Find the true weight!

To solve this problem, we shall assume that the scales add or subtract a constant amount from the true weight. The boy's weight, including the error, is recorded as 52 kg. When his sister steps on to the scales, the increase in weight, recorded by the scales, will be equal to her true weight. Hence, the girl weighs 111 52 = 59 kg, and we deduce that the scales subtracts 3 kg from the true weight.

Let $X,Y$ be the actual respective weights, (technically, masses), of the boy and his sister. Now suppose that the scale reading is of the linear form $W = ax + b$ for some real values $a,b$ where $x$ is the actual weight of whoever is on the scale and $W$ is the scale reading.

We then have $3$ equations to work with, namely

(i) $aX + b = 52$ ,

(ii) $aY + b = 56$ , and

(iii) $a(X + Y) + b = 111$ .

Subtracting (iii) from ((i) + (ii)) gives us that $b = 108 - 111 = -3$ .

So now we have that $W = ax - 3$ , and thus

$X = \dfrac{55}{a}, Y = \dfrac{59}{a}$ and $X + Y = \dfrac{114}{a}$ .

If we were to assume that $a = 1$ then we could conclude that $Y = 59$ kg, but this is an assumption not indicated in the wording of the problem. So in fact there is not a unique solution to this problem. For example, we could have $a = \frac{9}{10}$ , which would make $X = \frac{550}{9}, Y = \frac{590}{9}$ and $X + Y = \frac{1140}{9}$ , all in kg..

@abdulrahman khaled Good question, but given the discussion above I believe that the solution is not unique. You may need to add the condition that the scale readings are off by some constant value, (which in this case would be $-3$ kg), and that there is no 'multiplicative scaling factor', which would indicate that in fact $a = 1$ .

I think that this is what you meant by the word "faulty", but this is somewhat vague and requires us to make some assumptions as to your intent.

Let X be the weight of the boy

Let Y be the weight of the girl...

Let Z be the change in the weight that the scale adds or subtracts

So, X+Z=52....(i)

Y+Z = 56

X+Y+Z=111.......(ii)

Subtracting equation (i) from (ii),we get X= $\boxed{59}$

True weigh of boy =52+x

True weigh of girl =56+x

True weigh of both=108+2x ---(1)

The scale express the true weigh as 111+x ---(2)

Solve: x =3

Since the scale is faulty, when the sister scale her weight, the record was wrong. So, just ignore the 56 kg and focus to boy's weight and total of their weights. Let the sister's weight as x and then solve it. It is :

52 + x = 111 <=> x = 111-52 <=> x = 59

So, the true weight is 59 kg.