Governor Abubakar Atiku Bagudu has been declared winner of the governorship election in Kebbi State.
The presiding officer, professor Armu Yau Hamisu, said Bagudu of the All Progressives Congress (APC) emerged winner in the election, having scored the highest number of votes 673,717 with a difference of 567,084 against the 106,633 votes scored by the Peoples Democratic Party’s candidate, Senator Isa Galaudu, in the election.
The returning officer said the total registered voters is 1,789,075 while a total of 820078 were accredited for the election.
He said the total votes cast in the election was 814,084 while a total of 20,696 were rejected.
He said the “APC candidate, Abubakar Atiku Bagudu, having scored the highest number of votes in the election is hereby declared winner and returned”.
The APC candidate defeated his PDP opponent in all the 21 local government areas of the state.
The presiding officer, professor Armu Yau Hamisu, said Bagudu of the All Progressives Congress (APC) emerged winner in the election, having scored the highest number of votes 673,717 with a difference of 567,084 against the 106,633 votes scored by the Peoples Democratic Party’s candidate, Senator Isa Galaudu, in the election.
The returning officer said the total registered voters is 1,789,075 while a total of 820078 were accredited for the election.
He said the total votes cast in the election was 814,084 while a total of 20,696 were rejected.
He said the “APC candidate, Abubakar Atiku Bagudu, having scored the highest number of votes in the election is hereby declared winner and returned”.
The APC candidate defeated his PDP opponent in all the 21 local government areas of the state.
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