Averages

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richs_ca: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Fri Mar 28, 2008 6:01 am; Thanked: 1 times

by richs_ca » Sun May 04, 2008 10:21 am

What is the average heigh of the n people in a certain group?

1) The average heigh of the n/3 tallest people in the group is 6 feet 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 n people is 178 feet 9 inches.

I'm a bit confused, because 1) doesn't seem to give the total number of people in the group, yet that's the answer.

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Source: Beat The GMAT — Data Sufficiency | Sun May 04, 2008 10:21 am

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

Re: Averages

by lunarpower » Sun May 04, 2008 11:24 am

richs_ca wrote:What is the average heigh of the n people in a certain group?

1) The average heigh of the n/3 tallest people in the group is 6 feet 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 n people is 178 feet 9 inches.

I'm a bit confused, because 1) doesn't seem to give the total number of people in the group, yet that's the answer.

statement (1) is a weighted average: it's just like a normal average, except in that you must account for the # of times each data point appears. so here's how you set it up:
average = (sum of all heights) divided by (number of people)
= ( (n/3)(74.5 inches) + (2n/3)(70 inches) ) divided by (n)
no need to do the arithmetic here; just look at it,** and notice that you can pull a common factor of 'n' out of everything on top, after which you can strike the 'n's and leave a purely numerical expression.

**you might need to write it out on a piece of paper to see this, because fractions and forums just don't mix well.

--

there's also a cool shortcut for weighted averages: if a certain fraction of a population has statistic x and the rest has statistic y, then the average is the same fraction of the way from y to x.

so, for instance, in this problem, 1/3 of the population is 74.5 inches tall, and the remaining 2/3 of the population is 70 inches tall. so, the average is 1/3 of the way from 70 to 74.5 (or, if you want, 2/3 of the way from 74.5 to 70).

one consequence of the foregoing fact is that any weighted average involving definite fractions of a population will work out to the same number, regardless of the number of individuals in the population. this is a truly valuable observation for data sufficiency.

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you probably already figured this out, but (2) is insufficient because you need to know how many people there are.

so answer = a

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richs_ca: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Fri Mar 28, 2008 6:01 am; Thanked: 1 times

by richs_ca » Sun May 04, 2008 12:24 pm

Thanks! But I'm still a little confused.

I can see that it's a weighted average. I see that one group has an average weight of 74.5 inches and the other 70 inches. But I don't know what the weighting is. Where does the 2(n/3) come from?

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by lunarpower » Sun May 04, 2008 10:49 pm

richs_ca wrote:Thanks! But I'm still a little confused.

I can see that it's a weighted average. I see that one group has an average weight of 74.5 inches and the other 70 inches. But I don't know what the weighting is. Where does the 2(n/3) come from?

you have a total of n people.
n/3 of those people are 74.5 inches tall. this means that everyone else is 70 inches tall, so that's n - n/3, which is 2n/3 people.

if you like fractions better than symbols, consider this interpretation: n/3 is one-third of everyone. that means the other people (who are 70 inches tall) are the other two-thirds of everyone, or 2n/3.

in any case, just remember the general result: if you have the fractions of the entire population that correspond to your data points, then you can find the weighted average, even if you don't have the actual #s of data points. the fractions are all that's necessary.

hth.

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mbaapplicant2008: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Thu Oct 30, 2008 1:21 am

by mbaapplicant2008 » Thu Oct 30, 2008 4:23 am

The first statement is sufficient to answer the question. Therefore, it is A.

MBA Applicant

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California4jx: Master | Next Rank: 500 Posts; Posts: 227; Joined: Thu Aug 14, 2008 10:43 amFollowed by: 1 member; Thanked: 7 times; GMAT Score:650

by California4jx » Sat Jun 27, 2009 11:48 am

For weighted avg problems try to visualize a line on which these two averages exist to balance the data points.

so here the data points are heights that are balanced by averages.

Now, there is a further sub grouping (very typical of W-A problems) and for each group there exists an average. So visualize these averages as pivot points on a line.

Now, the next task would be to find out where most of the data points (here the height of all) is grouping together. How do you find that ? by finding the ratios of distribution.

Here they have already given the ratios 1/3 of n and the left overs are 2/3 of n. Clearly, the average height of all people will be more clustered (tilted) towards 5.10

Find the difference between these two average heights, multiple that differece with 2/3 and then add that result in 6 ft 2.5 inches height -- ofcourse, you do not need to do all these computations ...

Answer is A

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laxmiraghu: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Nov 07, 2011 2:37 pm

by laxmiraghu » Mon Nov 07, 2011 2:40 pm

lunarpower wrote:
richs_ca wrote:What is the average heigh of the n people in a certain group?

1) The average heigh of the n/3 tallest people in the group is 6 feet 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 n people is 178 feet 9 inches.

I'm a bit confused, because 1) doesn't seem to give the total number of people in the group, yet that's the answer.
statement (1) is a weighted average: it's just like a normal average, except in that you must account for the # of times each data point appears. so here's how you set it up:
average = (sum of all heights) divided by (number of people)
= ( (n/3)(74.5 inches) + (2n/3)(70 inches) ) divided by (n)
no need to do the arithmetic here; just look at it,** and notice that you can pull a common factor of 'n' out of everything on top, after which you can strike the 'n's and leave a purely numerical expression.

**you might need to write it out on a piece of paper to see this, because fractions and forums just don't mix well.

--

there's also a cool shortcut for weighted averages: if a certain fraction of a population has statistic x and the rest has statistic y, then the average is the same fraction of the way from y to x.

so, for instance, in this problem, 1/3 of the population is 74.5 inches tall, and the remaining 2/3 of the population is 70 inches tall. so, the average is 1/3 of the way from 70 to 74.5 (or, if you want, 2/3 of the way from 74.5 to 70).

one consequence of the foregoing fact is that any weighted average involving definite fractions of a population will work out to the same number, regardless of the number of individuals in the population. this is a truly valuable observation for data sufficiency.

--

you probably already figured this out, but (2) is insufficient because you need to know how many people there are.

so answer = a

Hi i m bit confused how u calculated
"( (n/3)(74.5 inches) + (2n/3)(70 inches) ) divided by (n)"
how u got 74.5 inches and 2n/3 of 70 inches.

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bpdulog: Master | Next Rank: 500 Posts; Posts: 105; Joined: Sun Jan 25, 2009 6:55 pmFollowed by: 1 member; Thanked: 2 times

by bpdulog » Wed Nov 09, 2011 1:10 pm

Someone please correct me if I'm wrong, but given statement 1, we can just ignore what the average height is for the rest of the group. The question is asking us if we can find n, which in this case, would be the number of the tallest people from a certain group.

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