Question

In one fortnight of a given month , there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280km*1km , show that the total rainfall was approxiamtely equivalent to the addition to the normal water of three rivers each of dimensions 1072km*75m*3m.

Answer 1

Task:

In one fortnight of a given month, there was a rainfall of $10\mathrm{cm}$ in a river valley. If the area of the valley is $7280\mathrm{km}^*1\mathrm{km}$, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each of dimensions $1072\mathrm{km}^*75\mathrm{m}^*3\mathrm{m}$.

Solution:

Please note that the area of the valley was corrected from $97280\mathrm{km}^*1\mathrm{km}$ to $7280\mathrm{km}^*1\mathrm{km}$, otherwise there is no equivalent to the water of three rivers.

Let's calculate the volume of water in the river in cubic meters:

$$V_{r} = 1072\mathrm{km} \cdot 75\mathrm{m} \cdot 3\mathrm{m} = 1072000\mathrm{m} \cdot 75\mathrm{m} \cdot 3\mathrm{m} = 241\,200\,000\mathrm{m}^{3}$$

Let's calculate the amount of rainfall during the night downpour in cubic meters:

$$V_{d} = 97280\mathrm{km} \cdot 1\mathrm{km} \cdot 10\mathrm{cm} = 7280000\mathrm{m} \cdot 1000\mathrm{m} \cdot 0.1\mathrm{m} = 728\,000\,000\mathrm{m}^{3}$$

Let's find the ratio between the calculated volumes:

$$\frac{V_{d}}{V_{r}} = \frac{728\,000\,000\mathrm{m}^{3}}{241\,200\,000\mathrm{m}^{3}} = 3.018\ldots \approx 3$$

Answer: Yes, indeed, rainfall was approximately equivalent to the addition to the normal water of three rivers.