India moves into the lead by 58 runs thanks to Thakur's seven wickets

Despite Shardul Thakur’s seven-wicket haul for India on Tuesday, the second test against South Africa remained tense at the Wanderers.

At conclusion, South Africa lead by 27 runs after being knocked out for 229 and India 85-2 in their second innings.

KL Rahul and Mayank Agarwal perished , but Cheteshwar Pujara and Ajinkya Rahane smashed the home bowling in the final half-hour to give India a 58-run advantage.

Rahul was caught in the slips by Marco Jansen for eight and Agarwal was caught leg before by Duanne Olivier for 23.

To help India set a challenging final innings goal for South Africa, Pujara (35 not out off 42 balls) and Rahane (11) continue on day three.

India led 1-0 after winning the first test by 113 runs.

After South Africa started the day on 35-1, Thakur, 30, bowled a career-best 7-61, the greatest return by an Indian bowler in a Test against South Africa.

Keegan Petersen and Temba Bavuma both hit a half-century as South Africa chased down India’s 202.

After the home captain had shrugged off an early bombardment of hostile bowling for 90 minutes, Thakur had Dean Elgar caught behind for 28.

Petersen reached his maiden test half-century but was caught in the slips by Thakur for 62.

Van der Dussen was out cheaply on the final ball of the first session, however television replays showed the ball may have slipped short of Rishabh Pant’s glove.

The umpires found no clear evidence that the catch was short and upheld the decision.

After lunch, Bavuma and Kyle Verreyne put on 60 runs until Verreyne was caught leg before for 21.

Thakur subsequently got his fifth wicket when Bavuma mishit a leg-side ball and was out for 51.

Thakur also grabbed the final two wickets of South Africa, yielding potentially valuable runs as the Wanderers track is predicted to favor seamers throughout the test.

 

Spread the love

Leave a Reply

Your email address will not be published.

WhizzPost
%d bloggers like this: