100 jellybeans were distributed to a group of 9 people such that the 3

100 beans, 9 people
3 people have most beans of all.
Sum=60
Remaining beans for 6 more people=40
the person with least beans among 6 have 5 beans.
Remaining people=5
Remaining beans=35
Average beans for 5 people=7 each.
The three people with most beans will have more than the 7 each.
Let's consider the second most and the third most having 8 each to make the first person having the max beans.
60 - 2 x 8 = 44 (total number of maximum beans with a person)
The ratio of max to min = 44/5
But they have not given the correct option for this question.

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given no one has <5 jelly beans
and for 3 people they have 60 among them
so balance 6 people are left with 40 jelly beans
considering that 1 amongst 6 people has 5 beans so other 5 would have ; 40-5/5 ; 7 each
we get person with max jelly beans
5+7*7+x=100
x= 46
ratio of highest to lowest ; 46/5
OPTION B

Is it wrong to simply choose B because the result has to be "something divided by 5" as the minimum jellybeans one can have will be equal to 5 in order to maximize the ratio ?

To maximize the ratio we need to assume the person with second most and third most beans also have 7 beans each.


The second most and third person should have 8 beans each because they have the most jelly beans. They cannot have any number equal to or less than 7

Then you are not maximizing the number of beans with the person having most beans. Sum of top 3 doesn't imply that the third maximum value has to be greater than the fourth maximum value. It can be the same as is in this case. The second and third maximum value can be assigned to anyone from rank 2 to rank 8 as they all have equal number of beans and sum of top 3 adds up to 60.

Hope it's clear.

chetan2u Sir can you please suggest a solution here. The above solutions are not understandable . many thanks.

Logical

To maximise ratio, the denominator should be minimum and it is given that the least is not less than 5.
So denominator should be a multiple of 5. But is it possible that 5 gets cut from both numerator and denominator??
NO, because numerator will then become 46*5, which is greater than 100.
Hence B


Method

100 jelly beans between 9 of them.
3 having the max have 60 within themselves
Remaining 100-60 or 40 is distributed amongst 9-3 or 6.
Maximise Ratio means maximise the numerator and minimise the denominator.
=> Minimise the denominator- Not <5, so minimum value 5.
=> Maximise the numerator- So, let the remaining 7 be minimum possible. For this let us take it into two parts
I) 40 distributed amongst 6 : we had to minimise the lowest so we took that as 5. Now we want the highest in these 5 also to be lower as it affects the lowest of top 3. So remaining 5 =(40-5)/5=7
II) 60 amongst top 3 : Let the lowest two be equal to 7 as they cannot be less than 7. So highest =(60-2*7)=46

Fraction =46/5

Simple approach:


100 jellybeans to be split into two groups ( 3 (having most), 6)

Group of 3 (has most = 60) - Group of 6

46_ 7 _ 7 _ (total 60) - 6_ 6_ 6_ 6_ 6_ 5_

The least number of jellybeans one can get is 5 ( since that it has to be least others can't be 5 but to be the least all others are given 6 each in second group.
since that the group of 3 is mentioned to have most we have to take the number greater than 6 ( to make it most) thus if we assume second and third highest to be 7 in group 1.
The highest number of jellybeans one posses = 46
The lowest number of jellybeans one posses = 5

Ratio = [46][/5]
Thus option B.

chetan2u thank you very much sir

I have a serious problem in dealing with such kind of problems. Can anybody please help me out how can I attack such problems. Any key strategies?

For example:-
In this problem this was my approach which of course did not match with the options

Scenario given

5 _ _ _ _ _( _ _ H)

(_ _ H) = 60

Hence my approach was to make H highest was

5 5 5 5 10 10 20 20 20

I took this approach because in other such questions, to get the highest of the list of numbers, if i replace the rest with the least numbers, it works.

But of course i couldnt find it out this way.

I saw the other discussons and realised this is the correct set of numbers

5 7 7 7 7 7 7 7 7 46

which was found out by keeping least number 5 one time.. and H being the highest equations were

5 + 7x + H = 100
2x + H = 60

H = 46, x=7

I need help desperately

Bunuel bb VeritasKarishma egmat chetan2u

This is a type of max-min question. Here are some posts on such questions:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... base-case/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -extremes/

In this question, the first thing I do is give 5 to each of the 9 people. 45 are gone and 55 are left.

5, 5, 5, ... 5, 5, 5

I need to maximise Greatest/Least so Greatest must be as large as possible and least must be as small as possible (5).

The largest 3 must have total 60. They already have 15 so another 45 need to be given to them. We are left with 10 beans. These 10 must be given to the other 6 people. Since I want the least to be smallest, I give it no beans and just give 2 each to the other 5 people.

5, 7, 7, 7, 7, 7, ??, ??, ??

The last 3 must be at least 7 and the greatest must be as big as possible. Let's give 7 each to two of them and 46 to the greatest one

5, 7, 7, 7, 7, 7, 7, 7, 46

Answer (B)

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