IPL 2022: Bhuvneshwar-led Sunrisers Hyderabad win toss,

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Stand-in Sunrisers Hyderabad skipper Bhuvneshwar Kumar has won the toss and elected to bat first against Punjab Kings in the final league match of IPL 2022 at the Wankhede Stadium on Sunday.

Both Hyderabad and Punjab are already out of the race for the playoffs and would look to finish off their IPL 2022 campaign on a high.

Bhuvneshwar further said that all-rounder Romario Shepherd and left-arm spinner Jagadeesha Suchith replace Kane Williamson and T Natarajan in the playing eleven.

“Same wicket as the last game so could be a bit sticky going into the second innings. We would love to win this match and go up a bit in the points table. It looks like a slow surface and we’re backing ourselves to defend a score,” stated Kumar.

Punjab Kings skipper Mayank Agarwal said pacer Nathan Ellis and lower-order batter Shahrukh Khan come into the playing eleven alongside debutant all-rounder Prerak Mankad in place of Bhanuka Rajapaksa, Rahul Chahar and Rishi Dhawan.

“We don’t want to make changes to the structure of the team, just giving chances to everyone. Every game is important and we’re looking forward to the two points. We are bowling first, so we’ll know what target we’ll be chasing tonight,” said Agarwal.

Playing XIs:

Punjab Kings: Jonny Bairstow, Shikhar Dhawan, Mayank Agarwal (captain), Liam Livingstone, Jitesh Sharma (wicketkeeper), Shahrukh Khan, Harpreet Brar, Nathan Ellis, Kagiso Rabada, Prerak Mankad and Arshdeep Singh

Sunrisers Hyderabad: Abhishek Sharma, Priyam Garg, Rahul Tripathi, Aiden Markram, Nicholas Pooran (wicketkeeper), Romario Shepherd, Washington Sundar, Bhuvneshwar Kumar, Fazalhaq Farooqi, Umran Malik and Jagadeesha Suchith

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