## Task 6 - 170836

Two plates of the same thickness are made of oak and steel, respectively. The area of the base of the oak plate is 20% larger than the area of the base of the steel plate. The density of oak is to the density of steel as 1:10.

Determine by what percentage the mass of the steel plate is greater than the mass of the oak plate!
Let the following designations be used:

|  | Oak Plate | Steel Plate |
| :--- | :---: | :---: |
| Mass | $m_{E}$ | $m_{S}$ |
| Density | $\rho_{E}$ | $\rho_{S}$ |
| Volume | $V_{E}$ | $V_{S}$ |
| Area of the Triangle | $A_{E}$ | $A_{S}$ |
| Thickness | $h$ | $h$ |

Then,

$$
\text { (1) } \quad m_{E}=A_{E} \cdot h \cdot \rho_{E} \quad, \quad \text { (2) } \quad m_{S}=A_{S} \cdot h \cdot \rho_{S}
$$

and by assumption (3) $\rho_{E}=\frac{1}{10} \rho_{S}$ and (4) $A_{E}=\frac{120}{100} A_{S}$. The mass of the steel plate is $x \%$ of the mass of the oak plate. Then $M_{s}=\frac{x}{100}=m_{E}$.

Consequently, due to (1), ..., (4):

$$
x=\frac{m_{s}}{m_{E}} 100=\frac{A_{S} \cdot h \cdot \rho_{S} \cdot 100}{A_{E} \cdot h \cdot \rho_{E}}=\frac{A_{S} \cdot h \cdot \rho_{S} \cdot 100}{\frac{120}{100} A_{S} \cdot h \cdot \frac{1}{10} \rho_{S}}=833 \frac{1}{3}
$$

The mass of the steel plate is $\left(833 \frac{1}{3}-100\right) \%=733 \frac{1}{3} \%$ greater than that of the oak plate.

Solutions of the III. Round 1977 adopted from [5]

### 5.20 XVIII. Olympiad 1978

### 5.20.1 I. Round 1978, Class 8