Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.

The required maximum capacity of a container that can measure the diesel in all three tankers exactly is the HCF of 403, 434 and 465.

First, find HCF of 403 and 434:

$\begin{array}{l} 403\overline{\smash{\big)}434\smash{\big(}}\phantom{}1 \ \phantom{x}\phantom{()}\underline{-403} \ \phantom{{x^2 } 1())}31\overline{\smash{\big)}403\smash{\big(}}\phantom{}13 \ \phantom{{x} +1xa}\underline{-403} \ \phantom{{x^2 a} + 2x+} 0 \ \end{array}$

So, HCF of 403 and 434 = 31.

Now, find HCF of 31 and 465:

$\begin{array}{l} 31\overline{\smash{\big)}465\smash{\big(}}\phantom{}15 \ \phantom{x}\phantom{}\underline{-465} \ \phantom{{x^2 } 1()}0 \ \end{array}$

So, HCF of 31 and 465 = 31.

Hence, the maximum capacity of the container is 31 litres.