New Delhi [India], May 1 (ANI): It was on this day, 26 years ago, that former Australia skipper Steve Waugh registered his highest Test score as he went on to play an innings of 200 against West Indies at Sabina Park in Kingston.
In the final Test of the four-match series, playing against a fearsome West Indies attack spearheaded by Courtney Walsh and Curtly Ambrose, Waugh stood firmly at one end and guided the side to post 531 runs in the first innings after the hosts were bundled out for 265.
Waugh's double ton was studded with 17 fours and one six. In the second innings, Paul Reiffel and Shane Warne scalped four wickets each to bundle Windies out for 213.
The visitors won the match by an innings and 53 runs under the leadership of Mark Taylor. Steve Waugh was awarded as the Player of the Match for his spectacular batting performance.
Australia also won the series 2-1 and with this, the team from Down Under became the first side to defeat the invincible Caribbean team on their own soil in 15 years.
Waugh played 168 Tests and scored 10,927 runs at an average of 51.06. He is at the eleventh spot in the all-time run-scorer list of the longest format.
In 325 ODIs, the middle-order batsman amassed 7,569 runs. He hammered 35 centuries and 95 fifties across all formats.
Under Steve Waugh, Australia also won the 1999 World Cup which was played in the UK. (ANI)