IPL 2020: Trent Boult traded to Mumbai Indians; Ankit Rajpoot traded to Rajasthan Royals

Left-arm fast bowler Trent Boult will play for Mumbai Indians in the upcoming season of the Indian Premier League (IPL) after being traded by Delhi Capitals (DC).

The seasoned New Zealand bowler made his IPL debut in 2014 and played for DC in the 2018 and 2019 season. He has 38 IPL wickets from 33 games.

MI has been chasing a left-arm pace bowler for quite some time now. They first signed Jason Behrendorff who either got injured or had to go return to represent Australia during his association with the team. They also tried Barinder Sran in the team last season who didn’t impress in his few outings. The ploy to sign Beuran Hendricks also didn’t work for the four-time IPL champions. They will be finally relieved after signing a renowned bowler like Boult for the upcoming season.

Fast bowler Ankit Rajpoot will represent Rajasthan Royals in the IPL 2020 after a successful trade by Kings XI Punjab. The right-arm fast bowler who joined KXIP in 2018 has played 23 IPL matches and has 22 wickets.

Rajpoot put in one of the most memorable bowling performances ever when he claimed 5/14 in a league game in the 2018 season against Sunrisers Hyderabad (SRH) at the latter’s home ground. His outstanding show earned him the ‘Man of the Match’ award. Rajpoot is the only uncapped player to pick a five-wicket haul in the IPL.

KXIP have traded Rajpoot for Rajasthan Royals’ allrounder K Gowtham. The IPLT20 website called the Boult deal a ‘trade’, but MI did not release any player in exchange.

Well, the trading window is set to close on November 14 as the franchises are busy finalizing the trades. The final list of released and retained players will be announced on November 15.

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CATEGORY: Delhi Capitals IPL Mumbai Indians Punjab Kings Rajasthan Royals T20 Trent Boult

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