Creative Method To Obtain All The Weights

Relevant wiki: Systems of Linear Equations - Problem Solving

Each of the boys is on the scale a total of 4 times, so the sum of the 10 weighings will be 4 times the weight of the 5 boys together. As the sum of the given values is $1056$ kg, we know that the sum of the weights of the 5 boys is $1056/4 = 264$ kg.

Letting $x_{k}, 1 \le k \le 5$ , be the weights of the 5 boys in ascending order, we have that

$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 264, x_{1} + x_{2} = 86, x_{4} + x_{5} = 127$ and $x_{3} + x_{5} = 119$ .

Then $x_{3} + x_{4} + x_{5} = x_{1} + x_{2} + x_{3} + x_{4} + x_{5} - (x_{1} + x_{2}) = 264 - 86 = 178$ ,

and so $x_{4} = x_{3} + x_{4} + x_{5} - (x_{3} + x_{5}) = 178 - 119 = \boxed{59}$ kg.

Let the weights of the boys be $a<b<c<d<e$ . From the question ,we have

$a+b=86\\a+c=92\\c+e=119\\d+e=127$

This implies that

$e=a+27\\c=b+6\\d=c+8$

Now, consider the sum resulting in 96. It is either $a+d$ or $b+c$ . Clearly it cannot be $a+d$ as $d=c+8$ . Hence $b+c=96 \Rightarrow b=a+4\Rightarrow e=d+9$ . Thus, we have $d = a+18$ .

Solving the very first equation, $a+b=86$ , we have $2a+4=86\Rightarrow a=41\Rightarrow d = a+18=59$ .

Now, let the weights of the boys be $a,b,c,d,e$ . The weights here are arranged in ascending order.

Each boy will be weighed $4$ times within the $10$ weighings. This means that if we add up all the $10$ measurements, we get

$4(a+b+c+d+e) = 86+92+96+100+104+109+110+113+119+127\\ a+b+c+d+e = \dfrac{1056}{4} = 264$

Now, we know that the smallest measurement is for $a+b$ , and the largest measurement is for $d+e$ . This gives:

$a+b = 86\quad\quad\quad d+e=127$

Substitute these two into the equation above:

$86+c+127 = 264 \implies c = 51$

Notice that the second last sum must be $c+e$ . This is because $c+e > c+d$ and $c+e > b+e$ . Therefore,

$c+e = 119\\ 51 + e = 119\\ e= 68$

Now, we can find the weight of the second heaviest boy, which is $d$ :

$d+e = 127\\ d+68=127\\ d=\boxed{59}$

Note: In a similar method, you can use the first two sums to find $a$ and $b$ . The answer would be $a=41,\;b=45$

I guess I took a shortcut. Total weight of the four boys is 264 (add the 10 numbers together and divide by 4, since each boy is part of 4 pairs). Subtract out the top two (127) and the bottom 2 (86 - NOT 82 as shown in the graphic!). This means the middle boy weighs 51.

The maximum weight for the heaviest is 70 (none can weigh more than 70, or they would be weighed on their own and eliminated from the problem). This means the minimum weight of the second heaviest is 127 - 70 = 57. The maximum weight for the second heaviest would be 63, since 127/2 = 63.5.

The possible weights of the second heaviest or then 57. 58. 59, 60, 61, 62, and 63. Knowing that the middle one weighs 51 and knowing all the pairs of weights, we can whittle the possibilities down to 58, 59, and 62. There is no other numbers on the list of possibilities that when combined with 51 equal one of the 10 results.

This means the heaviest boy weighs 65, 68, or 69. Only 68 will work as neither of the other two possibilities when combined with 51 equal one of the ten results.

So the second heaviest weighs 127 - 68 = 59. Basic arithmetic that can all be done in the head - didn't even need a pencil.