Sundar finished with the figures of 1/12 in his 4 overs against MI.
Indian cricket team head coach Ravi Shastri heaped praise on RCB all-rounder Washington Sundar for his brilliant bowling performance in the 10th game of the ongoing IPL 2020 against Mumbai Indians (MI) on Monday (September 28).
At the Dubai International Stadium on Monday, both MI and RCB managed to score 201 runs in their stipulated 20 overs and the game eventually went into Super Over.
Later, RCB edged past MI in the super over, as Navdeep Saini conceded just 7 runs, dismissing MI’s Kieron Pollard – which AB de Villiers and Virat Kohli easily scored against Jasprit Bumrah in the one-over eliminator to win their second game in the ongoing IPL 2020.
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Well, in the original MI vs RCB IPL 13 encounter, Sundar conceded just 12 runs in his quota of 4 overs, and managed to get rid of MI captain Rohit Sharma – the excellent performance left Shastri in awe for the young bowler as he described his spell as the best IPL performance so far in 2020 season.
Shastri wrote on Twitter after the game: “In a batsman's world - from Chennai to Washington. Best IPL performance so far in 2020. Special.”
Noteworthy, RCB has played three matches and lost only one game in the tournament so far in the UAE. The Virat Kohli-led side will next take on the Rajasthan Royals on October 3 in the IPL 13.