Candies

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Candies

by manik11 » Sat Apr 09, 2016 4:46 am
If a jar of candies is divided among 3 children, how many candies did the child that received the fewest pieces receive?
(1) The two children that received the greatest number of pieces received a total of 13 pieces.
(2) The two children that received the fewest number of pieces received a total of 11 pieces

OA : C
Source : Bell Curves

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by [email protected] » Sat Apr 09, 2016 5:16 am
Statement 1: The two children that received the greatest number of pieces received a total of 13 pieces.

This tells us that the minimum number of pieces received by the the child who received the greatest number is 7, in which case the one who got the second most would have gotten 6 pieces. The one who got the least could also have gotten 6 or could have gotten fewer than 6.

Insufficient.

Statement 2:

Since 6 + 5 = 11, this tells us that the minimum the child who received the second fewest pieces could have received is 6, because the second fewest pieces has to be greater than the fewest pieces. However other combinations are possible, combinations such as 7 and 4 and 8 and 3.

Insufficient.

Combined the statements tell us the following.

From Statement 1, if the minimum number of pieces received by the child who got the most is 7, because 6 + 7 = 13 and because the second most has to be fewer than the most, the maximum received by the child who got the second most is 6.

From Statement 2 we know that the minimum value of the second most is 6.

So the value of the second most has to be 6.

From Statement 2 we know that the second most + the least = 11. 11 - 6 = 5 = the number received by the child who received the fewest.

Sufficient.

The correct answer is C.
Last edited by [email protected] on Sat Apr 09, 2016 7:01 am, edited 1 time in total.
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by didieravoaka » Sat Apr 09, 2016 6:36 am
Yes! I got it right Marty. They are in the order 5 (the fewest)-6-7

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by [email protected] » Sat Apr 09, 2016 9:03 am
Hi manik11,

Sometimes the easiest way to answer a Quant question is to 'play around' with it and use 'brute force' to get to the solution. This type of approach is often applicable when the number of 'possibilities' is limited.

Here, we're told that a certain number of candies is divided among 3 children. We're asked to figure out the number candies that went to the child who received the LEAST number of candies.

1) The two children that received the greatest number of pieces received a total of 13 pieces.

Based on this Fact, the number of pieces received by these 2 children could have been...
11 and 2
10 and 3
9 and 4
8 and 5
7 and 6

No other options are possible (remember that there's a 3rd child who received at least one piece of candy, so this list CAN'T have 12 and 1 on it, otherwise the 3rd child - the one who received the LEAST - would receive 0 pieces and that's NOT an option). Given this list of options, the child who received the least number of pieces would have gotten from 1 - 5 pieces, but we don't know exactly how many from this information.
Fact 1 is INSUFFICIENT.

2) The two children that received the fewest number of pieces received a total of 11 pieces

From this Fact, we can create a similar list, but now we're tracking the two children who received the 'lower 2' numbers:

1 and 10
2 and 9
3 and 8
4 and 7
5 and 6

From these options, the child who received the least would have received form 1 - 5 pieces, but we still don't know how many pieces exactly.
Fact 2 is INSUFFICIENT

Combined, we need to 'match' one of the options from the second list with an option from the first list that allows for both "sub-totals" (the 13 and the 11). Starting from the 2nd list - the one that contains the two SMALLER numbers...

1 and 10 doesn't work (the biggest would be "3" and that's no mathematically possible).
2 and 9 doesn't work (the biggest would be "4" and that's no mathematically possible).
3 and 8 doesn't work (the biggest would be "5" and that's no mathematically possible).
4 and 7 doesn't work (the biggest would be "6" and that's no mathematically possible).

5 and 6 DOES work though (the biggest would be "7" and that IS mathematically possible. This is the ONLY possible solution.
Combined, SUFFICIENT.

Final Answer: C

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by [email protected] » Mon Apr 11, 2016 1:30 pm
Let's say that the kids got x, y, and z candies each, where x ≥ y ≥ z.

S1:

x + y = 13

S2:

y + z = 11

Working with the two together, we know that x - z = 2, so the kid who got the most candies got TWO MORE than the kid who got the least.

We also know that the three kids must each have gotten a different number of candies, since

Odd + Even = 13

Even + Odd = 11

That means that y must = x - 1, leaving us with

x = x
y = x - 1
z = x - 2

and three consecutive integers, with a unique solution of x = 7, y = 6, and z = 5.