math revolution certainly loves candies & children 2 continuous questions !
120 candies have to be split among 25 kids and no kid can get mora than 7 candies.
but we have to maximize the no. of kids who only get 1 candy each and that is possible by maximizing kids with 7 candies each.
using the options is better as we ll know what are the limits of possibilities so taking the biggest value from option E. 9
if 9 kids have 1 candy each then
remaining candies= 111
remaining kids = 16
nearest multiple of 7 to 111 is 7×16=112 which is greater than 111 this solves our problem
so we can give 15 kids 7 each = 105 candies
1 kid gets 6 candies
and remaining 9 get 1 each
everything adds up
we don't need to check other options as we needed highest no. and we already started with the highest option and it met the requirements.
so option E is correct
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