NEW DELHI: Indian wrestler Reetika Hooda defeated Hungary's Bernadett Nagy on Saturday to advance to the quarterfinals of the women's 76 kg freestyle competition at the Paris Olympics. Hooda demonstrated both force and finesse in equal measure.
With 29 seconds remaining in the second round, the match was called off by the referee as Reetika took a 10-point lead; the Indian ultimately won 12-2, as per PTI.
In the opening round, the powerful Reetika took the upper hand with a 4-0 lead after securing an early leg-hold and flip.
The Indian proved unstoppable in the second round with a string of two-pointers, although the Hungarian did manage a handful of points. Later today, she will play the top-seeded Aiperi Medet Kyzy of Kyrgyzstan in the quarterfinal.
