Reetika Hooda, an Indian wrestler, defeated Bernadett Nagy of Hungary to get to the quarterfinals of the women’s 76 kg freestyle event at the Paris Olympics.
On Saturday, Indian wrestler Reetika Hooda showed both force and skill by defeating Hungary’s Bernadett Nagy in the quarterfinals of the women’s 76 kg freestyle event at the Paris Olympics. The referee halted the game with 29 seconds left in the second round, giving Reetika a 10-point advantage and a final score of 12-2 in favour of the Indian. The burly Reetika took a 4-0 lead in the opening round after landing an early leg-hold and flip.
The Hungarian did get a few points, but the Indian proved unstoppable in the second round with a string of two-pointers. She will face Kyrgyzstan’s top seed, Aiperi Medet Kyzy, in the quarterfinals later today.
