The problem on the image states:

"A water tank was filled with $\frac{5}{6}$ of its capacity. If $\frac{2}{3}$ of the water is used, how much water is left in the tank?"

Let's solve this step-by-step:

1. Initial water in the tank: The tank is filled with $\frac{5}{6}$ of its capacity.

2. Water used: $\frac{2}{3}$ of the water that was initially in the tank is used. So, we need to calculate $\frac{2}{3}$ of $\frac{5}{6}$.

   $\text{Water used} = \frac{2}{3} \times \frac{5}{6} = \frac{10}{18} = \frac{5}{9}$

3. Water left in the tank: To find the water left in the tank, we subtract the amount used from the initial amount.

   $\text{Water left} = \frac{5}{6} - \frac{5}{9}$

   To subtract these fractions, we need a common denominator. The least common denominator of 6 and 9 is 18. Rewriting the fractions:

   $\frac{5}{6} = \frac{15}{18}, \quad \frac{5}{9} = \frac{10}{18}$

   Now, subtract:

   $\frac{15}{18} - \frac{10}{18} = \frac{5}{18}$

So, $\frac{5}{18}$ of the tank's capacity is left after using the water.

Would you like more details or have any questions?

Related Questions:

1. What is the tank's remaining capacity if 1/2 of the initial water had been used instead?
2. If the tank had been filled to 3/4 capacity instead, how much water would be left after using 2/3 of it?
3. How much water was used in terms of the total tank capacity?
4. If the tank had a total capacity of 180 liters, how much water is left in the tank?
5. What would happen if 100% of the water in the tank was used?

Tip:

When working with fractions, always ensure to find a common denominator before adding or subtracting!