IPL 2020: Trent Boult to play for Mumbai Indians, Ankit Rajpoot moves to Rajasthan Royals

By: |
November 13, 2019 7:33 PM

Ankit 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.

Trent Boult made his IPL debut in 2014 and played for Delhi Capitals in the 2018 and 2019 season.

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 they were 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 in his kitty.

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.

Get live Stock Prices from BSE, NSE, US Market and latest NAV, portfolio of Mutual Funds, calculate your tax by Income Tax Calculator, know market’s Top Gainers, Top Losers & Best Equity Funds. Like us on Facebook and follow us on Twitter.

Financial Express is now on Telegram. Click here to join our channel and stay updated with the latest Biz news and updates.

Next Stories
1IPL 2020: Virat Kohli-led RCB beat Mumbai Indians in thrilling Super Over game
2FIFA bans soccer club president Marco Trovato for life for fixing matches
3Sensational! Rahul Tewatia hits 5 sixes in an over to help Rajasthan Royals beat Kings XI Punjab