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 Pumpkins 2 (Posted on 2005-09-08)
Five pumpkins are weighed two at a time in all possible combinations, similar to the first pumpkins puzzle. The results of the weighings gives nine different values: 52, 56, 60, 68, 72, 76, 80, 84, and 88. One of the values was repeated, but which value was not written down.
Find out the weights of the pumpkins and which value is the repeat.

 See The Solution Submitted by Brian Smith Rating: 3.0000 (2 votes)

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 Puzzle Solution Comment 5 of 5 |

Let the five weights be P, Q, R, S, T with P <= Q <= R <= S <= T.

Then, from the given weight pair sums, it follows that;

(P+Q, P+R, S+T, T+R) = (52, 56, 88, 84) ...........(i)

Let the weight sum that is duplicated be x. Then, adding all the weight sums, we have:

4(P+Q+R+S+T) = 636+x
-> P+Q+R+S+T) = 159+x/4  .......(ii)

Then, S = (159+x/4) -(84+52) = 23 + x/4

Similarly, (Q, R) = (15+x/4, 19+x/4), giving:

P+Q = 52 -> P = 37 - x/4

T+R = 84 -> T = 65 - x/4

Hence: (P,Q,R,S,T) = (37-x/4, 15+x/4, 19+x/4, 23+x/4, 63-x/4)

Now, Q>= P -> 15+ x/4 >= 37-x/4 -> x>= 44
Also, S< = T -> 123 x/4 >= 65-x/4 -> x<= 84

So, 44<= x <= 84. Since x must be divisible by 4, the possible vaues of x are:

x = 44,48,52,56,60,64,68,72,76,80,84

Form (iii), the pisr sum weights in some order are:

(52,56,60,102-x/2, 34+x/2, 38+x/2, 80, 42+x/2, 84, 88)

Substituting x = 44,48,52,56,60,64,68,72,76,80,84 in turn, we observe that only x=68 in fully in consonance with the given observations, and the  pair sums for x=68, arranged in increasing order of magnitude are: (52,56,60,68,68,72,76,80,84,88), with 68 being the observation that is repeated twice.

Now, x= 68 in (iii) gives: (P, Q, R, S, T) = (20, 32,36,40,48)

Consequently the required weights in increasing order of magnitude are: 20, 32,36,40 and 48.

 Posted by K Sengupta on 2008-11-12 03:47:05

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