a) 2
b) 4
c) 46
d) 47
Answer: d
Explanation: Let the original average weight of the class be x kg and let there be n students.
Sum of weights of n students = (nx) kg
$$\eqalign{
& \frac{{nx + 50}}{{n + 1}} = x + 1 \cr
& nx + 50 = \left( {n + 1} \right)\left( {x + 1} \right) \cr
& nx + 50 = nx + x + n + 1 \cr
& x + n = 49 \cr
& 2x + 2n = 98…..(i) \cr
& {\text{And,}} \cr
& \frac{{nx + 100}}{{n + 2}} = x + 1.5 \cr
& nx + 100 = \left( {n + 2} \right)\left( {x + 1.5} \right) \cr
& nx + 100 = nx + 1.5n + 2x + 3 \cr
& 2x + 1.5n = 97…..(ii) \cr} $$
Subtracting (ii) from (i), we get: 0.5n = 1 or n = 2
Putting n = 2 in (i), we get: x = 47
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