n1 = half of riders with average weight of 180 pounds
n2 = half of riders with average weight of 215 pounds
n = greatest numbers of riders who can ride safely

n1 = n2
n1 + n2 = n = 2n1 = 2n2

2000 = 180n1 + 215n2 = 395n1 = 395n/2

then n = 4000/395 > 4000/400 => n is just over 10, which is the nearest integer
therefore answer is D

n = Number of Safe riders

2000 = (180*n/2) + (215*n/2)
=> 2000 = 395n/2
=> n = 10 (nearest integer)
Hence D.

I found it easier just to use a thought process on this one. We want the most people possible at the lesser weight. So, the average of 215 lbs could be just one person. So, subtract 2000 - 215 and get 1785. Then divide 1785/180 = 9. 9+1=10.

smclean23 wrote:A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

We can eliminate 3 answers doing almost 0 work, with a bit of common sense.

Since the question talks about the average weight of half the riders, we know that the total number of riders must be divisible by 2: eliminate (a), (c) and (e). If you're stuck, you have a 50/50 shot.

To actually solve, common sense again comes to the rescue of the algebra-challenged. If the average weight of half the members is 180 and the average weight of the other half is 215, that means that if we look at 1 person from each group their total weight is 395 lbs - let's round that up to 400.

The total limit is 2000 lbs. 2000/400 = 5. So, we can fit 5 pairs of riders on the elevator, for a max total of 10 people: choose (d).

As a side note, :

I found it easier just to use a thought process on this one. We want the most people possible at the lesser weight. So, the average of 215 lbs could be just one person. So, subtract 2000 - 215 and get 1785. Then divide 1785/180 = 9. 9+1=10.

is NOT a safe solution, since the question clearly states that half of the riders belong to each group. If the weight limit of the elevator had been, for example, 2100 lbs, (d) would still be the correct answer, but your method would have given you (e) 11 instead.

Thanks Stuart for pointing that out. I obviously wasn't reading it for clarity.

Stuart Kovinsky wrote:

smclean23 wrote:A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

We can eliminate 3 answers doing almost 0 work, with a bit of common sense.

Since the question talks about the average weight of half the riders, we know that the total number of riders must be divisible by 2: eliminate (a), (c) and (e). If you're stuck, you have a 50/50 shot.

To actually solve, common sense again comes to the rescue of the algebra-challenged. If the average weight of half the members is 180 and the average weight of the other half is 215, that means that if we look at 1 person from each group their total weight is 395 lbs - let's round that up to 400.

The total limit is 2000 lbs. 2000/400 = 5. So, we can fit 5 pairs of riders on the elevator, for a max total of 10 people: choose (d).

As a side note, :

I found it easier just to use a thought process on this one. We want the most people possible at the lesser weight. So, the average of 215 lbs could be just one person. So, subtract 2000 - 215 and get 1785. Then divide 1785/180 = 9. 9+1=10.

is NOT a safe solution, since the question clearly states that half of the riders belong to each group. If the weight limit of the elevator had been, for example, 2100 lbs, (d) would still be the correct answer, but your method would have given you (e) 11 instead.

Stuart, great strategies to solve them without actually following the conventional way of solving the question. I will try ot remember them, just incase I get stuck with the conventional approach, I know these techniques can atleast help me narrowing the choices.

Thanks Stuart... sometimes common-sense is what is uncommon...

I was sure that the answer should have been a C or a D, taking the sum total in the same way you have done...

(215+ 180) but did not eliminate C till I could actually calculate the answer.