Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 19 Jul 2010
Posts: 17

Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
05 Dec 2010, 13:17
Question Stats:
71% (01:05) correct 29% (01:17) wrong based on 1008 sessions
HideShow timer Statistics
Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 64318

Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
05 Dec 2010, 13:26
Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Three boxes have an average weight of 7kg > \(a+b+c=3*7=21\); Three boxes have a median weight of 9kg > median of a set with odd terms is middle term, hence \(b=9\); So we have a, 9, c. Question: \(a_{max}=?\) General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others. So to maximize \(a\) we should minimize \(c\) > minimum value of \(c\) is 9 (\(c\), the third term, can not be less than median, the second term) > so \(a_{max}+9+9=21\) > \(a_{max}=3\). Answer: C. Hope it's clear.
_________________




Intern
Joined: 19 Jul 2010
Posts: 17

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
05 Dec 2010, 17:11
Thanks a lot Bunuel. Your reasoning and approach to Quant problems is the best. Too bad I didn't study well enough in Avg/Medians/SD. MGMAT isn't as elaborate in their treatment. But I promptly went through GMATClub topics on Avg/Medians/SD and solved some problems, without spending too much time as you rightly highlighted



Director
Joined: 29 Nov 2012
Posts: 673

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
05 Mar 2013, 07:22
we can't use 11 in this question as maximum value can you please exaplin this part. Thanks. I assumed 11 as the max value and got the answer as 1.



Manager
Joined: 24 Sep 2012
Posts: 79
Location: United States
Concentration: Entrepreneurship, International Business
GPA: 3.2
WE: Education (Education)

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
05 Mar 2013, 14:22
Hello Fozzy, Hopefully I can help you with this one. The question asks us for the maximum weight of the lightest box. Let us take an example here. Suppose three of your friends got a total of 30 marks in an exam. Now, what would be the maximum possible mark you got? Well, you could calculate the maximum possible mark you got if both of your friends scored 0 in the test(I pity the poor friends!). This would mean that you would score about 30 marks in the test. Any other arrangement would make them score more and consecutively, you would have to score less , right? Does this make sense? Similarly, if you need to find the maximum possible weight of the lightest box, you would have to minimize the weight of the bigger boxes. Now, we know that one box weighs 9 lb for sure. What is the minimum weight that a box heavier than that must weigh so that it can appear at the end when arranged in ascending order based on weight. Well, the answer is 9kg. If you consider the weight of the heaviest box to be 11, you would be minimizing the weight of the lightest box. For example, let x be the lightest box and y the heaviest box. x+9+y=21 implies, x+y=12. For x to be largest, y=9. x=3. If y=11, then x=1 which is lighter than the maximum possible weight. Hope this clears your doubt! Let me know if I can help you further. fozzzy wrote: we can't use 11 in this question as maximum value can you please exaplin this part. Thanks. I assumed 11 as the max value and got the answer as 1.



Intern
Joined: 26 Feb 2013
Posts: 49
Concentration: Strategy, General Management
WE: Consulting (Telecommunications)

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
07 May 2013, 07:58
Option C. 3Kg.
Since we need to maximize the minimum weight , let the weights of the 2 heaviest boxes be equal to the median = (kg.
So total weight of all boxes is 3*7 =21 and the combined weight of the 2 heaviest boxes = 9+9 = 18Kg.
so the maximum weight of the remaining box can be 3 KG



Veritas Prep GMAT Instructor
Joined: 11 Dec 2012
Posts: 312

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
07 May 2013, 09:32
mdbharadwaj wrote: Option C. 3Kg.
Since we need to maximize the minimum weight , let the weights of the 2 heaviest boxes be equal to the median = (kg.
So total weight of all boxes is 3*7 =21 and the combined weight of the 2 heaviest boxes = 9+9 = 18Kg.
so the maximum weight of the remaining box can be 3 KG This solution is correct, and the whole question hinges only on whether you have the correct definitions for median and average. If the average is 7 then the total weight must be 7x3=21 kg. The median of a 3term set is the middle term, so the middle term must be 9. The biggest term can be no smaller than 9, so at a minimum it's 9, leaving only 3 kg for the smallest entry. My question here is then about the difficulty level. I'm not sure I'd put this as a 600700 level question as this is easily solvable in ~1 minute for most people. Thoughts? Thanks! Ron
_________________



Intern
Joined: 22 Jul 2014
Posts: 21
WE: Information Technology (Computer Software)

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
29 Nov 2014, 10:07
my approach was , Total weight 21 a+9+c=21 a+c=12 c must be >= 9 (as 9 is the median) so options will be (1,11),(2,10),(3,9) So, max could be 3 . Is this correct? Thanks,
_________________
Failures are stepping stones to success !!!



