HomeIPL 2020: Kings XI Punjab Defeat SunRisers Hyderabad By 12 Runs

IPL 2020: Kings XI Punjab Defeat SunRisers Hyderabad By 12 Runs

King’s XI Punjab Had Posted A Target Of 127 Runs For Sunrisers Hyderabad:

In the 43rd game of IPL 2020 at the Dubai International Cricket Stadium, Kings XI Punjab (KXIP) beat SunRisers Hyderabad (SRH) by 12 runs. On the back of a brilliant effort from the bowling attack, KXIP managed to defend a paltry total of 127. With some tight bowling from Mohammed Shami, Arshdeep Singh and Chris Jordan accounted for three wickets each.

In the powerplay, SRH got off to a good start with openers David Warner and Jonny Bairstow smashing 56 runs. However, with some fast wickets, the KXIP bowlers pulled things back in their favour, finally bowling out SRH for 114. KXIP had previously been limited to 126 runs in total for the loss of 7 wickets in 20 overs.

Against a disciplined SRH bowling attack, the batsmen did not make much of an impact. Even a big-hitter like Nicholas Pooran scored just 32 out of 28 deliveries, finishing as the top run-scorer. Two wickets each were accounted for by Sandeep Sharma, Rashid Khan and Jason Holder.

Neelam Shaw
Neelam Shawhttps://stumpsandbails.com/
I am a journalist carrying experience in Crime journalism. And I am passionate about cricket which has driven me to pick up Sports reporting and writing. I bring you the latest happening in the cricket world both at National and International level.

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