On this day in 1998, 25 years ago, India’s batting maestro Sachin Tendulkar scored an unbeaten 155 versus Australia to inspire one of the greatest comebacks in the history of Test cricket. Tendulkar recorded the feat during the opening Test match of the three-match series during Australia’s tour of India in 1998.
Batting first on the spin-friendly pitch of Chennai, India’s top three Nayan Mongia, Navjot Sidhu, and Rahul Dravid scored 58, 62, and 52 runs, respectively. However, despite their efforts, India could manage only 257 in the first innings due to a middle-order batting collapse.
Australian great Shane Warner picked four wickets for 85 runs, including that of Tendulkar (4 runs off 5 balls).
In reply, propelled by Ian Healy’s gritty 90-run knock, Australia posted 328/10 in 130.3 overs to take 71 runs lead. Anil Kumble bagged four wickets while Venkatapathy Raju picked three.
Tendulkar led India’s fightback during the second innings as they posted a mammoth total of 418 runs for the loss of four wickets before declaring their innings.
Tendulkar hammered a stunning 155 runs off 191 balls with the help of 14 fours and four sixes. The Mumbaikar spent 4.7 hours at the crease and single-handedly scored 37% runs of the team score.
Sidhu scored 64 runs off 127 balls while Dravid contributed with 56 runs.
Then, Indian skipper, Mohammad Azharuddin smashed 64 runs off 89 balls and Sourav Ganguly remained unbeaten at 30.
Chasing 348 runs for a win in the final innings, Australia were bowled out for 168 runs in 67.5 overs to lose the match by 179 runs. Kumble and Raju shared seven wickets between them while Rajesh Chauhan claimed two wickets.
Tendulkar was awarded player of the match for his magnificent efforts with the bat. The second match of the series ended in a draw while Australia won the third Test to level the series 1-1.
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