Business Standard

7 Indian-American kids win 2019 Spelling Bee


IANS Washington
Seven Indian-American children were among the eight winners of this year's edition of the Scripps National Spelling Bee, that saw more than one champion for the first time in its 92-year history.
The winners on Thursday night were Erin Howard, Rishik Gandhasri, Saketh Sundar, Navneeth Murali, Shruthika Padhy, Sohum Sukhatankar, Abhijay Kodali, Christopher Serrao and Rohan Raja, the American Bazaar reported.
Howard's victory marks the first time since Sameer Mishra won the contest in 2008.
Thursday night's finals went on more than an hour and half past the scheduled time with words like omphalopsychite, Geeldikkop and auftaktigkeit, CNN reported.
Five rounds and 47 words in a row were spelled perfectly. Each of the eight champions will receive a $50,000 prize.
This year's competition started on Sunday with 562 spellers -- all of whom are 15 or younger but have not passed eighth grade -- who made it to the national stage.
Contestants came from all 50 US states, as well as several territories and other countries including the Bahamas, Canada, Germany, Ghana, Jamaica, Japan and South Korea.
Last year, Karthik Nemmani, a 14-year-old eighth-grader from Texas, was named the winner.
In 1985, Balu Natarajan became the first Indian-origin child to win the Spelling Bee.
The Spelling Bee is not merely a memory test of exotic words as it also requires learning about the origins of words and the languages they are derived from.

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First Published: May 31 2019 | 11:56 AM IST

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