Al, Bert, and Carl are the winners of a school drawing for a pile of

Question

Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of 3 : 2 : 1, respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the first to arrive. If each takes what he believes to be his correct share of candy, what fraction of the candy goes unclaimed?

(A)  \frac{1}{18}

(B)  \frac{1}{6}

(C)  \frac{2}{9}

(D)  \frac{5}{18}

(E)  \frac{5}{12}

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Rucha.Shukla · Answer

Is the answer D, 5/18 ?

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Nidzo · Answer

As they each come at different times thinking they are the first, they each assume that the amount of sweets present is the full amount and thus will take the fraction value of the amount present.

The ratio of the amount of sweets Al, Bert, and Carl will each get of 3 : 2 : 1 in terms of fractions will be:  \frac{1}{2} ,  \frac{1}{3}  and  \frac{1}{6} . However, as we want to work out what is left we should look at it rather as Al will leave  \frac{1}{2}  of wwhat is infront of him, Bert will leave  \frac{2}{3}  of what is infront of him and Carl will leave  \frac{5}{6}  of what he sees infront of him.

Multiplying these values together will yield the fractional value of the amount of sweets which remain, regardless of the order of who comes first, second or third.

\frac{1}{2}*\frac{2}{3}*\frac{5}{6} = \frac{5}{18}

ANSWER D

Alternatively one could assign a value to the number of sweets available. As we are dealing with the fractions of  \frac{1}{2} ,  \frac{1}{3}  and  \frac{1}{6}  a good number to pick is  2*3*6 = 36

Once again, order is irrelevant. Let's assume Al comes first, he will take 18 and leave 18. Then Bert comes and he takes 6 and leaves 12. Finally Carl comes and takes 2 sweets and leaves 10.

The amount left over as a fraction will be:  \frac{10}{36} , simplified to  \frac{5}{18}

ANSWER D

Purnank · Answer

I think its D.

candies are divided into the ratios of 3:2:1 for A:B:C (Al, Bert and Carl) so total candies must be in multiple of 6. 
Therefore assuming total candies 216.
Now here we will check for cases assuming first who came is A after that B and C is the last one. Similarly we will also check the case where C is first B is second and A is third.

1st Case - 1st A 2nd B and 3rd C
No. of candies picked by A = 3 * 216 / 6 = 108. Now Remaining Candies = 216-108 = 108.
No. of candies picked by B = 2 * 108 / 6 = 36.  Now Remaining Candies = 108-36 = 72.
No of candies picked by C = 1 * 72 / 6 = 12.  Now Remaining Candies = 72-12 = 60.

Therefore fraction of the candy goes unclaimed = 60/216 = 5/18.

Lets check for one more case - 1st C 2nd B and 3rd A.
No of candies picked by C = 1 * 216 / 6 = 36.  Now Remaining Candies = 216-36 = 180.
No. of candies picked by B = 2 * 180 / 6 = 60.  Now Remaining Candies = 180-60 = 120.
No. of candies picked by A = 3 * 120 / 6 = 60. Now Remaining Candies = 120-60 = 60.

Therefore fraction of the candy goes unclaimed = 60/216 = 5/18

and now if you check for another case (like 1st B 2nd A 3rd C) you will end up getting same result.

Hence D.

Themightyknight · Answer

Let's assume the no of candies be 36
And each takes half of whats leftover since every one thinks he is first 
So each assumes he is first to arrive thus 
1st one - takes 18 
18 are left out 
2nd one thinks he is frst and takes 
9 out of it

3rd one thinks he is frst thus he takes 
4.5 out of it ????
So left out is 4.5 ??

Can someone explain 
 KarishmaB   bb   Bunuel   GMATNinja

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KarishmaB · Answer

They have to take candies in the ratio 3:2:1 so Al gets 1/2 of all, Bert gets 1/3rd of all and Carl gets 1/6th of all.

If there are 36 candies, when Al arrives, he takes 18 and leaves 18.
When Bert arrives and thinks he is the first one to arrive, he thinks that total candies are 18 so he takes 1/3rd of them i.e. 6 so he leaves 12.
When Carl arrives and he thinks he is the first one to arrive, he takes 1/6th of the 12 candies i.e. 2 candies. So 10 candies are left unclaimed.

So 10 out of 36 i.e. 5/18 go unclaimed.

Answer (D)

Kinshook · Answer

Given: Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of 3 : 2 : 1, respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the first to arrive.

Asked: If each takes what he believes to be his correct share of candy, what fraction of the candy goes unclaimed?

Let us assume that total candies are x.

Ideal condition: 
Al's share = 3x/6 = x/2
Bert's share = 2x/6 = x/3
Carl's share = x/6

Non-ideal condition: 
Al's share = x/2; Remaining candies = x/2
Bert's share = x/2*1/3 = x/6; Remaining candies = x/2 - x/6 = 2x/6 = x/3
Carl's share = x/3*1/6 = x/18; Remaining candies = x/3 - x/18 = 5x/18

The fraction of the candy goes unclaimed = (5x/18)/x = 5/18

IMO D

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=88298 --- Harsha GMAT-Club-Forum-df2uiy59.png

Al, Bert, and Carl are the winners of a school drawing for a pile of

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