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Home  » Cricket » IPL: Boult to play for Mumbai; Rajpoot for Rajasthan

IPL: Boult to play for Mumbai; Rajpoot for Rajasthan

November 13, 2019 19:28 IST
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Ankit Rajpoot

for 14 for Kings XI Punjab against Sunrisers Hyderabad in IPL 2018. Photograph: BCCI

New Zealand fast bowler Trent Boult will play for Mumbai Indians in next year's Indian Premier League while domestic pacer Ankit Rajpoot will represent Rajasthan Royals after being successfully traded by their respective franchises.

 

According to an IPL statement, Boult will play for Mumbai Indians after being traded by his current team Delhi Capitals, while Rajpoot has been traded successfully by Kings XI Punjab.

Boult made his IPL debut in 2014 and played for Delhi Capitals in the 2018 and 2019 season. He has 38 IPL wickets from 33 games.

Rajpoot, a 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 for 14 in a league game in the 2018 season against Sunrisers Hyderabad. He is the only uncapped player to claim a five-wicket haul in the IPL.

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