IPL 2026: RCB Beat DC in Just 6.3 Overs, Register Record Win

New Delhi.

Royal Challengers Bengaluru (RCB) defeated Delhi Capitals (DC) by 9 wickets in the 39th match of the Indian Premier League (IPL) 2026, played at Arun Jaitley Stadium on Monday. Chasing a modest target of just 76 runs, RCB completed the victory in only 6.3 overs. With this, RCB became the second team in IPL history to win a match with the most balls remaining.

RCB won the match with 81 balls to spare. Mumbai Indians (MI) hold the top position in this category, having chased down a target of 68 runs against Kolkata Knight Riders (KKR) in 2008 with 87 balls remaining.

In the match, Delhi Capitals, who were put in to bat first after losing the toss, were bowled out for just 75 runs in 16.3 overs. This was DC’s third-lowest total in IPL history. The team had previously been dismissed for 66 runs against MI in 2017 and 67 runs by Kings XI Punjab in the same year.

This was also the third-lowest total by any opposition team against RCB. In 2009, Rajasthan Royals (RR) were bowled out for just 58 runs against RCB, while in 2023, RR were dismissed for 59 runs.

With this win, RCB moved to joint second place for the most wins against a single team in IPL history. RCB registered their 21st win against DC, equaling records such as Chennai Super Kings vs RCB (21 wins), KKR vs Punjab Kings (21 wins), MI vs Chennai Super Kings (21 wins), MI vs DC (21 wins), and Chennai Super Kings vs KKR (21 wins). MI leads this list with 25 wins against KKR.

At Arun Jaitley Stadium, DC lost 6 wickets for just 8 runs while batting first. From there, Abishek Porel (30) and David Miller (19) somehow helped the team reach 75 runs. For RCB, Josh Hazlewood picked up 4 wickets, while Bhuvneshwar Kumar claimed 3 wickets.

In reply, RCB chased down the target in 6.3 overs. Devdutt Padikkal remained unbeaten on 34 off 13 balls, while Virat Kohli scored an unbeaten 23 off 15 balls.

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