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Three boxes of supplies have an average (arithmetic mean)
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05 Dec 2010, 14:17
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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
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Re: Three boxes avg vs median
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05 Dec 2010, 14:26
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?
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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 \(a_{max}+9+9=21\) > \(a_{max}=3\). Answer: C.
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Re: Three boxes avg vs median
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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



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Re: Three boxes avg vs median
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19 May 2011, 01:08
for max possible value of lowest, make the highest = median.
9*2 + x = 21 x=3



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Re: Three boxes of supplies have an average (arithmetic mean)
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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.



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Re: Three boxes of supplies have an average (arithmetic mean)
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05 Mar 2013, 15:22
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.



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Re: Three boxes of supplies have an average (arithmetic mean)
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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,
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Re: Three boxes of supplies have an average (arithmetic mean)
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01 Jun 2016, 02:26
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



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Re: Three boxes of supplies have an average (arithmetic mean)
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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=129 = 3
Hence C
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Re: Three boxes of supplies have an average (arithmetic mean)
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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
Therefore the answer is C



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Re: Three boxes of supplies have an average (arithmetic mean)
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19 Dec 2017, 06:40
What if the Boxes where Five (5) instead of three (3)? What would then be the appropriate approach regarding this problem?



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Three boxes of supplies have an average (arithmetic mean)
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20 Dec 2017, 21:49
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?
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Three boxes of supplies have an average (arithmetic mean)
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24 Nov 2018, 14:49
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.
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Re: Three boxes of supplies have an average (arithmetic mean)
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11 Aug 2019, 19:28
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
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Three boxes of supplies have an average (arithmetic mean)
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25 Aug 2019, 01:55
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
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Re: Three boxes of supplies have an average (arithmetic mean)
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25 Aug 2019, 11:52
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 a+9+c=21 let c=9 a=2118 ; 3 IMO C
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