Three cans contain 23.265 ml, 60.5 ml and 71.486 ml of kerosene. What should be the minimum capacity of a can that can hold the total quantity of kerosene rounded off to 2 decimal places?

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Solution

All the units of measurement are same. After converting the numbers to like decimals, we can add all three quantities.

$23.265$
$+ 60.500$
$+ 71.486$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$
$155.251 m l$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$

Rounding off to two decimal places we get 155.25 ml which is the minimum capacity that the can should have.

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