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# A child had 5 friends at her birthday party. The children

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A child had 5 friends at her birthday party. The children [#permalink]  23 Oct 2008, 19:21
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A child had 5 friends at her birthday party. The children opened a box containing 21 pieces of candy. Each piece of candy was received by a child. There were no other pieces of candy received by the children at the party. Did each child at the party receive at least 1 piece of candy from the box?

(1) Each child received a different number of candies.

(2) The birthday girl received 6 pieces of candy, which was more than any other child.
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Re: DS [#permalink]  23 Oct 2008, 19:30
study wrote:
A child had 5 friends at her birthday party. The children opened a box containing 21 pieces of candy. Each piece of candy was received by a child. There were no other pieces of candy received by the children at the party. Did each child at the party receive at least 1 piece of candy from the box?

(1) Each child received a different number of candies.

(2) The birthday girl received 6 pieces of candy, which was more than any other child.

there are totally 6 children (including the bday girl).
total no of candies = 21

(1) each child recd diff no of candies --- not sufficient info to determine whether each child received at least a candy

(2) the b day girl recd 6 candies which is more than any other child got. Still not suff info.

(1) and (2)

b day gal recd 6 candies and everyone recd diff no of candies.
so no one else can get 6 or more than 6.

even if we try and give maximum number of candies to each child, we end up giving at least one candy to each child.

Say there are 6 kids a,b,c,d,e and f. say a is the b day gal

a - got 6
b - cant have 6 or more. lets give b the next maximum possible number of candies, that is 5
c - cant have 5 or more, max no. of candies possible is 4
d - 3
e - 2
f - 1

totally 6+5+4+3+2+1 = 21
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Re: DS [#permalink]  23 Oct 2008, 20:20
study wrote:
A child had 5 friends at her birthday party. The children opened a box containing 21 pieces of candy. Each piece of candy was received by a child. There were no other pieces of candy received by the children at the party. Did each child at the party receive at least 1 piece of candy from the box?

(1) Each child received a different number of candies.

(2) The birthday girl received 6 pieces of candy, which was more than any other child.

C.

1+2 tells that each of the girl's friends received some number of candy and that number needs to be less than 6 and that they received different number of candy.

So, 5 children, different numbers and choices we have are 1, 2, 3, 4, 5.
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Re: DS [#permalink]  23 Oct 2008, 21:57
I think A is sufficient. Each child received a different number of candies. That means, no child received 0 candies and in order for the sum to be 21, these will be 1,2,3,4,5,6. Hence, sufficient.
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Re: DS [#permalink]  23 Oct 2008, 22:05
scthakur wrote:
I think A is sufficient. Each child received a different number of candies. That means, no child received 0 candies and in order for the sum to be 21, these will be 1,2,3,4,5,6. Hence, sufficient.

scthakur ... it can be 7,5,4,3,2,0 or some other combination ?? or am i missing something as usual ?
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Re: DS [#permalink]  23 Oct 2008, 22:06
Thanks all - I overlooked the basc assumption - different no of candies!!

OA is C
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Re: DS [#permalink]  24 Oct 2008, 13:10
C

1 - In Suff
2 - In Suff

Together
6,5,4,3,2,1 = 21
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Re: DS [#permalink]  25 Oct 2008, 03:15
study wrote:
Thanks all - I overlooked the basc assumption - different no of candies!!

OA is C

What is the source? I still go for A. If 21 candies are divided among six children and each of them receives different number, how can one receive zero candies?
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Re: DS [#permalink]  25 Oct 2008, 03:16
amitdgr wrote:
scthakur wrote:
I think A is sufficient. Each child received a different number of candies. That means, no child received 0 candies and in order for the sum to be 21, these will be 1,2,3,4,5,6. Hence, sufficient.

scthakur ... it can be 7,5,4,3,2,0 or some other combination ?? or am i missing something as usual ?

How can 0 be the number of candies? May be I am missing something. I saw the OA and somehow I do not seem to agree with OA.
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Re: DS [#permalink]  25 Oct 2008, 04:24
scthakur wrote:
amitdgr wrote:
scthakur wrote:
I think A is sufficient. Each child received a different number of candies. That means, no child received 0 candies and in order for the sum to be 21, these will be 1,2,3,4,5,6. Hence, sufficient.

scthakur ... it can be 7,5,4,3,2,0 or some other combination ?? or am i missing something as usual ?

How can 0 be the number of candies? May be I am missing something. I saw the OA and somehow I do not seem to agree with OA.

Maybe one of the kids got no candy .... the question does not mandate that EVERY kid has to get a candy, in fact the question is whether every kid got at least a candy or not
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Re: DS [#permalink]  25 Oct 2008, 05:25
I am also getting A.

Because of 2 conditions.
1) every one gets atleast 1 piece of candy
2) Every one gets different number of candies.

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Re: DS [#permalink]  25 Oct 2008, 06:02
Twoone wrote:
I am also getting A.

Because of 2 conditions.
1) every one gets atleast 1 piece of candy
2) Every one gets different number of candies.

How did you get the first condition ? The question is asking if everyone gets at least one candy, it does not state that everyone gets one candy ....
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Re: DS [#permalink]  25 Oct 2008, 09:01
amitdgr wrote:
Twoone wrote:
I am also getting A.

Because of 2 conditions.
1) every one gets atleast 1 piece of candy
2) Every one gets different number of candies.

How did you get the first condition ? The question is asking if everyone gets at least one candy, it does not state that everyone gets one candy ....

A " Each child received a different number of candies " and the Q is asking Did each child get at least one candy. So what is a good assumption here? A child received 0 candies makes sense mathematically but does not make sense logically. I believe the Q wants us to consider the option of 0 candies. If A is ignoring that case, we have an easy answer.

Where is this Q from? I could not decide either way and was stuck on the interpretation and then figured out that I have enough company.
Re: DS   [#permalink] 25 Oct 2008, 09:01
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