Question

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?

• A. 2
• B. 1
• C. 5
• D. 4
• E. 3

Answer 1

Answer: (E)

3

Answer 2

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.