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

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.

M36-86

Bunuel · Accepted 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

AnubhutiSharma · Answer

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

AshutoshB · Answer

Imo Answer will be :E

Correct mei id im woring

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mohit67890 · Answer

Answer is C, I think.

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RooIgle · Answer

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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dileeprk · Answer

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)

sp2018 · Answer

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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chetan2u · Answer

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

Bunuel · Answer

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

PierTotti17 · Answer

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.

ksrutisagar · Answer

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

Kinshook · Answer

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

AkshayBhushan · Answer

Hello Bunnel,

Kindly explain why no other student can have weight more than 70 ?
As we dont have any other information on the last two students of the group?

Bunuel · Answer

The stem says that the average weight of   any  two students  must be less than 70 kg. This means that no matter which two students we pick, their average weight must be less than 70 kg.

Statement (1) says that  one student weighs more than 70 kg . So no other student can weigh 70 kg or more, because then the average weight of those two students would be more than 70 kg, contradicting the stem. Therefore,  there must be only one student who weighs 70 kg or more .

ohhno · Answer

Let the five students' weights in increasing order be:
w1≤w2≤w3≤w4≤w5w_1\le w_2\le w_3\le w_4\le w_5
We are told the average weight of  any two students  is less than 70 kg.
That means for every pair:
wi+wj2<70\frac{w_i+w_j}{2}<70
or
wi+wj<140w_i+w_j<140
In particular, the  two heaviest students  must satisfy:
w4+w5<140w_4+w_5<140
Therefore,  both cannot weigh 70 kg or more . So the number of students weighing 70 kg or more can only be  0 or 1 .
We need to determine which.
Statement (1)
One student weighs  more than 70 kg .
Since at most one student can weigh 70 kg or more, this tells us there is  exactly 1  such student.
✅  Statement 1 alone is sufficient. 
 
Statement (2)
Median weight = 68 kg.
So:
w3=68w_3=68
This tells us the first three students are at most 68 kg, but w4w_4 and w5w_5 could be:
 
 69, 69 →  0  students ≥70 
 70, 70 → impossible because 70+70=14070+70=140, not less than 140 
 70, 69 →  1  student ≥70 
 75, 60 → but ordering requires w5≥w4w_5\ge w_4, so possibilities can still give  1  student ≥70.  
For example:
60,65,68,69,6960,65,68,69,69
gives 0 students ≥70.
But:
60,65,68,69,7060,65,68,69,70
also satisfies the pairwise condition, giving 1 student ≥70.
❌  Statement 2 alone is insufficient. 
Answer:  (A)  ✅
 Statement 1 alone is sufficient, but statement 2 alone is not.

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

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

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615 to 715 in 15 Days (Ria's Strategies)

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