Rohit needed four runs to get to the landmark and the senior India opener got to it in the final ball of the first over of India’s innings.
(Photo: Twitter/@BCCI)
India vice-captain Rohit Sharma became the third-fastest batsman to reach 9,000 ODI runs during the series-deciding third and final game against Australia in Bengaluru on Sunday.
Rohit needed four runs to get to the landmark and the senior India opener got to it in the final ball of the first over of India’s innings to sit behind skipper Virat Kohli and South Africa’s AB de Villiers.
Kohli achieved the feat in 194 innings and he is followed by de Villiers (205 innings).
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Rohit (216 innings) beat batting greats Sourav Ganguly (228 innings), Sachin Tendulkar (235 innings) and Brian Lara (239 innings) to the feat.
The series is locked 1-1 and here the Aussies rode Steve Smith’s 131 to post 286/9 in 50 overs after winning the toss and electing to bat first.
India, in reply, opener with KL Rahul and Rohit after Shikhar Dhawan had to be taken for an x-ray. Dhawan was taken off the field after he hurt his left shoulder in the series-deciding tie after he dived to save an Aaron Finch shot in the cover region before hurting his left shoulder in the fifth over.
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