Indian tennis stars Ramkumar Ramanathan and Ankita Raina both won their first round of qualifying with consummate ease at the Australian Open 2019 in Melbourne on Tuesday. It wasn’t, however, an entirely flawless day for India as Karman Kaur Thandi bowed out.
The 21st seeded Ramanathan needed 1 hour 10 minutes to see off the challenge of the 211th ranked Sergio Gutierrez-Ferrol 6-3, 6-2. Ramanathan produced five aces and had 20 winners in this match. The World No. 132 managed to convert four out of the seven break points that he earned.
The Indian next faces World No. 207 Rudolf Molleker of Germany for a place in the final round of qualifying at the season’s first Major.
The 204th ranked Ankita Raina was dominant in her 6-2, 6-2 win over France’s Myrtille Georges, ranked 35 places below her. The 25-year-old had a good show on her serve, winning 82% of her first serves. Raina thundered 18 winners past her opponent and was able to break the Frenchwoman six times.
This is the third time in her career that Raina is playing in the qualifying rounds of a Slam after the French Open and Wimbledon last year. She now goes on to meet the 29th seed and World No. 142 Paula Badosa Gibert of Spain.
World No. 206 Karman Kaur Thandi had an uphill task against the 16th seed and World No. 125 Jennifer Brady of the USA. After being blanked by the American in the first set, she made a strong comeback in the second set before conceding the match 0-6, 5-7.
The 20-year-old Indian dropped her serve four times and failed to earn a break point in this match, lasting 1 hour 17 minutes.
On Wednesday, the India No. 1 singles player Prajnesh Gunneswaran will start his campaign. The World No. 112 has been seeded sixth and his first opponent is the 228th ranked Viktor Galovic of Croatia.