the 10 weights recorded were: 86,92,96,100,104,109,110,113,119,127.

what is the weight of the second heaviest boy?

(explain your logic)

on
30-Aug-2016 18:51

67.5

67.5

on
9-Sep-2016 22:16

59

59

on
9-Sep-2016 22:25

Sum of weights of all boys = (86+92+96+....+127)/4 = 264. Let A be weight of lightest boy, B be weight of 2nd lightest boy and so on (upto E) A+B=86; D+E=127 Weight of 3rd heaviest/lightest boy = C = 264 - (127+86) = 51. Second smallest weight combination is 92. This must be sum of weights of lightest and 3rd lightest boys. (i.e) A+C=92 Since C=51, A =92. Since A+B=86, B=45 Third smallest weight combination may be B + C or A + D Clearly B=45 and C=51, therefore B + C =96 Thus B + C is the 3rd smallest weight combination. Fourth smallest weight combination must be A+D A+D=100 Since A=41, it follows D=59. Since D+E=127, it follows E=68. The values can be verified using the given weight combinations. Answer: Weight of second heaviest boy is 59.

Sum of weights of all boys = (86+92+96+....+127)/4 = 264. Let A be weight of lightest boy, B be weight of 2nd lightest boy and so on (upto E) A+B=86; D+E=127 Weight of 3rd heaviest/lightest boy = C = 264 - (127+86) = 51. Second smallest weight combination is 92. This must be sum of weights of lightest and 3rd lightest boys. (i.e) A+C=92 Since C=51, A =92. Since A+B=86, B=45 Third smallest weight combination may be B + C or A + D Clearly B=45 and C=51, therefore B + C =96 Thus B + C is the 3rd smallest weight combination. Fourth smallest weight combination must be A+D A+D=100 Since A=41, it follows D=59. Since D+E=127, it follows E=68. The values can be verified using the given weight combinations. Answer: Weight of second heaviest boy is 59.

on
24-Mar-2017 11:22

Answer of VIDYANATHAN T IS correct!

Answer of VIDYANATHAN T IS correct!