Reetika Hooda in action© AFP
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 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.
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