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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{fror} \text{'}\mathrm{n}\text{'} \mathrm{obesrevations} {\mathrm{x}}_1, {\mathrm{x}}_2, {\mathrm{x}}_3,\dots \dots , {\mathrm{x}}_{\mathrm{n}} \\ \sum _{\mathrm{i}=1}^{\mathrm{n}} \left({\mathrm{x}}_{\mathrm{i}}-50\right)=-10 \mathrm{and} \sum _{\mathrm{i}=1}^{\mathrm{n}} \left({\mathrm{x}}_{\mathrm{i}}-46\right)=70 \\ \mathrm{This} \mathrm{implies} \mathrm{that}, \\ \underset{\mathrm{i}=1}{\overset{\mathrm{n}}{ \sum }} \left({\mathrm{x}}_{\mathrm{i}}-50\right)=-10 \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}- \sum _{\mathrm{i}=1}^{\mathrm{n}}50=-10 \\ \mathrm{Note} \mathrm{that}, \sum _{\mathrm{i}=1}^{\mathrm{n}}50=\underset{\mathrm{n} \mathrm{times}}{\underbrace{50+50+50+\cdots \cdots +50}}=50\mathrm{n} \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-50\mathrm{n}=-10 .....\left(1\right) \\ \mathrm{From} \mathrm{second} \mathrm{relation}, \\ \underset{\mathrm{i}=1}{\overset{\mathrm{n}}{ \sum }} \left({\mathrm{x}}_{\mathrm{i}}-46\right)=70 \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}- \sum _{\mathrm{i}=1}^{\mathrm{n}}46=70 \\ \mathrm{Note} \mathrm{that}, \sum _{\mathrm{i}=1}^{\mathrm{n}}46=\underset{\mathrm{n} \mathrm{times}}{\underbrace{46+46+46+\cdots \cdots +46}}=46\mathrm{n} \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-46\mathrm{n}=70 .....\left(2\right) \\ \mathrm{Subtracting} \left(1\right) \mathrm{from} \left(2\right) \\ \left(\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-46\mathrm{n}\right)- \left(\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-50\mathrm{n}\right)=70-\left(-10\right) \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-46\mathrm{n}-\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}+50\mathrm{n}=70+10 \\ 4\mathrm{n}=80 \\ \mathbf{n}\mathbf{=}\mathbf{20} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(1\right) \mathrm{to} \mathrm{get}, \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-50\left(20\right)=-10 \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-1000=-10 \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}=-10+1000 \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}=990 \\ \mathrm{Thus} \mathrm{the} \mathrm{mean} \mathrm{of} \mathrm{the} \mathrm{observations} \mathrm{is}, \\ \overline{\mathrm{x}}=\frac{1}{20} \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}=\frac{990}{20}=49.5 \\ \mathbf{So}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{correct}\mathbf{ }\mathbf{option}\mathbf{ }\mathbf{is}\mathbf{ }\left(\mathbf{C}\right) \end{gathered}$$

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