, 2003

Solution:

The answer is $70$. Let $P_{A}$ be the weight of nougat $A$, $V_{A}$ the volume of nougat $A$, $P_{B}$ the weight of nougat $B$, $V_{B}$ the volume of nougat $B$. We have that $30\%\,P_{A}$ occupies $40\%\,V_{A}$, so the specific weight of the hazelnuts is $\frac{3}{4} \frac{P_{A}}{V_{A}}$. Considering the weight and the space occupied by the white paste in nougat $A$, we instead find that its specific weight is $\frac{7}{6} \frac{P_{A}}{V_{A}}$. So to calculate the volume occupied by the hazelnuts and by the white paste in nougat $B$ we divide their weight, that is $60\%\,P_{B}$ for the hazelnuts and $40\%\,P_{B}$ for the white paste, by their specific weight, obtaining that the former occupy $\frac{4}{5} \frac{P_{B}}{P_{A}} V_{A}$ and the latter $\frac{12}{35} \frac{P_{B}}{P_{A}} V_{A}$. Adding the two volumes we obtain the total volume, that is
$$V_{B} = \frac{4}{5} \frac{P_{B}}{P_{A}} V_{A} + \frac{12}{35} \frac{P_{B}}{P_{A}} V_{A} = \frac{8}{7} \frac{P_{B}}{P_{A}} V_{A}.$$
To calculate the percentage occupied by the hazelnuts we only need to divide the volume occupied by them by the total volume of the nougat:
$$\frac{\frac{4}{5} \frac{P_{B}}{P_{A}} V_{A}}{\frac{8}{7} \frac{P_{B}}{P_{A}} V_{A}} = 70\%$$
So the hazelnuts occupy $70\%$ of the volume of nougat $B$.

Solution:

In the first nougat, with $30\%$ of the weight the hazelnuts made up $40\%$ of the volume, and therefore the paste with $70\%$ of the weight made up $60\%$ of the volume. Thus if the "voluminosity" of the hazelnuts is $4/3$, that of the paste turns out to be $6/7$. Therefore the share of volume of the hazelnuts in the second nougat is
$$\frac{60 \cdot 4/3}{60 \cdot 4/3 + 40 \cdot 6/7} = \frac{80}{80 + 240/7} = \frac{560}{800} = 70\%$$