In a hotel , 60% had vegetarian lunch while 30 % had non-vegetarian lunch and 15% had both types of lunch. If 96 people were present ,how many did not eat either type of lunch ?
n(A) = `(60/100 xx 96) = 288/5`, n(B) = `(30/100 xx 96) = 144/5` n`(A nn B)` = `(15/100 xx 96)` = `72/5`
`:.` n`(A uu B)` = n(A) + n(B) - n`(A nn B)`= `288/5 + 144/5 - 72/5` = `360/5` = 72
So, people who had either or both types of lunch = 72
Hence ,people who had neither type of lunch = 96 - 72 = 24.
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