It is currently 17 Oct 2017, 11:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Three boxes of supplies have an average (arithmetic mean)

Author Message
TAGS:

### Hide Tags

Intern
Joined: 19 Jul 2010
Posts: 24

Kudos [?]: 55 [2], given: 1

Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

05 Dec 2010, 14:17
2
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

66% (00:37) correct 34% (00:40) wrong based on 1027 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

Kudos [?]: 55 [2], given: 1

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128577 [10], given: 12180

Re: Three boxes avg vs median [#permalink]

### Show Tags

05 Dec 2010, 14:26
10
KUDOS
Expert's post
16
This post was
BOOKMARKED
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?

1

2

3

4

5

As the median value of 3 (odd) # of boxes is 9 then weights of the boxes in ascending order are: {a, 9, b}. Also as the mean equals to 7 then a+9+b=3*7=21;

Now, general rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

So we need to maximize the weight of the lightest box, so we want to maximize $$a$$ --> to maximize $$a$$ we should minimize $$b$$ --> min value of $$b$$ is 9 (it cannot be less than the median value), so we'll have $$a_{max}+9+9=21$$ --> $$a_{max}=3$$.

_________________

Kudos [?]: 128577 [10], given: 12180

Intern
Joined: 19 Jul 2010
Posts: 24

Kudos [?]: 55 [0], given: 1

Re: Three boxes avg vs median [#permalink]

### Show Tags

05 Dec 2010, 18:11
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

Kudos [?]: 55 [0], given: 1

VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1286

Kudos [?]: 281 [0], given: 10

Re: Three boxes avg vs median [#permalink]

### Show Tags

19 May 2011, 01:08
for max possible value of lowest, make the highest = median.

9*2 + x = 21
x=3
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Kudos [?]: 281 [0], given: 10

Director
Joined: 29 Nov 2012
Posts: 868

Kudos [?]: 1409 [0], given: 543

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

05 Mar 2013, 08:22
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.
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Kudos [?]: 1409 [0], given: 543

Manager
Joined: 24 Sep 2012
Posts: 90

Kudos [?]: 166 [1], given: 3

Location: United States
GMAT 1: 730 Q50 V39
GPA: 3.2
WE: Education (Education)
Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

05 Mar 2013, 15:22
1
KUDOS
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.

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.

Kudos [?]: 166 [1], given: 3

Intern
Joined: 22 Jul 2014
Posts: 22

Kudos [?]: 2 [0], given: 13

GMAT 1: 450 Q38 V12
WE: Information Technology (Computer Software)
Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

29 Nov 2014, 11:07
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 !!!

Kudos [?]: 2 [0], given: 13

Intern
Joined: 28 Dec 2015
Posts: 42

Kudos [?]: 3 [1], given: 62

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

01 Jun 2016, 02:26
1
KUDOS
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

Kudos [?]: 3 [1], given: 62

BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2212

Kudos [?]: 836 [0], given: 595

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

15 Dec 2016, 17:28
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=12-9 = 3

Hence C

_________________

Give me a hell yeah ...!!!!!

Kudos [?]: 836 [0], given: 595

Intern
Joined: 14 May 2016
Posts: 6

Kudos [?]: [0], given: 4

Re: Three boxes of supplies have an average (arithmetic mean) [#permalink]

### Show Tags

15 Jul 2017, 14:22
Just want to mention that in a GMAT course I took, the professor thought us to plug in the answers and always start by trying out C to avoid wasting time in questions like this one.

B1 + B2 + B3/3 = 7
Median = 9
Find B1

Let 7 be the target

Try A:

Let B1 be 1
Let B2 be 9
Let B3 be 9

1 + 9 + 9/3 = 6.3

Try C:

Let B1 be 3
Let B2 be 9
Let B3 be 9

3 + 9 + 9 /3 = 7

Kudos [?]: [0], given: 4

Re: Three boxes of supplies have an average (arithmetic mean)   [#permalink] 15 Jul 2017, 14:22
Display posts from previous: Sort by