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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
pranrasvij wrote:
A is easily the right choice since it gives info about the rest (1-n/3) and n/3 => we dont need the final value of "n" in that case.

B gives us total height but no way to get the mean without more on "n".



so the mean height is the heights given at a ration 1:2 ????

please explain....
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
zisis wrote:
pranrasvij wrote:
A is easily the right choice since it gives info about the rest (1-n/3) and n/3 => we dont need the final value of "n" in that case.

B gives us total height but no way to get the mean without more on "n".



so the mean height is the heights given at a ration 1:2 ????

please explain....


yep... all you have to do is the add the 2 heights and divide by 2 to get the mean height => easy to calculate (IMHO!)
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
Bunel - I need one help. This is not a difficult question and it's easy to see that A is sufficient. However I thought B is suf too because the stimulus says N. Now N could be anything. Is it not sufficient to say avg height = sum of heights (given in the second option)/N - N is mentioned in the stem. Why do we have to concern with the actual value of N?
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
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dwivedys wrote:
Bunel - I need one help. This is not a difficult question and it's easy to see that A is sufficient. However I thought B is suf too because the stimulus says N. Now N could be anything. Is it not sufficient to say avg height = sum of heights (given in the second option)/N - N is mentioned in the stem. Why do we have to concern with the actual value of N?


Official Guide:

In data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.

Hope it helps.
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
Bunuel wrote:
(1) The average height of \(\frac{n}{3}\) people is 74.5 inches and the average height of \(\frac{2n}{3}\) people (the res of the people in the group \(n-\frac{n}{3}=\frac{2n}{3}\)) is 70 inches --> \(average \ height=\frac{sum \ of \ heights}{# \ of \ people}=\frac{74.5*\frac{n}{3}+70*\frac{2n}{3}}{n}\) --> \(n\) cancels out --> \(average \ height=74.5*\frac{1}{3}+70*\frac{2}{3}\). Sufficient.

(2) Sum of heights equals to 178 feet 9 inches --> only nominator is given. Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel,
The fact that confused me is the word tallest in statement 1.
It says "average height of \(n/3\) [highlight]tallest[/highlight] people in the group is 6 feet 2.5 inches.

Isn't this bit ambiguous? We have no clue how many people to consider in tallest category?
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
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Orange08 wrote:
Bunuel wrote:
(1) The average height of \(\frac{n}{3}\) people is 74.5 inches and the average height of \(\frac{2n}{3}\) people (the res of the people in the group \(n-\frac{n}{3}=\frac{2n}{3}\)) is 70 inches --> \(average \ height=\frac{sum \ of \ heights}{# \ of \ people}=\frac{74.5*\frac{n}{3}+70*\frac{2n}{3}}{n}\) --> \(n\) cancels out --> \(average \ height=74.5*\frac{1}{3}+70*\frac{2}{3}\). Sufficient.

(2) Sum of heights equals to 178 feet 9 inches --> only nominator is given. Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel,
The fact that confused me is the word tallest in statement 1.
It says "average height of \(n/3\) [highlight]tallest[/highlight] people in the group is 6 feet 2.5 inches.

Isn't this bit ambiguous? We have no clue how many people to consider in tallest category?


It means that if we order these n people from shortest to tallest and consider \(\frac{n}{3}\) tallest people, then their average height would be 74.5 feet.
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
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trivikram wrote:
Guys a basic conceptual question.

