Three boxes of supplies have an average (arithmetic mean) weight of 7

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.
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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.

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,

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

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.

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

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
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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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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 3-term 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 600-700 level question as this is easily solvable in ~1 minute for most people. Thoughts?

Thanks!
-Ron

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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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
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A video explanation can be found here:
https://www.youtube.com/watch?v=NdB-JOq2ch4

After 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.

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, 9,12-s}
To maximise s, 12 -s is to be minimised.
12-s=9
s = 3
the max possible weight, in kg, of the lightest box = 3 kg

IMO C

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