In a family of 5 sisters, the average weight of the 4 sisters weighing the least is 50 kg, and the average weight of the 4 sisters weighing the most is 55 kg. What is the difference between the maximum and minimum possible overall average weight?
- 2
- 1
- 5
- 4
- 3
The Correct Option is
Solution and Explanation
Let the weights of the five sisters in increasing order be a, b, c, d, and e kg.
So, the sum of the weights of the 4 sisters weighing the least = \(50\times 4 = 200\) kg
The sum of the weights of the 4 sisters weighing the most = \(55\times 4 = 220\) kg
Overall, sum of the weights of 5 sisters = \(200 + e = 220 + a\)
\(\Rightarrow e-a = 20\)
Now, in order to have the maximum overall average weight, the value of e in the sum value should be maximum and that happens when the value of a is maximum.
So, the maximum value of a can be taken as \(50\).
Thus, the value of e will be \(70\).
Hence, the overall sum will be \(200 + e = 200 + 70 = 270\) kg
So, the maximum overall average weight = \(\frac{270}{5} = 54\) kg
Now, to find the minimum overall average weight, the value of a should be minimum, which will happen when e is minimum.
So, the minimum possible value of \(e = 55\) kg
So, \(a = 35\) kg
Thus, the overall sum will be = \(220 + 35 = 255\) kg
So, the minimum overall average weight = \(\frac{255}{5} = 51\) kg
Thus, the required average = \(54-51 = 3\) kg
Hence, option E is the correct answer.
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