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Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?
(1) No two friends purchased the same amount of fudge.
(2) The minimum increment in which the chocolatier sells fudge is one pound.
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]
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23 Jan 2015, 08:12
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Bunuel wrote:
Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?
(1) No two friends purchased the same amount of fudge.
(2) The minimum increment in which the chocolatier sells fudge is one pound.
Kudos for a correct solution.
Let's calculate the average pound per person --> 3,2.
1) set up two sets S=(a; b; c; d; e) and set Q=(f; g; h; i; j). No two friends purchased the same amount of fudge. We expect \(a+b+c+d+e = f+g+h+i+j=16\). Assume S (1; 2; 3; 4; 6) the answer to our question is yes. Assume Q (3,2; 3,3; 3,4; 3,5; 2,6) the answer to our question is no.
2) set up two sets S and Z. Assume S (1; 2; 3; 4; 6) the answer to our question is yes; assume Z (3; 3; 3; 3; 4) the answer to our question is no.
1+2) minimum increment is one. Set Q is not acceptable, thus the answer is C.
hope it helps _________________
learn the rules of the game, then play better than anyone else.
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]
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23 Jan 2015, 10:07
The two statement is contradict. So the answer must A, B, D, E.
S1/ The average fudge/people is 16/5 = 3.2 All can less than 5, regardless of any equal value.
s2/ Suppose we don't have any person buy more than 5. So the most fudge bought by a person must be 5. The increment is at least 1. So the other value could be 4, 3, 2, 1. Total is 15 => Supposing statement is incorrect. There must be one person bought more than 5 pounds.
The two statement is contradict. So the answer must A, B, D, E.
S1/ The average fudge/people is 16/5 = 3.2 All can less than 5, regardless of any equal value.
s2/ Suppose we don't have any person buy more than 5. So the most fudge bought by a person must be 5. The increment is at least 1. So the other value could be 4, 3, 2, 1. Total is 15 => Supposing statement is incorrect. There must be one person bought more than 5 pounds.
Some thing like Dirichlet distribution.
Answer: B
ans C ...you are missing the point where some of them buy same no of fudge for eg 4,4,4,2,2, STATEMENT 2 ALONE WILL NOT BE SUFFICIENT
_________________
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]
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24 Jan 2015, 20:18
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Sum of 5 friends=16 pounds Question: Did any one of the friends buy >5 pounds?
(1) No two same amounts: Answer to the question is YES: (6, 1,2,3,4) Answer to the question is NO: (1.5, 2.5, 3.5, 4, 4.5) Two possible answers, hence insufficient.
(2) The minimum amount is 1 pound. YES: (6, 1,2,3,4) NO: (4,4,4,2,2) Insufficient.
(1)+(2) The only case that satisfy both statements is (6, 1,2,3,4) Answer C
Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?
(1) No two friends purchased the same amount of fudge.
(2) The minimum increment in which the chocolatier sells fudge is one pound.
In this Min/Max problem, your goal is to determine whether the minimum value of the highest total can or must be greater than 5. Statement 1 is not sufficient largely because the question (as statement 2 hints) does not prohibit fractional pounds, so the 16 total pounds can be distributed with all values under 5 (for example, 1, 2, 3, 4.4, and 4.6) or with one value well above 5 (0.1, 0.2, 0.3, 0.4, and the rest all going to the large value).
Statement 2 serves as a check against statement 1 ("Why Are You Here?") but is not sufficient on its own. But taken together, the statements tell you that the five values must be unique integers. To get to 15 total (one less than the question's mandated 16 pounds), there are really two ways:
Minimize the low value at 0, which guarantees that one value has to be above 5 (0, 1, 2, 3 for the lower 4, or 0, 2, 3, 4 all require that the high value is over 5).
or
Maximize the low value at 1, which gives the distribution 1, 2, 3, 4, 5. But since there is a 16th pound, and by adding one to any of the values 1-4 you'd match one of the higher values, the extra pound must go to the high value, meaning that even if you maximize the lower values, the minimum high value must be 6. Accordingly, the statements together guarantee the answer "yes" and the answer is C.
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05 Mar 2016, 04:14
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