AKB48 Watanabe Mayu hopes to win Japan Record Awards for Takamina

16 Dec

AKB48 Watanabe Mayu, Mukaichi Mion, HKT48 concurrent member Miyawaki Sakura appeared at a press conference for “57th Japan Record Award” (30th December on TBS) on the 15th December.

This year, their 40th single “Bokutachi wa Tatakawanai” has won the “Execellent Work Award” and the group will be up against strong contenders Sandaime J Soul Brothers for the “Japan Record Award”.

Watanabe Mayu talked about winning the award for exiting member Takahashi Minami. “It’s been 10 years since we were formed, she’s the one who thinks about the members most. I really respect her. This is her last Japan Records ceremony. For all her hard work so far, we want to gift the award to her.” She also said the winning the award would make it a memorable year for AKB48’s 10th anniversary.

AKB48 previously won the “Japan Record Award” in 2011 for “Flying Get” and 2012 for “Manatsu no Sounds Good!”. This year they will be up Sandaime J Soul Brothers, who won last year for their smash hit “R.Y.U.S.E.I”, as well as other artistes.

Source

One Response to “AKB48 Watanabe Mayu hopes to win Japan Record Awards for Takamina”

  1. Denny Sinnoh December 20, 2015 at 12:24 am #

    Off topic: Sakura looks grown up now.

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