India 'A' battle to ward off innings defeat

India 'A' battle to ward off innings defeat

 India ‘A’ blew away a strong opening partnership by losing four quick wickets for an addition of just 16 runs as they ended the third day of their four-day game against Australia ‘A’ at 158 for four, still needing 108 runs to avoid the innings defeat, here on Saturday.

After Australia ‘A’ were all out for 435 in reply to India ‘A’s first innings total of 169, openers Akhil Herwadkar and Faiz Fazal stitched 84 runs in 30 overs.

But then India ‘A’ lost their first four wickets in a cluster — three to Jon Holland, the left-arm spinner, and one to a run out — to finish the day on 158 for four in their second innings.

Herwadkar was unbeaten on 82 while Sanju Samson was giving him company on 34 at the end of third day’s play.

The Indian batting collapse began with Fazal being run out for 29. From 84 for no loss in the 30th over, India were reduced to 100 for four in the 37th over, with Holland taking three wickets.

First, Karun Nair was trapped lbw, and then Manish Pandey was caught by Beau Webster, and finally, Naman Ojha was caught by Bird.

Herwadkar and Samson then came together for an unbroken fifth-wicket stand of 58. Samson faced 76 balls for his unbeaten 34, while Herwadkar had faced 188 balls and struck six fours and three sixes.

Earlier, Australia ‘A’, who began the day on 319 for five in their first innings, piled on 435 for a first-innings lead of 266 runs, with Hilton Cartwright, who was unbeaten on 99 overnight, completing his century.

He eventually fell for 117, caught behind off Shardul Thakur, who completed his 10th first-class five-wicket haul and first outside India. He ended with figures of 5 for 101.

Brief scores: India A: 169 & 158/4 (Akhil Herwadkar 82 batting, Sanju Samson 34 batting; Jon Holland 3/59) vs Australia A: 435 (Hilton Cartwright 117, Nic Maddinson 81; Shardul Thakur 5/101, Jayant Yadav 3/95).

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