Brisbane: Washington Sundar and Shardul Thakur on Sunday stitched together the highest seventh-wicket partnership for India at The Gabba as they helped the visitors remain in the game on Day Three of the fourth and final Test against Australia.
After losing set batsman Mayank Agarwal and Rishabh Pant early in the session, Sundar and Thakur shared an unbeaten stand of 67 runs to help India reach 253/6 at Tea, still trailing by 116 runs.
Kapil Dev and Manoj Prabhakar had earlier held the record of the highest seventh-wicket partnership for India in Brisbane, scoring 58 runs in 1991.
Resuming the session at 161/4, India lost Agarwal on the second delivery as the right-handed batsman was caught in the slips by Steve Smith against Josh Hazlewood. Agarwal scored 38 runs, with the help of three fours and a six, in the 75 balls he faced during the course of his innings.
Pant (23) then shared a 25-run partnership with Sundar before he became the third scalp of Hazlewood who had the India wicketkeeper-batsman caught at gully by Cameron Green.
However, Thakur and Sundar scripted a brilliant fightback as they not only allowed Australia to take any further wickets but also made sure that the runs kept coming at a decent rate.
India scored 92 runs in 27 overs bowled in the post-Lunch session as they chase Australia’s first innings score of 369. Sundar and Thakur remained unbeaten on 38 and 33 respectively at the Tea Break.
Earlier, in the morning session, India had lost Cheteshwar Pujara and skipper Ajinkya Rahane on their respective scores of 25 and 37 respectively.
Brief scores: India 253/6 at Tea (Washington Sundar 38*, Shardul Thakur 33*; Josh Hazlewood 3/43), Australia 369 all out
(IANS)