Question

a group of Children has 120 chocolates with them.eight of each eat 7 chocolate , each of the remaining eats 1 chocolate less than the average number of chocolate eaten by all of them . what could be the maximum number of children in the group?

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Answer 1

Answer:

In first case, you can choose the range [1 , 5] as 1+2+3 + 4 + 5 = 20 , each box will have 5 chocolates. In second case, you can choose the range [3 , 5] as 3 + 4 + 5 = 12 , each box will have 3 chocolates. In third case, there is no way to choose any range such that 8 boxes can be filled equally.

Step-by-step explanation:

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Answer 2

Here

given no of chocolates = 120

let no of students = x

Now

since 8 students ate 7 chocolates each then 56(8*7) chocolates are already eaten →1

we are also given that the remaining students, ie (x-8) students each had one chocolate less than the average

Here

Average = Total no of chocolates/Total no of students

= 120/x

Therefore the remaining students ate

(x-8)*((120/x)—1) chocolates in total → 2

From 1 and 2

56 + (x-8)*((120/x)—1) = 120

=> (x-8)*((120/x)—1) = 64

Upon simplifying we get the following quadratic

x^2–64x-960=0

and solving using Bhaskara's Formula

x = (64 + 16)/2 = 40

or

x= (64–16)/2 = 24

Hence the answer could be 40 or 24, but since the maximum number of students is being asked then the answer would be 40.