India beat Zimbabwe by nine wickets in Harare, take 1-0 lead

After impressive show with the ball, India rode on KL Rahul-Ambati Rayudu stand to chase down Zimbabwe's 168.

By: Express Web Desk | Updated: June 11, 2016 9:04 pm
Indian batsman Lokesh Rahul, left, and Ambati Rayudu chat on the pitch during the One Day International cricket match against Zimbabwe at Harare Sports Club, Saturday, June, 11, 2016. The Indian cricket team is in Zimbabwe for One Day International and T20 matches.(AP Photo/Tsvangirayi Mukwazhi) KL Rahul and Ambati Rayudu anchored the visitors home with a solid stand. (Source: AP)

An unbeaten century from debutant opener Lokesh Rahul steered India to an easy nine- wicket victory over hosts Zimbabwe in the first one-day International on Saturday.

Rahul smashed a six off the last ball of the innings to finish on 100 not out from 115 balls and complete a dream debut.

Also Read: Rahul’s debut ton, Harare’s rare feat

India won the toss and sent Zimbabwe in to bat, restricting the home side to 168 all out one ball short of their 50 overs.

They were measured in their response but always in front of the rate and reached 173 for one wicket with 45 balls to spare.

The tourists lost another debutant, Karun Nair (seven), early on but an unbeaten 162-run second-wicket stand between Rahul and Ambati Rayudu (62 not out) made it an easy chase.

Zimbabwe had earlier struggled from the start of their innings as a cautious approach failed to stop them losing wickets at regular intervals.

Former captain Elton Chigumbura top-scored with 41 and seamer Jasprit Bumrah proved the pick of the Indian bowlers with four for 28 from 9.5 overs.

The second game in the three-match series is at the same venue on Monday.

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