Suppose we have a set S = {10,20,30,40,50}

If we take individual values as 10,20 and take their av = 15

and the rest 3 values 30,40,50 and avg of these is = 40

So the group average of S {10,20,30,40,50} = 30 is not equal to the sum of individual averages 15+40 = 55 or even the av of the values 55/2 =27.5

Whats the deal here? Explanations needed please


Remember, it needs to be 'weighted average', not the simple average.
Say you have two groups. Average height of one group is 5 feet and of the other group is 6 feet. What is the average of both groups combined? It will depend on how many people each group has. Hence, it is 'weighted'. If both groups have equal number of people, the average height will be 5.5 feet. If the first group has more people, the average will be closer to 5 than to 6.
In your question, the average of 2 elements is 15 and of 3 elements is 40. If you calculate their weighted average, it will be 30.
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
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Stmt 1 : Average = Sum/Total number of items

Hence, Average * Total number of items = Sum

1 foot = 12 inches, therefore the average height of n/3 people is 74.5 inches

Sum of heights of n/3 people is 74.5n/3

Rest of the people = total – n/3 = (n – n/3) = 2n/3

Similarly, the average height of the 2n/3 people is 70 inches

Hence, the sum of the heights of 2n/3 people is 140n/3

Sum of both groups = (74.5n/3)+ (140n/3) = 214.5n/3

Average = Sum/Total number of items

therefore, (214.5n/3)/n = 214.5n/3n = 71.5 inches

Hence, stmt 1 is sufficient


Stmt 2 : This statement gives us the sum of the heights of all the people in the group but the exact value of n is unknown

Since we can not determine the average height of the group, this statement is insufficient

Answer : A
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What is the average (arithmetic mean) height of the n people in a cert [#permalink]
Bunuel wrote:
What is the average (arithmetic mean) height of the n people in a certain group?

(1) The average height of the n/3 tallest people in the group is 6 feet \(2 \frac{1}{2}\) inches, and the average height of the rest of the people in the group is 5 feet 10 inches.


The average height of the n/3 tallest people is 74.5 inches;
The average height of the remaining n - n/3 = 2n/3 people is 70 inches.

    (The average height of the group) = (The sum of the heights)/n \(= \frac{74.5*\frac{n}{3}+70*\frac{2n}{3}}{n}\)

Reduce by n:

    \(= 74.5*\frac{1}{3}+70*\frac{2}{3}\).

Sufficient.


(2) The sum of the heights of the n people is 178 feet 9 inches.

We don't know the number of students, in the group, n, so cannot get the average. Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel,
I don't understand why this is the case. I think there might be a small error in the question or in my understanding of the question.

So lets assume there are 10 people. Heights are as follows: {20,19,18,7,6,5,4,3,2,1}. Then we need to know the average height of 6 people. so n=6.

1st tells us that average height of 2 tallest people is 19.5. and the average of other 8 people is 5.75. How can you find out the average height of 6 people from this group? and by the way, which 6 people are we talking about?

I feel that the question confuses language a little bit. Instead of "the average height of the rest of the people in the group is 5 feet 10 inches" it probably should say "the average height of the rest of the people in the group n is 5 feet 10 inches".

Your response would be highly appreciated.
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
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Paagrio wrote:
Bunuel wrote:
What is the average (arithmetic mean) height of the n people in a certain group?

(1) The average height of the n/3 tallest people in the group is 6 feet \(2 \frac{1}{2}\) inches, and the average height of the rest of the people in the group is 5 feet 10 inches.


The average height of the n/3 tallest people is 74.5 inches;
The average height of the remaining n - n/3 = 2n/3 people is 70 inches.

    (The average height of the group) = (The sum of the heights)/n \(= \frac{74.5*\frac{n}{3}+70*\frac{2n}{3}}{n}\)

Reduce by n:

    \(= 74.5*\frac{1}{3}+70*\frac{2}{3}\).

Sufficient.


(2) The sum of the heights of the n people is 178 feet 9 inches.

We don't know the number of students, in the group, n, so cannot get the average. Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel,
I don't understand why this is the case. I think there might be a small error in the question or in my understanding of the question.

So lets assume there are 10 people. Heights are as follows: {20,19,18,7,6,5,4,3,2,1}. Then we need to know the average height of 6 people. so n=6.

1st tells us that average height of 2 tallest people is 19.5. and the average of other 8 people is 5.75. How can you find out the average height of 6 people from this group? and by the way, which 6 people are we talking about?

