pair of students in the class  must have an average weight of less than 70 kg.

So in your example, if one student weighs 90 kg and another weighs 50 kg, their average is 70 kg. Since the stem requires the average to be  less than 70 , that pair is not allowed.

oshin29 · Answer

exactly. so in this case even statement 1 fails

Bunuel · Answer

No! That example does not make statement (1) fail. It simply means that  90 kg and 50 kg cannot be two of the students , because that would violate the condition in the stem.

Statement (1) tells us that one student weighs more than 70 kg. Since every pair must average less than 70 kg, none of the other four students can weigh 70 kg or more. Therefore, exactly one student weighs 70 kg or more. Statement (1) is sufficient.

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