RRB Assistant Loco Pilot and Technician Model Paper 3 with solutions

(b) Given that, mean of 100 values is 45
∴ Sum of 100 values, i.e. $\sum _{\mathit{i}=1}^{100}\mathit{x}$ = 45 × 100 = 4500
According to condition,
$\sum _{\mathit{i}=1}^{40}\left({\mathit{x}}_{\mathit{i}}+15\right)$ + $\sum _{\mathit{i}=41}^{100}\left({\mathit{x}}_{\mathit{i}}-5\right)$
= $\sum _{\mathit{i}=1}^{40}{\mathit{x}}_{\mathit{i}}+15\times 40$ + $\sum _{\mathit{i}=41}^{100}{\mathit{x}}_{\mathit{i}}-5\times 60$
= $\left(\sum _{\mathit{i}=1}^{40}{\mathit{x}}_{\mathit{i}}+\sum _{\mathit{i}=41}^{100}{\mathit{x}}_{\mathit{i}}\right)+600-300$
= $\sum _{\mathit{i}=1}^{100}{\mathit{x}}_{\mathit{i}}+300$
= 4500 + 300 = 4800 [from equation (i)]
∴ New mean = $\frac{4800}{100}$ = 48