Nishikori & Medvedev to clash in Brisbane International final

Speed Post News Network

Patna : World No 9 in ATP rankings Kei Nishikori, 29, Japan, and World No 16 Daniil Medvedev, 22, Russia, will clash in the final of Brisbane International, an ATP 250 point tournament, at Brisbane, Queensland, Australia, on January 6, 2019.

Nishikori, the 2014 US Open finalist, beat World No 40 Jeremy Chardy, 31, France, 6-2, 6-2 in the semi-finals on January 5, 2019. Chardy fired 10 Aces in the match to just 2 by Nishikori. Chardy committed 3 double faults to 2 by Nishikori. But, Nishikori’s winning percentage on both his first serve ( 79%) and second serve ( 75%) was much better than that of Chardy. Nishikori also won 4 break points besides winning 5 games in a row.

Medvedev beat World No 239 Jo-Wilfried Tsonga, 33, France, 7-6, 6-2 in the semi-finals. Tsonga hit 17 Aces in the match to 10 by Medvedev. But, Medvedev won 3 break points to 1 by Tsonga. Medvedev won 5 games in a row in the match to 3 by Tsonga. Tsonga won 43 service points to 42 by Medvedev. Medvedev’s winnings winning percentage on both his first and second serve was much higher than that of Tsonga.