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What is the average (arithmetic mean) height of the n people in a cert

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New post 20 Dec 2005, 09:03
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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.

(2) The sum of the heights of the n people is 178 feet 9 inches.
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New post 14 Jul 2010, 15:40
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zisis 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 and 2.5 inches and the average height of the rest of the people in the group is 5 feet 10 inches

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


pranrasvij wrote:
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!)


Answer to the question is A, but you shouldn't divide the "sum" by 2, you should divide by \(n\).

\(Weighted \ average=\frac{sum \ of \ weights}{# \ of \ data \ points}\), or in our case \(average \ height=\frac{sum \ of \ heights}{# \ of \ people}\).

(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.
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New post 14 Jul 2010, 13:35
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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 and 2.5 inches and the average height of the rest of the people in the group is 5 feet 10 inches
(2) the sum of the heights of the n people is 178 feet 9 inches
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New post 20 Dec 2005, 10:27
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I go for A.

1) 1/3 * 6 feet 2 1/2 inches + 2/3 * 5 feet 10 inches.... SUFF

2) Doesn't give us the value of n (number of elements/people)... INSUFF
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New post 20 Dec 2005, 20:13
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From (1), (n/3*(6'2.5") + 2n/3*(5'10"))/n -> average height. Sufficient.

From (2), we do not know n. Insufficient.

Ans: A
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New post 02 Apr 2007, 16:25
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
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New post 21 Mar 2010, 11:35
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nifoui 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(1/2) inches, and the average height of the rest of the people in the group is 5 feet 10 inches.

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


OA =

Stmt1: Avg of n/3 people= 6feet 2(1/2) inches = n/3 * 6feet 2(1/2) inches
Avg of 2n/3 people = 5feet 10 inches = 2n /3*5feet 10 inches
If we add both above quantities and divide it by n we will get the avg since n will cancel out. So, sufficient.

Stmt2: if sum is given to calculate avg we have to divide it by n. But we dont know the value of n so insufficient.
Hence A is the answer
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New post 14 Jul 2010, 13:43
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".
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New post 14 Jul 2010, 13:50
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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New post 14 Jul 2010, 14:06
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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New post 14 Jul 2010, 16:33
thanks Bunuel.... this is a v good explanation. Sorry- I messed up the divide part!....I knew i was wrong somewhere and came back to fix it !!
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Re: What is the average (arithmetic mean) height of the n people in a cert  [#permalink]

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New post 15 Jul 2010, 01:20
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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New post 15 Jul 2010, 08:06
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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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New post 09 Oct 2010, 12:46
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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New post 09 Oct 2010, 12:54
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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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New post 18 Jun 2011, 08:55
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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New post 03 Jun 2013, 05:37
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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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New post 12 Feb 2017, 01:24
Looking at statement (2) first, we see that it is not sufficient, because the average (arithmetic mean) of a group of numbers is defined as (sum of data) / (# of data points). With statement (2), we only have the numerator of this expression (the # of people in the group is unknown), so we can't figure out the average.

Looking at statement (1) alone, we can set up the average as follows:
Average = (sum of data points) / (# of data points)
= [(n/3)(74.5) + (2n/3)(70)] / (n) <-- note that I used inches here, so I won't have to write in more fractions than necessary.
= [(1/3)(74.5) + (2/3)(70)]
There's no need to simplify further, because the 'n' is gone: you get one number. Therefore, this statement is sufficient.

Answer = A

Note that, if you have the averages of all the FRACTIONS or PERCENTAGES of a group, then you'll be able to calculate the overall average of the group. This is a worthwhile fact to memorize for the data sufficiency problems.
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Re: What is the average (arithmetic mean) height of the n people in a cert  [#permalink]

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New post 24 Nov 2018, 16:46
A video explanation can be found here:
https://www.youtube.com/watch?v=kyRmAjIIiNI

Average = (SUM of terms) / (# of terms)

Need to know something about the numerator and denominator.

(1) gives us information to find a weighted average. Don't do any more math than necessary!

Average = [n/3(6ish feet) + 2n/3(5ish feet)] / n

Factor n out of the two expressions in the numerator. Then n in the numerator and n in the denominator cancel each other.

We're left with Average = 1/3(6ish) + 2/3(5ish), so statement (1) is sufficient - no need to find the exact average!

Statement (2) tells us the numerator (SUM), but not the denominator. Insufficient.

Answer A.
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New post 20 Oct 2019, 07:57
Hello,

1. If you have a DS question and it involves area and you solve the solution, but it has two answers: one positive and one negative. Since you have a negative answer, that can't be correct as area cant be negative. Therefore, you have one solution and it is SUFFICIENT to solve the question.

2. If you solve the question and you have a quadratic equation like 2x^2+7x+20. Since you can't solve for this using (X+@)(X+$), it is not sufficient to solve the question? We don't use quadratic formula for GMAT, as this can get us decimals too.

Thanks!!!
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Re: What is the average (arithmetic mean) height of the n people in a cert   [#permalink] 20 Oct 2019, 07:57
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