Paris: Indian wrestler Reetika Hooda displayed power and skill in equal measure while outclassing Bernadett Nagy of Hungary to enter the quarter-final of the women's 76 kg freestyle event at the Paris Olympics here on Saturday.
The referee stopped the contest with 29 seconds left in the second round as Reetika gained a 10-point lead with the final score reading 12-2 in favour of the Indian.
The burly Reetika gained upper-hand with 4-0 lead in the first round after getting an early leg-hold followed by a flip.
The Hungarian did get a couple of points but the Indian was unstoppable in the second round with a series of two pointers. She will face top-seed Aiperi Medet Kyzy of Kyrgyzstan in the quarterfinal later today.
Published 10 August 2024, 11:02 IST