The volume of a cylindrical container is $10001212 \text{ cm}^3$. Water is poured into the container until it fills 70% of the container's volume. What is the approximate volume of the water in base 8?

1. Problem Description

The volume of a cylindrical container is $10001212 \text{ cm}^3$. Water is poured into the container until it fills 70% of the container's volume. What is the approximate volume of the water in base 8?

2. Solution Steps

First, we need to find the volume of the water in cubic centimeters.

Since the water fills 70% of the container, we can calculate the volume of the water as follows:

Volume of water = 70% of the volume of the container

Volume of water $= 0.70 \times 10001212$

Volume of water $= 7000848.4 \text{ cm}^3$

Next, we need to convert the volume of water from base 10 to base

8. To do this, we repeatedly divide the number by 8 and record the remainders.

$7000848 \div 8 = 875106$ remainder $0$

$875106 \div 8 = 109388$ remainder $2$

$109388 \div 8 = 13673$ remainder $4$

$13673 \div 8 = 1709$ remainder $1$

$1709 \div 8 = 213$ remainder $5$

$213 \div 8 = 26$ remainder $5$

$26 \div 8 = 3$ remainder $2$

$3 \div 8 = 0$ remainder $3$

Reading the remainders from bottom to top, we get $32551420_8$.

So the volume of water in base 8 is approximately $32551420_8$.

3. Final Answer

$32551420_8$