We have two types of screws. The combined weight of 3 smaller and 2 larger screws was found to be 319 grams on a scale that measures with gram accuracy. By replacing one smaller screw with a larger one, the scale reads 351 grams. What is the average weight of each screw and what is the accuracy of the results?
Let the average mass of a smaller screw be $x$, and that of a larger one be $y$ grams. Then, from the system of equations

$$
\begin{aligned}
& 3 x+2 y=319 \\
& 2 x+3 y=351
\end{aligned}
$$

we get

$$
x=\frac{3 \cdot 319-2 \cdot 351}{3 \cdot 3-2 \cdot 2}=51\left(=x_{0}\right), \quad y=\frac{3 \cdot 351-2 \cdot 319}{3 \cdot 3-2 \cdot 2}=83\left(=y_{0}\right)
$$

Considering the precision achievable with a balance, the numerical values of the two measurement results should be any $d$, and $e$ such that

$$
318.5 \leqq d \leqq 319.5 \quad \text { or } \quad 350.5 \leqq e \leqq 351.5
$$

because our balance indicates 319 grams for any mass between 318.5 and 319.5 grams, and 351 grams for any mass between 350.5 and 351.5 grams.

Instead of (I) and (II), we thus have the system

$$
3 x+2 y=d, \quad 2 x+3 y=e
$$

from which

$$
x=\frac{3 d-2 e}{5}, \quad y=\frac{3 e-2 d}{5}
$$

To characterize the accuracy of our results (1) by the largest deviation that can occur between (1) and (3) as $d$ and $e$ take on all possible pairs of values, similar considerations to those in problem 617 would show that the numerator of $x$ will be largest when $d$ takes its largest value and $e$ its smallest value:

$$
x \leqq \frac{3 \cdot 319.5-2 \cdot 350.5}{5}=51.5, \quad \text { or } \quad x_{\max }-x_{0}=0.5 \text { grams }
$$

Similarly, we get

$$
x_{\min }=\frac{3 \cdot 318.5-2 \cdot 351.5}{5}=50.5
$$

$$
y_{\max }=\frac{3 \cdot 351.5-2 \cdot 318.5}{5}=83.5,
$$

$$
y_{\min }=\frac{3 \cdot 350.5-2 \cdot 319.5}{5}=82.5
$$

The largest $|x-x_{0}|$ and $|y-y_{0}|$ deviations are 0.5 grams in all cases. We can say that the accuracy of the results (1) is the same as that of the individual measurements, as if we had measured one small and one large screw separately.

Tamás Görbe (Budapest, Bem J. g. I. o. t.)

Remarks. 1. The deviations can also be examined in the following form:

$$
x-x_{0}=\frac{3(d-319)-2(e-351)}{5}, \quad y-y_{0}=\frac{3(e-351)-2(d-319)}{5}
$$

2. The final conclusion of the solution is surprising, as it is natural to assume that the accuracy can be increased by measuring several identical objects at once, i.e., the upper limit of the possible error can be made smaller. Indeed, if we measured 5 small screws as 255 grams and 5 large screws as 415 grams, we would have obtained the values of $x_{0}$ and $y_{0}$ with an accuracy of 0.1 grams. - The apparent contradiction is resolved by considering that the result of a single measurement can deviate either above or below the true value; in this problem, however, the calculation of $x_{0}$ and $y_{0}$ is burdened by a double possibility of error from the two independent measurement readings. The direction of the two deviations also has 4 different pairs to consider: up-up, up-down, down-up, down-down. We saw that two deviations of opposite signs reinforce each other; however, two deviations in the same direction distort the result less, e.g., from $d=319.4$ and $e=351.3$, we get $x=51.12$ and $y=81.02$.
3. The accuracy of the results is equal to the accuracy of the individual measurements in every case where the number of small screws in one measurement is one more than the number of large screws in the other. Indeed, from the system of equations

$$
k x+(k-1) y=d, \quad(k-1) x+k y=e
$$

where $k$ is the number of small screws,

$$
x=\frac{k d-(k-1) e}{2 k-1}, \quad y=\frac{k e-(k-1) d}{2 k-1}
$$

and here, for example, in the numerator of $x$, an increase of $0.5 k$ can occur with the largest value of $d$ and $0.5(k-1)$ with the smallest value of $e$, totaling $0.5(2 k-1)$, so the largest possible increase in $x$ is 0.5 grams.

4. In a graphical solution similar to the second solution of problem 617, the strips corresponding to equations I and II determine a parallelogram instead of the intersection point $M$. The lines of both strips slope from left to right, because, for example, in I, increasing $x$ decreases $y$; furthermore, they move away from the intersection of the axes with the increase of $d$ and $e$. From this, it is clear that $x_{\min }$ and $y_{\max }$ are given by $d_{\min }$ and $e_{\max }$, and $x_{\max }$ and $y_{\min }$ by $d_{\max }$ and $e_{\min }$.

![](https://cdn.mathpix.com/cropped/2024_05_02_32fe3222c950c95f68dfg-2.jpg?height=656&width=715&top_left_y=379&top_left_x=694)

5. It is intuitively plausible - although its proof would lead far afield - that if the ratio of the coefficients of $x$ and $y$ in I and II, instead of $3 / 2$ and $2 / 3$, is more different from 1, then the lines representing them will slope at a smaller angle to one of the axes, and the strips will be narrower, and therefore the "long" diagonal of the parallelogram will be shorter, and its projections on the axes will also be shorter, making the inaccuracy of $x$ and $y$ smaller.