Lok Sabha Elections 2019: BJP Releases First List Of 184 Candidates, Smriti Irani Vs Rahul Gandhi In Amethi Again

News World India | 0
| March 22 , 2019 , 09:46 IST

The Bharatiya Janata Party (BJP) released its first list of candidates for the Lok Sabha elections. Prime Minister Narendra Modi will contest the polls from Varanasi, which he represents in the current Lok Sabha, the party said.

BJP National President Amit Shah will contest from Gujarat’s capital Gandhinagar, which is represented by senior party leader LK Advani.

Union Minister Smriti Irani has been renominated from Amethi and will take on Congress president Rahul Gandhi.

Senior party leader J P Nadda told a press conference that the party also decided to field Home Minister Rajnath Singh from Lucknow and another Union minister Nitin Gadkari from Nagpur. Both leaders had won from these seats in the last general election.

The BJP announced its candidates from many states, including Uttar Pradesh, Uttarakhand, Tamil Nadu, Gujarat, Chhattisgarh, West Bengal, Rajasthan, Arunachal Pradesh, and Tripura.

Union ministers VK Singh, Mahesh Sharma, Kiren Rijiju, Jitendra Singh will fight from Ghaziabad, Noida, Arunachal West, and Udhampur respectively. Poonam Mahajan will seek re-election from Mumbai North Central constituency.

In Kerala, the BJP has pitted Kummanam Rajasekharan against Congress veteran Shashi Tharoor from Thiruvananthapuram.

On Sunday, the party had declared candidates for the 123 Assembly seats in Andhra Pradesh and 54 in Arunachal Pradesh. Chief minister Pema Khandu will contest the elections from the Mukto Vidhan Sabha constituency.

The polling will be held on April 11, April 18, April 23, April 29, May 6, May 12 and May 19 for 543 Lok Sabha seats across the country in which nearly 90 crore voters would be eligible to vote.