In a certain class of 5 students, the average (arithmetic mean) weight

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
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Answer should be E as none of the statements even if combined together is sufficient to answer the question.
[Given] : 5 students, average weight of any 2 students is less than 70 kg.
[Required] : Weight of all students

1. One of the students weight exceeds 70 kg. But we do not know if only one student exceeds 70 kg weight . Hence weight of other students cannot be derived . The only inference is, one student weights 70+ kg so one other student must weight less than 70 kg.
2. The median weight of all 5 students is 68 kg. So maybe all students weight 68 kg or maybe the sequence is 66, 67, 68, 69 ,70 or maybe the sequence is 66.5 67, 68, 69, 70.5 or maybe the sequence is 66 66 68 70 70 or maybe 65 65 68 71 71 .. many sequences are possible. So we do not get any definite answe.

Even if we combine the two statements, we do not achieve any concrete conclusion on how many students exceed 70 kg weight.

Hence, option E should be correct.
Please correct me if I am wrong

Imo Answer will be :E

Correct mei id im woring

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Answer is C, I think.

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Only two cases:

1. None of the students is over 70. In that case the average of ANY two students would be assured to be below 70.
2. ONLY ONE of the students is over 70. For example, the average of 71 and 35 is clearly below 70, BUT the average of 71 and 71 is 71 what is not allowed.



S1.


Case 2. Only 1.


SUFF.



S2.

Two students on the right of 68 can be equal to 68 or greater. It can be either case.


INSUFF.




AC: A



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Statement 1 is sufficient.
Even if one of the student is more than 70, all the other students must weight lesser than 70. (as the average weight of any two students is 70)

Question says mean of any 2 students less than 70. So sum of weights of any 2 students is less than 70. Statement 1 says 1 student is 70 and more. So if we compare other students with this student their weight should be less than 70. So sufficient. Statement 2 says median is 68. So if arrange is ascending order there will be 2 students before and 2 students after 68. Of 2 students who has weight beyond 68 both of them cannot be greater than 70. But we cannot say that either is one is more than 70 or both is between 68 and 70. So not sufficient. So answer is A

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Half the problem is solved by the main statement..


If there are two numbers equal to or above 70, their average will surely be MORE than or equal to 70..
so MAX number above 70 can be ONLY one ..
second possibility is NONE ..

Our statement should be able to give us some info to eliminate any one of the two..

lets see the statements..
(1) One of the students weighs more than 70 kg.
so it gives us ONE as answer
suff
(2) The median weight of all 5 students is 68 kg.
nothing much..
all 68 ... ans 0
66,66,68,70,70.... ans 2
insuff

A
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Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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Hi,

Sorry to bring up a post from a year ago, but I am working through this New Year set and couldn't help but notice something that seems out of place.

Chetan, you gave as an example for Statement 2:

66, 66, 68, 70, 70 --> this would not fit with the restrictions of the problem. If we take ANY TWO students, their average weight should be LESS THAN 70, not less or equal to. If we take the last 2 students, (70 + 70) / 2 = 70 which is not less than 70. Hence, the second to last student should be less than 70 (either 69 or 68).

Statement 2 is insufficient because of the following 2 cases:
1) 68, 68, 68, 68, 68 whereby the answer to the target question is 0 (0 students over 70kg).
2) 66, 66, 68, 68, 71 whereby the answer is 1 student is over 70kg.

St 1 : Even in the following case

<70, <70, <70, 70, >70

The average of the highest two is greater than 70, which violates the condition given

So , the only arrangement possible is

<70, <70, <70, <70, >70

Only one student weighs 70 or more

Sufficient

St 2 : The arrangement could be

68,68,68,68,68 or 68,68,68,68,70

Zero students or one student

Not Sufficient

Choice A

There are only 2 possibilities for students weighing more than 70 kgs = 1 or 0 since for 2 or greater average of 2 students will be greater than 70 kgs.

(1) One of the students weighs more than 70 kg.
This gives us the answer = 1
SUFFICIENT

(2) The median weight of all 5 students is 68 kg.
There may be students equal to 1 or 0 having weight greater than 70 kgs
NOT SUFFICIENT

IMO A

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