Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 05 Sep 2007
Posts: 124
Location: New York

A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
11 Mar 2008, 16:20
Question Stats:
74% (01:57) correct 26% (02:13) wrong based on 367 sessions
HideShow timer Statistics
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 Project PS Butler : Question #63 Subscribe to get Daily Email  Click Here  Subscribe via RSS  RSS
Official Answer and Stats are available only to registered users. Register/ Login.



Director
Joined: 10 Sep 2007
Posts: 708

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
11 Mar 2008, 16:54
Lets assume there are 2X people. Half of them have average weight of 180 and other half has 215. Maximum Weight is = 2000 So 180*X + 215*X = 2000 => 395X = 2000 => X is approximately equal to 5. So total people is 2*5 = 10
We are not taking 11 as answer because say 11th person has minimum of 180 weight then 180*6 + 215*5 = 2155 (Which is more than 2000)
Answer D.



Manager
Status: Preparing Apps
Joined: 04 Mar 2009
Posts: 84
Concentration: Marketing, Strategy
GMAT 1: 650 Q48 V31 GMAT 2: 710 Q49 V38
WE: Information Technology (Consulting)

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Dec 2010, 08:29
A certain elevator has a safety weight limit of 2000 pounds. Whats is the greatest possible number of people who can safely ride on the elevator at one time with the average (aritmetic 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



Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 339
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross)  Class of 2014
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Dec 2010, 09:17
.5X(180)+.5X(215)<=2000 X<=~10.125 Greatest possible integer is 10 SoD
_________________



Intern
Joined: 30 Nov 2010
Posts: 22
Location: Boston
Schools: Boston College, MIT, BU, IIM, UCLA, Babson, Brown

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Dec 2010, 10:07
x/2(180) + x/2(215) = 2000
That gives x = (2000 * 2)/395, x ~ 10.1
So it has to be D



SVP
Joined: 06 Sep 2013
Posts: 1566
Concentration: Finance

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
16 Dec 2013, 12:25
el1981 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 2000 pounds total 180*x/2 + 215*x/2 = 2000 Max is 10 Hence D Hope it helps Cheers! J



Intern
Joined: 13 Dec 2013
Posts: 35

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
18 Jul 2014, 18:12
Since it is 50%50% it has to be an even number. Only 8 and 10 are even.
Average weight = (180+215)/2 = 197.5
We can then easily see that 197.5*10 = 1,975 pounds, so answer = 10.



Manager
Joined: 22 Feb 2009
Posts: 156

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
18 Jul 2014, 23:21
Enael wrote: Since it is 50%50% it has to be an even number. Only 8 and 10 are even.
Average weight = (180+215)/2 = 197.5
We can then easily see that 197.5*10 = 1,975 pounds, so answer = 10. wow, you saves a lot of time with your brilliant method
_________________
......................................................................... +1 Kudos please, if you like my post



Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 357

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
18 Oct 2016, 10:28
10 people.
Equation to use: > (#/2)(180)+(#/2)(215) = CANNOT EXCEED 2,000 5(180)+5(215) = 1975 [We've maxed out on this option, thus this is correct]



RC Moderator
Joined: 24 Aug 2016
Posts: 789
GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
08 Dec 2018, 00:02
Alternative Approach : People can not be divided in half..... so we remove odd options . Remaining with 8 & 10. Start with the largest :10: so 5*180 +5 * 215=900+1075=1975<2000. Hence, Ans is 10 (D)
_________________
Please let me know if I am going in wrong direction. Thanks in appreciation.



VP
Joined: 09 Mar 2016
Posts: 1230

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Dec 2018, 02:58
walkerme13 wrote: x/2(180) + x/2(215) = 2000
That gives x = (2000 * 2)/395, x ~ 10.1
So it has to be D Gladiator59how did we get this ? \(\frac{180x}{2}\)and this \(\frac{215x}{2}\) I understand that average weight of half group is 180 , and average weight of another one is 215 but why do we divide by 2 ; we dont know number of persons in each group right thanks!



Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 737
GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Dec 2018, 05:00
dave13, So let n folks have an average weight of 180 pounds ... total weight will be 180n. The other n will have average wt of 215, so total weight will be 215n Now, the total weight of the entire group will be 215n + 180n = 395n whereas total no. of folks will be 2n... so the new average will be = 395n/2n = 395/2 = 197.5 That's how we get the divide by 2Hope it is clear. best, Gladi dave13 wrote: walkerme13 wrote: x/2(180) + x/2(215) = 2000
That gives x = (2000 * 2)/395, x ~ 10.1
So it has to be D Gladiator59how did we get this ? \(\frac{180x}{2}\)and this \(\frac{215x}{2}\) I understand that average weight of half group is 180 , and average weight of another one is 215 but why do we divide by 2 ; we dont know number of persons in each group right thanks!
_________________
Regards, Gladi
“Do. Or do not. There is no try.”  Yoda (The Empire Strikes Back)



Director
Joined: 09 Mar 2018
Posts: 994
Location: India

Re: A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Feb 2019, 21:56
el1981 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
Now as we see that average weight of two groups is divided in half, then only even values will work. Said that out of 8 and 10,only 10 will work (180 + 215)/2 = 197.5 Since we will get 1975 as the sum, which when divided by 10, gives us the average as 197.5
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.



Target Test Prep Representative
Status: Chief Curriculum and Content Architect
Affiliations: Target Test Prep
Joined: 24 Nov 2014
Posts: 619

A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
Show Tags
10 Feb 2019, 23:34
el1981 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 Since the correct answer will include two equal numbers of people, and since people do not come in fractional quantities and, thus, may be represented only by integer values, the correct answer will be the sum of two equal integer values, and thus, must be an even number. So, the only two choices that work are (B) 8 and (D) 10. While the riders may have various weights, we can ignore the variation in the weights because we know that the average weight of half of the riders is 180 pounds and the average weight of the other half of the riders is 215 pounds. So, we can proceed as we would if half the riders weighed exactly 180 pounds each and the other half weighed exactly 215 pounds each. Since the average rider weight of half of the riders is 180 pounds, and the average rider weight of other half of the riders is 215 pounds, the average rider weight of all the riders is the average of these two numbers. We can see that, because 180 is 20 less than 200, 215 is 15 greater than 200, and 20 > 15, the average rider weight is under 200 pounds. Thus, 10 riders can ride, as 10 x (a number < 200) < 2000. The correct answer is (D).
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.




A certain elevator has a safe weight limit of 2,000 pounds.
[#permalink]
10 Feb 2019, 23:34






