After clinching a PSA Challenger Title in Canada, Indian squash player Vikram Malhotra has won the Texas Open competition in Houston on Sunday. He defeated Daniel Mekbeb of Czech Republic 11-6, 2-11, 5-11, 15-13, 11-1 to register his eighth PSA title overall in his career thus far.
The fourth-seeded Indian, who moved ahead after a first-round bye, got past three players en route to the final. His second round victim was Syed Hamzad Bukhari of Pakistan, followed by Reuben Phillips of England in the quarterfinals and another Pakistani Shahjahan Khan, the eighth seed in the semi-final 11-6, 9-11, 11-5, 11-6.
Incidentally, the top seed in this event was Ramit Tandon but he conceded his match in the quarter-final to Shahjahan Khan, Vikram's semi-final victim.
The Mumbai-born squash player started taking part in PSA World Tour in 2015 and also won his maiden title in the same year in Betty Griffin Memorial Florida Open. Malhotra is ranked 72 in the world rankings and has won three major events this year which includes Atlanta Open as well which was held in the
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