Recruits undergoing basic training are currently housed in t

Question

Recruits undergoing basic training are currently housed in three dormitories. The first two dormitories house 20 and 50 recruits, respectively. On average, each recruit shares a dormitory with 49 others. Next week, the first dormitory will be torn down, and the recruits currently housed there will move into the second dormitory. With approximately how many other recruits on average will each recruit then share a dormitory?

(A) 62
(B) 63
(C) 64
(D) 65
(E) 74

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Official Answer

C

(A) 62
(B) 63
(C) 64
(D) 65
(E) 74

kevincan · Answer

Suppose the third dorm houses n recruits . What equation can we write?

officiisimpedit · Answer

Hii Kevincan,

Wont it be like this:

50 = (20 + 50 + n)/3
150 = 70 + n
n = 80 (recruits in the third dormitory)

According to the question, after the first dormitory is demolished:

no. of recruits in the 2nd dorm. = 70
 no. of recruits in the 3rd dorm. = 80

so wont the new average be 75, which will make the final answer 74?

kevincan · Answer

Let the number of recruits in the third dormitory be n.

Since each recruit shares a dormitory with an average of 49 other recruits, each recruit is in a dormitory containing an average of 50 recruits.

\frac{20^2+50^2+n^2}{20+50+n}=50

\frac{400+2500+n^2}{70+n}=50

2900+n^2=3500+50n

n^2-50n-600=0

(n-60)*⁠￼(n+10)=0

Since  n>0 ,

n=60

After the first dormitory is demolished:

* Second dormitory: 20+50=70 recruits
* Third dormitory: 60 recruits

The new average dormitory size per recruit is

\frac{70^2+60^2}{70+60}=\frac{4900+3600}{130}=65.38\ldots

Therefore, the average number of other recruits each recruit shares a dormitory with is

65.38\ldots-1=64.38\ldots

Approximately,

\boxed{64}

officiisimpedit · Answer

Hi, I still don't understand why we are squaring the number of recruits?

egmat · Answer

Hi officiisimpedit,

Great question, and your instinct isn't crazy. The reason you got n =  80  is that you averaged the  three dorm sizes  ( 20 ,  50 , n) equally, as if each dorm counts once. But re-read what the question measures: it averages over  recruits , not over dorms.

Why that changes everything

The phrase "each recruit shares a dormitory with  49  others" means we take the average  across all the people , and there are far more people in a big dorm than a small one. So the big dorm's number should count more times.

Here's the clean way to see the squaring:

- Every recruit in a dorm of size  s  experiences a dorm size of  s  (they share with s-1 others, so they're in a group of s).
- There are  s  such recruits in that dorm - each contributing  s  to the total.
- So that one dorm contributes  s × s = s2  to the top of the average.

Add this up across dorms and divide by the total number of recruits:

Average dorm size = (202 + 502 + n2) / (20 + 50 + n) = 50

The square isn't a special rule - it's just "size, counted once for each of the size-many people in that dorm." Weighting by headcount is exactly what your equal-weight version left out, and that's why  80  was too high.

Lock it in with tiny numbers

Two dorms holding  2  and  4  recruits.

- Your original method (average the sizes): ( 2  +  4 )/ 2  =  3 .
- The per-recruit method: ( 2 2 +  4 2)/( 2  +  4 ) = ( 4  +  16 )/ 6  =  3.33 .

List the six people by hand: two of them feel a dorm of  2 , four of them feel a dorm of  4 . Their average experience leans toward  4  - because more people live in the bigger dorm. That lean is the squaring at work.

Answer: C

Recruits undergoing basic training are currently housed in t

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Recruits undergoing basic training are currently housed in t

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