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Hooda posted a 21-16, 19-21, 21-13 triumph over the seasoned Sindhu in a matchup that lasted just over an hour and fifteen minutes in Changzhou.
Unnati Hooda. (X)
Indian teenage badminton sensation Unnati Hooda registered a resounding triumph over her senior compatriot PV Sindhu on Thursday in the Round of 16 encounter between the two at the China Open 2025.
Hooda posted a 21-16, 19-21, 21-13 triumph over the seasoned Sindhu in a matchup that lasted just over an hour and fifteen minutes in Changzhou.
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Hooda opened the fixture on the front foot as she clinched the first game before Sindhu pegged her junior back in the second. But, the youngster held her nerve to eke out the fixture in the deciding third set to further pile on Sindhu’s miseries in an already challenging year for the seasoned campaigner.
The 30-year-old star shuttler, Sindhu, posted a 21-15, 8-21, 21-17 victory over the teenage Japanese shuttler Tomoka Miyazaki in her tournament opener, marking her first win against a top-10-ranked opponent this year and advancing to the Ro16.
Sindhu started the match strong, winning the first game with a six-point margin. However, the Japanese star retaliated in the second game, levelling the score with a significant gap between them.
Despite her recent struggles with form, Sindhu ensured her campaign in China continued by closing out the match in the decisive third set, moving into the next round.
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The 17-year-old Hooda defeated Scot Kirsty Gilmour in straight games, 21-11, 21-16, earning her place in a clash against compatriot PV Sindhu, making history as the teenager progressed further in the tournament.
With her win on over the two-time Commonwealth Games medallist, Gilmour, Hooda’s became only the third Indian, after Saina Nehwal and PV Sindhu, to win a Superseries/Superseries Premier/World Tour 1000 before turning 18.
At the end of the day, it was the youngster who prevailed in the battle between the two Indians on Chinese soil as the teenager advanced to the quarterfinal of the event.
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