Intern
Joined: 28 Dec 2015
Posts: 37

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
01 Jun 2016, 01:26
Three boxes Average=7 kg So total weight=21kg
Now various combinations are available 1,9,11 2,9,11 3,9,9 but we cannot go beyond 3,9,9 to 4,9,8 As numbers right to median must be equal or greater than the median value.
So,at max the weight of lightest box can be 3 Kg



Current Student
Joined: 12 Aug 2015
Posts: 2522

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
15 Dec 2016, 16:28
Great Official Question. Here is what i did in this Question => Let the boxes be => w1 w2 w3
Mean = 7 Sum(3)=7*3=21
Hence w1+w2+w3=21
Median=w2=9 Hence w1+w3=12
Now to maximise w1 we must minimise w3 minimum value of w3=median =9 Hence maximum w1=129 = 3
Hence C
_________________



Intern
Joined: 03 Dec 2017
Posts: 13

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
19 Dec 2017, 05:40
What if the Boxes where Five (5) instead of three (3)? What would then be the appropriate approach regarding this problem?



Senior SC Moderator
Joined: 22 May 2016
Posts: 3858

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
20 Dec 2017, 20:49
sameerdrana wrote: Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9kg. What is the max possible weight, in kg, of the lightest box?
A. 1 B. 2 C. 3 D. 4 E. 5 Anestists wrote: What if the Boxes where Five (5) instead of three (3)? What would then be the appropriate approach regarding this problem? Anestists ,the approach would be similar: minimize everything other than the lightest box in order to maximize that very box. I assume you intended a mean of 7, and a median of 9. As before, we assume that the weights must be in integers. 1) Median of 9, five numbers: we have A, B, 9, D, E 2) Find total weight of all 5 boxes. A*n=S: (5 * 7) = 35 3) Minimize D and E. We have to subtract 9, D, and E from 35 first Smaller numbers, subtracted from 35, will leave a larger amount left for boxes A and B Let D and E both = 9 (they cannot be smaller than the median) 4) Now we have A, B, 9, 9, 9 5) Amount left for boxes A and B: 35  27 = 8 kg 6) What is the max possible weight, in kg, of the lightest box? Minimize box B to maximize box A A and B  Possibilities 1 and 7 2 and 6 3 and 5 4 and 4 I suspect most people would pick A = 3 and B = 5 The question asks for THE lightest box, implying there is only one. In ordinal and ordinary terms, least implies lesser: A must weigh less than B to qualify as the "lightest." ANSWER: "The maximum possible weight, in kg, of the lightest box, is 3." Is that what you were wanting to know?
_________________
Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter.  Dr. Martin Luther King, Jr.



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2799

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
02 Jul 2018, 08:49
kwhitejr wrote: Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?
A. 1 B. 2 C. 3 D. 4 E. 5 The sum of the weights of the boxes is 3 x 7 = 21 kg, and the median is 9 kg. To make the smallest box as heavy as possible, we can make the largest box also 9 kg. So the heaviest possible weight of the smallest box is 21  18 = 3 kg. Answer: C
_________________
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.



Intern
Joined: 20 Aug 2018
Posts: 25

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
24 Nov 2018, 13:49
A video explanation can be found here: https://www.youtube.com/watch?v=NdBJOq2ch4After writing out the average formula A = (SUM of terms)/(# of terms) A = (a + 9 + c)/3 = 7 then SUM = a + 9 + c = 21 The numbers to the left and right of the mediam must add to 12 Whenever a math question asks you about maximums, always thing about minimums, as well  and vice versa. In this question, "What is the max possible weight of the lightest box?" we'll also ask ourseleves, "What is the minumum possible weight of the heaviest box?" Recalling that the median number doesn't necessarily have to be the ONLY number in the set with a value of 9, we'll see that the minumum weight of the heaviest box is also 9, therefore the maximum possible weight of the lighest box is 3.
_________________



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10684
Location: United States (CA)

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
11 Aug 2019, 18:28
sameerdrana wrote: Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9kg. What is the max possible weight, in kg, of the lightest box?
A. 1 B. 2 C. 3 D. 4 E. 5 We are given that three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. We must determine the maximum weight of the lightest box. Since the average weight of the 3 boxes is 7, the sum of the weights of the 3 boxes is 3 x 7 = 21. We can also define a few variables. x = lightest box y = second heaviest box z = heaviest box We can create the following equation: x + y + z = 21 Since the median is 9, y must be 9. So we now have: x + 9 + z = 21 x + z = 12 Remember, we need the value of x to be as large as possible, so we want to minimize the value of z. Since 9 is the median weight of the boxes, the smallest value of z is also 9. Thus, the maximum value of x is 12 – 9 = 3. Answer: C
_________________
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.



CEO
Joined: 03 Jun 2019
Posts: 2951
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
25 Aug 2019, 00:55
sameerdrana wrote: Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9kg. What is the max possible weight, in kg, of the lightest box?
A. 1 B. 2 C. 3 D. 4 E. 5 Given: Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9kg. Asked: What is the max possible weight, in kg, of the lightest box? Let the weight of 3 boxes be {s, 9, l} s + 9 + l = 21 l = 12  s Weights = {s, 9,12s} To maximise s, 12 s is to be minimised. 12s=9 s = 3 the max possible weight, in kg, of the lightest box = 3 kg IMO C
_________________
Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com



NonHuman User
Joined: 09 Sep 2013
Posts: 15104

Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
Show Tags
09 Oct 2019, 04:18
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Three boxes of supplies have an average (arithmetic mean) weight of 7
[#permalink]
09 Oct 2019, 04:18




