?
In a class with a certain number of students, if one student weighting 50 kg is added then the average weight of the class increased by 1 kg. If one more student weighting 50 kg is added, then the average weight of the class increased by 1.5 kg over the original average. What is the original average weight (in kg) of the class?
Answer & Solution
Correct Answer:
Option
D
Let the original average weight of the class be x kg and let there be n students.
Then, sum of weights of n students = (nx) kg
$$\eqalign{ & \therefore \frac{{nx + 50}}{{n + 1}} = x + 1 \cr & \Rightarrow nx + 50 = \left( {n + 1} \right)\left( {x + 1} \right) \cr & \Rightarrow nx + 50 = nx + x + n + 1 \cr & \Rightarrow x + n = 49 \cr & \Rightarrow 2x + 2n = 98.....(i) \cr & {\text{And,}} \cr & \Rightarrow \frac{{nx + 100}}{{n + 2}} = x + 1.5 \cr & \Rightarrow nx + 100 = \left( {n + 2} \right)\left( {x + 1.5} \right) \cr & \Rightarrow nx + 100 = nx + 1.5n + 2x + 3 \cr & \Rightarrow 2x + 1.5n = 97.....(ii) \cr} $$
Subtracting (ii) from (i), we get: 0.5n = 1 or n = 2
Putting n = 2 in (i), we get: x = 47
Then, sum of weights of n students = (nx) kg
$$\eqalign{ & \therefore \frac{{nx + 50}}{{n + 1}} = x + 1 \cr & \Rightarrow nx + 50 = \left( {n + 1} \right)\left( {x + 1} \right) \cr & \Rightarrow nx + 50 = nx + x + n + 1 \cr & \Rightarrow x + n = 49 \cr & \Rightarrow 2x + 2n = 98.....(i) \cr & {\text{And,}} \cr & \Rightarrow \frac{{nx + 100}}{{n + 2}} = x + 1.5 \cr & \Rightarrow nx + 100 = \left( {n + 2} \right)\left( {x + 1.5} \right) \cr & \Rightarrow nx + 100 = nx + 1.5n + 2x + 3 \cr & \Rightarrow 2x + 1.5n = 97.....(ii) \cr} $$
Subtracting (ii) from (i), we get: 0.5n = 1 or n = 2
Putting n = 2 in (i), we get: x = 47
Join the Discussion
Login to post a comment or share your explanation.
Login