4. Unlucky Emelya was given several metal balls, from which he broke 3 of the largest ones (their mass was $35 \%$ of the total mass of all the balls), then lost 3 of the smallest ones, and brought the remaining balls (their mass was $8 / 13$ of the unbroken ones) home. How many balls were given to Emelya
Answer: 10 balls.

## Solution:

Let the total mass of all the balls be M. Of these, Emelya broke balls with a total mass of $\frac{35}{100} M = \frac{7}{20} M$, did not break balls with a total mass of $\frac{13}{20} M$, brought home balls with a total mass of $\frac{8}{13} \cdot \frac{13}{20} M = \frac{8}{20} M$, and lost balls with a total mass of $M - \frac{8}{20} M - \frac{7}{20} M = \frac{5}{20} M$.

Notice that Emelya brought home more than 3 balls, because the three heaviest balls together weigh $\frac{7}{20} M$, while he brought home already $\frac{8}{20} M$.

However, Emelya could not have brought home 5 or more balls, because otherwise their average mass would be no more than $\frac{1}{5} \cdot \frac{8}{20} M = \frac{2}{25} M$, while the average mass of the three lightest balls is $\frac{1}{3} \cdot \frac{5}{20} M = \frac{1}{12} M$, and $\frac{1}{12} M > \frac{2}{25} M$ (i.e., the average mass of the three lightest is greater than the average mass of the next few by weight).

Thus, the only possible case is that Emelya brought home 4 balls, which means he was given 10 balls in total.

## Criteria:

Correct answer with incorrect reasoning - 1 point.

Fractions of broken/lost/brought home balls are recorded - plus 1 point.

Proved why he could not bring home 3 or fewer balls - plus 2 points. Proved why he could not bring home 5 or more balls - plus 3 points.