India head coach Ravi Shastri has hailed Washington Sundar after his stunning performance for Royal Challengers Bangalore against Mumbai Indians on Monday. In a game where all the other bowlers conceded at almost 8 or more than 8 runs per over, the young spinner conceded at just three.
The match between RCB and MI became the highest ever tied score in the history of IPL as both the teams finished on 201 before the Bangalore-based outfit won the super over.
Asked to bat first, RCB rode on fifties from Aaron Finch, Devdutt Padikkal and AB de Villiers to post 201 for 3, their highest score in the tournament so far. With the ball, they started well and had Mumbai Indians reeling at 78 for 4 in the 12th over. At that stage, it looked all over for the reigning champions.
However, a 119-run stand between Ishan Kishan and Kieron Pollard helped Mumbai Indians bounce back and push the game to the super over. In the super over, Navdeep Saini bowled brilliantly to restrict Mumbai to 7. In reply, AB de Villiers and Virat Kohli sealed the game for their side on the final ball of the super over.
Washington Sundar special:
In a game where batsmen put up a power-hitting spectacle, Washington Sundar made his presence felt with the ball. He bowled three of his four overs in the powerplay when Mumbai were looking for a quick start and conceded just seven runs in addition to picking up the big wicket of captain Rohit Sharma.
— Royal Challengers Bangalore (@RCBTweets) September 28, 2020
And when he was reintroduced in the attack in the tenth over, he did not disappoint either as he conceded just 5 singles to finish with stunning figures of 12 runs for 1 wicket in a game where more than 400 runs were scored in just 40 overs. And following the game, Ravi Shastri took to Twitter to lavish praise on the young spinner, writing:
“In a batsman’s world – from Chennai to Washington. Best IPL performance so far in 2020.”
— Ravi Shastri (@RaviShastriOfc) September 28, 2020