what is the average (arithmetic mean) height of the n people

[urooj.khan]

what is the average (arithmetic mean) height of the n people of a certain group?

(1) the average height of the n/3 tallest people in the group is 6 feet 2 1/2inches ( in decimal this would be =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


correct answer is A.

is this the most efficient and fastest way to arrive at this answer:

(1) assume n = 10, then the average is [(3)(6 feet 2.5 inches) + (7)(5 feet 10 inches)]/10
sufficient

(2) not sufficient
??

[stock.mojo11]

You dont have to necessarily assume n=10.

As soon as you see A, you see that it is a classic weighted average. With weighted averages, to find the whole average, you do not need to know the total number of elements

A is saying { (n/3) (6 2.5) + (2n/3) (5 10) } / n

n's in numerator and denominator get canceled, in any weighted avg problem.

HTH

[RonPurewal]

post5774.html

[RonPurewal]

post5774.html

[christina.susie.wong]

I know it's a data sufficiency problem where you don't have to solve for the average but I don't understand how you can find the average from the information provided from the weighted average.

Using what you said that the n's cancel, you are left with

1/3 ( 6 2.5 ft) + 2/3 (5'10 feet) = Average. How does it equal the average?

Would you please clarify the weighted average to average rule that you were explaining?

Thanks

[akhp77]

Statement 1:
No of tallest people = n/3
avg weight of tallest people = 6 ft 2.5 inches = 74.5 inches
Total weight of tallest people = 74.5 * n/3

No people other than tallest = 2n/3
avg weight of people other than tallest = 5 ft 10 inches = 70 inches
Total weight of people other than tallest = 70 * 2n/3 = 140n/3

Total weight of the people = (140n + 74.5n) / 3 = 71.5n
Avg weight of the people = 71.5n/n = 71.5 inches

Sufficient

Statement 2:
is insufficient because n is unknown

[StaceyKoprince]

Christina:

In this problem, 1/3 of all of the people are one certain height (6 feet 2.5 in) and the remaining 2/3 are another certain height (5 feet 10 in).

With a "normal" average, you would have equal amounts of both groups -- half at 6 feet 2.5 in and 1/2 at 5 feet 10 in -- and then you would just take a "straight" average of those two numbers.

In this problem, we have to weight the two numbers differently. The easiest way to do this is to assume a certain real number of people. Let's say that one person is 6 feet 2.5 in. If that's true, then 2 other people have to be 5 feet 10 in, so that we fit the 1/3 and 2/3 parameters given in the problem. (Note that it doesn't matter how many people you pick, as long as you fit the requirements of the problem; the average will still be the same as long as the ratio of people in each group is the same.)

So, first find the sum of all of the heights:
(1)(6 feet 2.5in) + (2)(5 feet 10in) = 16 feet 22.5in (which equals 17 feet 10.5 in).

Then, divide by the number of people (3):
(17 feet 10.5 in) / 3 = 5 feet 11.5 in

Note: you might find the actual calculation easier to do if you convert everything to inches.