(PS-IPL/BCCI)
Jasprit Bumrah becomes the highest wicket-taker for Mumbai Indians (MI) in the IPL when he took the first wicket of the Lucknow Super Giants’ (LSG) innings by removing Aiden Markram — caught by Naman Dhir.
In the first over of his second spell, Bumrah took three wickets in the 16th over of the innings. On the second ball, David Miller flicked a full toss and was caught by Corbin Bosch at deep mid-wicket. On the fifth ball of that over, Bumrah deceived Abdul Samad with a slower ball. Samad had no clue about the delivery and was bowled.
Jasprit Bumrah has taken 174 wickets or MI, surpassing Malinga’s 170
In the very next ball, Bumrah delivered an inch-perfect yorker to Avesh Khan, which was too good for the tail-ender to tackle. In total, Bumrah now has 174 wickets from 139 matches, while Lasith Malinga comes in second with 170 wickets from 122 matches.
MI as a team displayed a great bowling performance at the Wankhede today. Trent Boult gave perfect support to Bumrah, picking up three wickets. He removed Mitchell Marsh (34 off 24 balls) and Ayush Badoni (35 off 22 balls), who were the top run-scorers for LSG.
More importantly, he dismissed them at crucial junctures of the match. Boult finished with figures of 3/20 in his four overs, at an economy rate of 5.0. Credit must also be given to Hardik Pandya for his overall captaincy, particularly for using Will Jacks the ‘bowler’ effectively. Jacks removed Nicholas Pooran and Rishabh Pant — both of whom have the ability to take the match away single-handedly.
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