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Eleven chocolates are distributed among 5 children: A, B, C, D & E

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Eleven chocolates are distributed among 5 children: A, B, C, D & E  [#permalink]

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New post Updated on: 24 Nov 2018, 18:19
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A
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C
D
E

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  35% (medium)

Question Stats:

58% (01:17) correct 42% (01:16) wrong based on 64 sessions

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Eleven chocolates are distributed among 5 children: A, B, C, D & E: such that everyone gets at least one chocolate. What is the number of chocolates that E gets?

(1) D & E get equal number of chocolates.

(2) Each of A, B, C, D get a distinct number of chocolates from each other.

Originally posted by amanvermagmat on 22 Nov 2018, 23:02.
Last edited by chetan2u on 24 Nov 2018, 18:19, edited 1 time in total.
Corrected the OA
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Re: Eleven chocolates are distributed among 5 children: A, B, C, D & E  [#permalink]

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New post 23 Nov 2018, 18:25
Eleven chocolates are distributed among 5 children: A, B, C, D & E: such that everyone gets at least one chocolate. What is the number of chocolates that E gets?

(1) D & E get equal number of chocolates.
Say both D and E get one each, so remaining 9 can be distributed between remaining
Say both get 2 each, again remaining (11-2*3) can be distributed amongst the remaining three.
Insufficient

(2) Each of A, B, C, D get a distinct number of chocolates from each other.
So each one gets atleast 1, which means the least possible amongst all four of them would be 1+2+3+4=10.
As there are only 11 chocolate, E could have got only one.
As there are no other possibility of distribution of chocolates, E would get one.
Sufficient

B
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Re: Eleven chocolates are distributed among 5 children: A, B, C, D & E  [#permalink]

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New post 24 Nov 2018, 17:03
amanvermagmat wrote:
Eleven chocolates are distributed among 5 children: A, B, C, D & E: such that everyone gets at least one chocolate. What is the number of chocolates that E gets?

(1) D & E get equal number of chocolates.

(2) Each of A, B, C, D get a distinct number of chocolates from each other.


amanvermagmat

From 1:
a+b+c+d+e=11
since d=e
so can be any value a+b+c<=9 as minimum of value of D & E would be 1..
not sufficient

From 2:
a,b,c,d get distinct no of chocolates
so 1+2+3+4+x=11
x=1
so E got 1 chocolate

IMO B should be correct ..

I am not sure why is answer given as C? when the solution to the question is coming from B...


By C option we can say that either of A,B,C distribution was between ( 4,3,2) and amongst D & E (1,1).. i think C would be correct option only if when the question would have asked us to find value for both D & E...

GMATinsight : sir could you please check why C is given as correct over option B..
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Re: Eleven chocolates are distributed among 5 children: A, B, C, D & E  [#permalink]

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New post 25 Nov 2018, 04:01
Archit3110 wrote:
amanvermagmat wrote:
Eleven chocolates are distributed among 5 children: A, B, C, D & E: such that everyone gets at least one chocolate. What is the number of chocolates that E gets?

(1) D & E get equal number of chocolates.

(2) Each of A, B, C, D get a distinct number of chocolates from each other.


amanvermagmat

From 1:
a+b+c+d+e=11
since d=e
so can be any value a+b+c<=9 as minimum of value of D & E would be 1..
not sufficient

From 2:
a,b,c,d get distinct no of chocolates
so 1+2+3+4+x=11
x=1
so E got 1 chocolate

IMO B should be correct ..

I am not sure why is answer given as C? when the solution to the question is coming from B...


By C option we can say that either of A,B,C distribution was between ( 4,3,2) and amongst D & E (1,1).. i think C would be correct option only if when the question would have asked us to find value for both D & E...

GMATinsight : sir could you please check why C is given as correct over option B..



Hello

Yes, the OA was written as C by mistake. The answer is indeed B.

And now OA has already been edited by Chetan.

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Re: Eleven chocolates are distributed among 5 children: A, B, C, D & E  [#permalink]

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New post 28 Nov 2018, 06:09
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Hello,

This problem involves using the minimization concept in a subtle way, but the core of the problem involves addition of positive integers, which is the main concept being tested here. So, as with most Data Sufficiency questions, it is always a good idea to try and figure out the concept/s which are being tested in the question, rather than just trying values.

This is a Data Sufficiency question which demands a definite value as an answer.

The data given in the question statement mentions that each of the 5 persons viz, A, B, C, D and E recieved at least 1 chocolate out of a total of 11 chocolates distributed among them. Hence, we can directly subtract 5 from 11 leaving us with 6 chocolates, which we can try and distribute among the 5 people in such a way as to answer the question.

Using statement I alone, we will not be able to say the exact value of E. As mentioned above, out of the 6 chocolates remaining, 3 can go to E and 3 can go to D. Therefore, E has 4 chocolates. Another combination is, out of the 6 chocolates, 2 can go to E and 2 can go to D (making their total 3) and the remaining 2 can be distributed in any way among the remaining 4 persons. This is evidence enough that, using the first statement alone, we will not be able to find a unique value of E.

Statement II alone - Each of A, B, C and D got distinct number of chocolates.

Let us approach this in a slightly different way. Since everyone has to get at least 1 chocolate, let us give E one chocolate. Now we are left with 10 chocolates, which have to be distributed among A, B C and D such that it satisfies the condition given in the second statement.

10 is an even number. When 4 numbers are added to get an even number as a result, the following combinations are possible

1) All 4 can be even
2) All 4 can be odd
3) 2 numbers can be odd and 2 numbers can be even

However, the first 2 combinations can be ruled out since the sum of the first four even numbers and the sum of the first four odd numbers will exceed 10. Therefore,

A + B + C + D = 10 can only be satisfied by substituting the values 1, 2, 3 and 4 in any order.

If E gets 2 chocolates (which he definitely can ),
then, A+ B + C + D = 9, for which the possible values could be 1,2,3,3 or 1,3,4,1 or 2,2,2,3 and so on. Hence, we see that, as the total gets smaller, some values are bound to repeat.

As such, we can safely say that the only combination of values that satisfies the conditions given in the question statement and statement II is (1,2,3,4). Hence, we can conclude that E gets 1 chocolate. Therefore, the second statement alone is sufficient to answer the question. So, answer option B.

A pertinent point to be noted is that, when you try to solve DS questions using values, you have to be able to back it up using concepts. If you can figure out the concept prior to plugging in values, there's nothing better, because then it means that you would not be plugging random values, but only those values that the concepts dictate.

Hope this helps!
Cheers,

CrackVerbal Academics Team
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Re: Eleven chocolates are distributed among 5 children: A, B, C, D & E &nbs [#permalink] 28 Nov 2018, 06:09
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