I feel that the question confuses language a little bit. Instead of "the average height of the rest of the people in the group is 5 feet 10 inches" it probably should say "the average height of the rest of the people in the group n is 5 feet 10 inches".

Your response would be highly appreciated.


Several things. The question asks to find the number of people in a certain group, which is given to be n. (1) gives the average height of the n/3 tallest people in the group, which indicates that n must be a multiple of 3. Hence, when considering (1), n cannot be 10, it must be some positive multiple of 3: 3, 6, 9, ... Assume, n = 9, then (1) gives the average height of the 3 tallest people in the group, and the average height of the remaining 9 - 3 = 6 people in the group. With this, we can in fact find the average height of the entire group, which will be: (3*74.5 + 6*70)/9 = 1/3*74.5 + 2/3*70. If you use n instead of 9, you'd get: (n/3*74.5 + 2n/3*70)/n. Here n can be reduced and we get the same expression: 1/3*74.5 + 2/3*70.

Does this make sense?

P.S. This is an official question so its wording is as good and as precise as it gets. :dontknow:
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Re: What is the average (arithmetic mean) height of the n people in a cert [#permalink]
Bunuel wrote:
Paagrio wrote:
Bunuel wrote:
What is the average (arithmetic mean) height of the n people in a certain group?

(1) The average height of the n/3 tallest people in the group is 6 feet \(2 \frac{1}{2}\) inches, and the average height of the rest of the people in the group is 5 feet 10 inches.


The average height of the n/3 tallest people is 74.5 inches;
The average height of the remaining n - n/3 = 2n/3 people is 70 inches.

    (The average height of the group) = (The sum of the heights)/n \(= \frac{74.5*\frac{n}{3}+70*\frac{2n}{3}}{n}\)

Reduce by n:

    \(= 74.5*\frac{1}{3}+70*\frac{2}{3}\).

Sufficient.


(2) The sum of the heights of the n people is 178 feet 9 inches.

We don't know the number of students, in the group, n, so cannot get the average. Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel,
I don't understand why this is the case. I think there might be a small error in the question or in my understanding of the question.

So lets assume there are 10 people. Heights are as follows: {20,19,18,7,6,5,4,3,2,1}. Then we need to know the average height of 6 people. so n=6.

1st tells us that average height of 2 tallest people is 19.5. and the average of other 8 people is 5.75. How can you find out the average height of 6 people from this group? and by the way, which 6 people are we talking about?

I feel that the question confuses language a little bit. Instead of "the average height of the rest of the people in the group is 5 feet 10 inches" it probably should say "the average height of the rest of the people in the group n is 5 feet 10 inches".

Your response would be highly appreciated.


Several things. The question asks to find the number of people in a certain group, which is given to be n. (1) gives the average height of the n/3 tallest people in the group, which indicates that n must be a multiple of 3. Hence, when considering (1), n cannot be 10, it must be some positive multiple of 3: 3, 6, 9, ... Assume, n = 9, then (1) gives the average height of the 3 tallest people in the group, and the average height of the remaining 9 - 3 = 6 people in the group. With this, we can in fact find the average height of the entire group, which will be: (3*74.5 + 6*70)/9 = 1/3*74.5 + 2/3*70. If you use n instead of 9, you'd get: (n/3*74.5 + 2n/3*70)/n. Here n can be reduced and we get the same expression: 1/3*74.5 + 2/3*70.

Does this make sense?

P.S. This is an official question so its wording is as good and as precise as it gets. :dontknow:


All of these makes sense if we assume that the group and n are the same. After this assumption, the question is very easy, probably 7th grade math question. But the question might be much clearer if the question was: "What is the average (arithmetic mean) height of all people in a certain group?"
I know that this is an OG question, and it is especially frustrating for a test-taker that even OG questions sometimes have this kind of confusion. On a test-day this may be an issue.

Anyway, thanks for the explanation.
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