Four packages have an average weight of 10.5 pounds. What is the minim

Question

Four packages have an average weight of 10.5 pounds. What is the minimum possible weight of the heaviest package in pounds if the median is 10 pounds?

A. 10
B. 10.5
C. 11
D. 12
E. 13

ManojReddy · Answer

For the heaviest package(x) to be the minimum possible weight, the weight of rest three packages has to be maximized.
As median is 10 pounds, the rest three packages weight can be at max 10 pounds

so we have 10,10,10,x
Mean = 10.5
(30+x)/4 = 10.5
x =12

so minimum possible weight of the heaviest package is 12

Ans:D

ENGRTOMBA2018 · Answer

Let the 4 packages be ABCD with A being the lightest and D being the heaviest.

Given : A+B+C+D =42 and B+C =20 ---> A+D =22

So to minimise the heaviest weight, you need to maximise the lightest.

1 scenario is 10,10,10,12. All other scenarios will violate median to be 10. Thus, D is the correct answer.

KS15 · Answer

To minimize the heaviest weight, we need to maximize the lightest.

Let four packages be a,b,c,d
a+b+c+d=42
b+c/2=10
b+C=20
Only combination possible is 10,10,10,12
Answer D

MBAHopeful99 · Answer

The sum of the packages is 42 (10.5*4).

For the median to be 10, the second and third package in weight must average to 10. Let's assume that both of these numbers are 10. Their sum is 20, which leaves the sum of the two unknowns as 22. If you pick 11 as the heaviest package, the median changes to 10.5. If you pick 12 as the heaviest package, the median is 10. (10, 10, 10, 12)

D

Bunuel · Answer

800score Official Solution:

Let us denote the weights of the packages in pounds by v, x, y, z naming from the lightest one to the heaviest one. The median is 10 pounds. Therefore (x + y) / 2 = 10.
x + y = 20

The average is 10.5 pounds. Therefore (v + x + y + z) / 4 = 10.5.
v + (x + y) + z = 42
v + 20 + z = 42
v + z = 22

The weight v must be no greater than 10, since 10 is the median. Therefore the minimum possible weight of the heaviest package is 22 – 10 = 12 pounds (all the other packages would weigh 10 pounds in this case).

The correct answer is (D).

gracie · Answer

let x=difference between median and package 2 and 3 weights
let y=sum of package 1 and 4 weights
y+(10-x)+(10+x)=4*10.5
y+20=42
y=22
heaviest package is at least 12 pounds
D

stonecold · Answer

Great Question.
Here is my Solution to this one ->

Let the packages be 
w1
w2
w3
w4

Mean=10.5
Hence Sum(4)=10.5*4=42

Now median = 10
Hence w2+w3=20

Now we have to minimise w4 => We must maximise 
w1=w2=Median = 10
Now as w2+w3=20
Hence w3=10

Now w1+w2+w3+w4=42 => 30+w4=42
Hence w4=12

Hence D

altairahmad · Answer

Pardon the ignorance, but why can't the number v be greater than median 10 ?How is median (center value) restricting the highest value ?

Bunuel · Answer

We have  v \leq x \leq y \leq z  and v + z = 22. We want to minimimze z. To minimize z we should maximize v. v cannot be more than 10 because if it is, then x and y will also be greater than 10 and the median (which is (x + y)/2) will be greater than 10.

ZahidPavel · Answer

As it is said about minimum weight for the heaviest package.
Total weight = average weight*number of packages that give 10.5*4 = 42 pounds
We have to maximize the three packages based on the median which gives 10+10 = 20
So, the first one is 10, and the last one must be 12

Ans: D

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Four packages have an average weight of 10.5 pounds. What is the minim

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Four packages have an average weight of 10.5 pounds. What is the minim

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