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Ashok had two vessels that contain $720 ml$ and $405$ ml of milk respectively. Milk in each vessel was poured into glasses of equal capacity to their brim. Find the minimum number of glasses which can be filled with milk.
$A)15$
$B)20$
$C)25$
$D)30$

Answer
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Hint: Find the capacity of the glass, so that it finishes both the vessels. Thus, it is equal to the highest common factor of both the quantities $720$ and $405$.

Complete step-by-step answer:
It is given in the problem that Ashok has two vessels that contain $720$ ml and $405$ ml of milk respectively. First find the capacity of the glass, which finishes both the vessels. Thus, it is equal to the highest common factor of both the quantities $720$ and $405$.
Finding the highest common factor of these quantities,
$720 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$
$405 = 3 \times 3 \times 3 \times 3 \times 5$
Therefore, the highest common factor of 720 and 405 is$3 \times 3 \times 5$ .
H.C.F$ = 45$
Therefore, the capacity of one glass is 45 ml.
Now, find the number of glasses of 45 ml, that can be filled from the first vessel having capacity 720ml.
Number of glasses from the first vessel$ = \dfrac{{720}}{{45}} = 16$
Therefore, 16 glasses are filled from the first vessel.
Now, find the number of glass that can be filled with the second vessel having a capacity of 405 ml.
Number of glasses from the second vessel$ = \dfrac{{405}}{{45}} = 9$
Therefore, 9 glasses are filled from the second vessel.
Now, find the total number of glass that can be filled from both the vessels.
Total number of glasses$ = \;{\text{Glasses from}}{1^{st}}{\text{vessel}} + {\text{Glasses from the}} {2^{nd}} {\text{vessel}}$
Total number of glasses$ = \;16 + 9 = 25$
So, the minimum number of glasses that can be filled with milk is $25$.

Therefore, option C is correct.

Note: The highest common factor of $720$ and$405$ can be obtained by using the factorization method. First, factorize both the numbers and then take out the common factors. The product of these common factors is the HCF of these numbers.