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There wheretwo bags which together contained 50 chocolates.from the first bag , 12 chocolates were given to a child. Then 8 chocolates were taken from the

first bag and put into the second bag to make both the bags have equal number of chocolates. How many chocolates were there originally in each bag??? Please answer fast!!!,



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Total chocolates = 50
let first bag has x chocolates.
second bag will have = (50-x) chocolates

when 12 chocolates were removed from first bag, 
number of chocolates in first bag = (x-12)

when 8 chocolates were taken from the first bag and put into the second bag
number of chocolates in first bag = (x-12)-8 = x-20
number of chocolates in second bag = (50-x)+8 = 58-x

now both bags have equal number of chocolates, so
x-20 = 58-x
⇒ x+x = 58+20
⇒ 2x = 78
⇒ x = 78/2 = 39 chocolates
50-x = 50-39 = 11 chocolates

So first bag had 39 chocolates and second bag had 11 chocolates.

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