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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

Given that the median is 9 that leaves us with the other two boxes with a combined weight of 12 kg. [Average * number = total --- 7 * 3 = 21. And total - median = sum of heaviest and smallest box = 12kg]

Now 9 should be the median and the smallest box has to have the max weight. To leave the median as 9, the heaviest box should be at least 9 kg ... which gives us 3 kg for the smallest box.

Hence the weight of the box should be 3, 9, 9. Hence answer is C.
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Thanks for your suggestion. That is actually what I did originally to solve the question, but I came up with the max weight for the smallest box to be 2. Thus the three boxes would be 2, 9 and 10. However, if the boxes are 3, 9 and 9 then the median is still 9? This is what throws me off.

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

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 \(a_{max}+9+9=21\) --> \(a_{max}=3\).

lets assume 3 boxes are X,Y,Z x+y+z= 21 Median = 9 ( middle number ) By back solving smallest maximum weight X Y(median) Z 1 9 11 2 9 10 3 9 9 4 9 8 ( it cant be the answer because Z value is below the Y (median value)) 5 9 7 (it cant be the answer because Z value is below the Y (median value))

Thanks for your suggestion. That is actually what I did originally to solve the question, but I came up with the max weight for the smallest box to be 2. Thus the three boxes would be 2, 9 and 10. However, if the boxes are 3, 9 and 9 then the median is still 9? This is what throws me off.

Hi kwhitejr,

Definitely - the median of 3,9 and 9 is indeed the 2nd term -> 9.

Just pick the middle term after sorting them in ascending order - no matter what.

14. Three boxes of supplies have an average (A.M.) weight of 7 kg and median of 9 kg. What is the maximum possible weight in kg of the lightest box? a)1 b)2 c)3 d)4 e)5

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?

14. Three boxes of supplies have an average (A.M.) weight of 7 kg and median of 9 kg. What is the maximum possible weight in kg of the lightest box? a)1 b)2 c)3 d)4 e)5

I also agree with md .... I got the answer as C..........
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14. Three boxes of supplies have an average (A.M.) weight of 7 kg and median of 9 kg. What is the maximum possible weight in kg of the lightest box? a)1 b)2 c)3 d)4 e)5

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