ujjagrawal Posted June 29, 2012 Report Share Posted June 29, 2012 Mr. Smith, a school teacher needs to weigh 5 of his students. In a puzzling mood, he weighted them in couples. The output weights are 114, 116, 118, 120, 121, 122, 123, 124, 125 and 129 pounds. After sometime of mathematical jugglery, he was able to tell individual weights of each of the students. Can you tell what are their individual weights? Quote Link to comment Share on other sites More sharing options...
0 MikeD Posted June 29, 2012 Report Share Posted June 29, 2012 I'm just posting my answer, will post reasoning and stuff later when I have more time. 56, 58, 60, 64, 65 Quote Link to comment Share on other sites More sharing options...
0 MikeD Posted June 29, 2012 Report Share Posted June 29, 2012 I had some spare time now so here are my workings. (a+b) + (a+c) + (a+d) + (a+e) + (b+c) + (b+d) + (b+e) + (c+d) + (c+e) + (d+e) = (114+116+118+120+121+122+123+124+125+129) 4a + 4b + 4c + 4d + 4d = 1212 a + b + c + d + e = 303 Assume that a is the lightest and e is the heaviest. So that a + b must be 114 and d + e must be 129 therefore a+b+d+e = 243 So c = 303 - 243 c = 60 Therefore a and b must be less than 60 but when added together equals 114. Possibilities: (Assuming no two students are the same weight) Case 1 55 + 59 Case 2 56 + 58 If we use a+c c = 60 a = 55 a+c = 115 (can't be no sure value weighed) C = 60 a = 56 a+c = 116 (value in list) Therefore a = 55 and b = 58. Now to find d and e. d and e must be more than 60 but when added together equals 129. Case 1 61 + 68 Case 2 62 + 67 Case 3 63 + 66 Case 4 64 + 65 Using c+e case 1 68 + 60 =128 (not in list) case 2 67 + 60 =127 (not in list) case 3 66 + 60 =126 (not in list) case 1 65 + 60 =125 (is in list) therefore d = 64 and e = 65. 2 Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted June 29, 2012 Report Share Posted June 29, 2012 Add them all and divide by 4 for the sum of the 5 weights (since each person is weighed with four others). This gives a total weight of 303. Subtracting away the lightest pairing (two lightest people) and heaviest pairing (two heaviest people) gives the middle person's weight. 303-114-129 = 60, so the middle person weighs 60. The difference between the two lightest pairings gives the difference between the second and third people. 116-114=2, so the second lightest person weighs 58. 114-58 = 56, so the lightest person weighs 56. The difference between the two heaviest pairings gives the difference between the third and fourth people. 129-125=4, so the second heaviest person weighs 64. 129-64 = 65, so the heaviest person weighs 65. So the weights of all the students are {56, 58, 60, 64, 65}. Quote Link to comment Share on other sites More sharing options...
0 ujjagrawal Posted July 2, 2012 Author Report Share Posted July 2, 2012 I'm just posting my answer, will post reasoning and stuff later when I have more time. 56, 58, 60, 64, 65 I had some spare time now so here are my workings. (a+b) + (a+c) + (a+d) + (a+e) + (b+c) + (b+d) + (b+e) + (c+d) + (c+e) + (d+e) = (114+116+118+120+121+122+123+124+125+129) 4a + 4b + 4c + 4d + 4d = 1212 a + b + c + d + e = 303 Assume that a is the lightest and e is the heaviest. So that a + b must be 114 and d + e must be 129 therefore a+b+d+e = 243 So c = 303 - 243 c = 60 Therefore a and b must be less than 60 but when added together equals 114. Possibilities: (Assuming no two students are the same weight) Case 1 55 + 59 Case 2 56 + 58 If we use a+c c = 60 a = 55 a+c = 115 (can't be no sure value weighed) C = 60 a = 56 a+c = 116 (value in list) Therefore a = 55 and b = 58. Now to find d and e. d and e must be more than 60 but when added together equals 129. Case 1 61 + 68 Case 2 62 + 67 Case 3 63 + 66 Case 4 64 + 65 Using c+e case 1 68 + 60 =128 (not in list) case 2 67 + 60 =127 (not in list) case 3 66 + 60 =126 (not in list) case 1 65 + 60 =125 (is in list) therefore d = 64 and e = 65. Add them all and divide by 4 for the sum of the 5 weights (since each person is weighed with four others). This gives a total weight of 303. Subtracting away the lightest pairing (two lightest people) and heaviest pairing (two heaviest people) gives the middle person's weight. 303-114-129 = 60, so the middle person weighs 60. The difference between the two lightest pairings gives the difference between the second and third people. 116-114=2, so the second lightest person weighs 58. 114-58 = 56, so the lightest person weighs 56. The difference between the two heaviest pairings gives the difference between the third and fourth people. 129-125=4, so the second heaviest person weighs 64. 129-64 = 65, so the heaviest person weighs 65. So the weights of all the students are {56, 58, 60, 64, 65}. I understand you have got it right but let me put my version of answer too. Let e > d > c > b > a be the student's weights. No two students have the same weights, otherwise there would have been duplicate weighing. e+d must be the heaviest and e+c the second heaviest weighing. a+b must be the lightest and a+c the second lightest weighing. The sum of the weighings is 1212 and each student is weighed four times => a+b+c+d+e = 1212/4 = 303 c = (a+b+c+d+e) - (a+b) - (d+e) = 303 - 114 - 129 = 60 a+c = 116 => a = 116 - 60 = 56 a+b = 114 => b = 114 - 56 = 58 c+e = 125 => e = 125 - 60 = 65 e+d = 129 => d = 129 - 65 = 64 So the student weights are: 56, 58, 60, 64 and 65 Quote Link to comment Share on other sites More sharing options...
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ujjagrawal
Mr. Smith, a school teacher needs to weigh 5 of his students. In a puzzling mood, he weighted them in couples.
The output weights are 114, 116, 118, 120, 121, 122, 123, 124, 125 and 129 pounds.
After sometime of mathematical jugglery, he was able to tell individual weights of each of the students.
Can you tell what are their individual weights?
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