Q20 of 35 Page 74

There were 800L, 725L and 575 L kerosene oil in 3 kerosine oil drums of the house. The oil of these 3 drums is poured into a cubical pot and for this, the depth of the drum becomes 7dcm.If the ratio of the length and breadth of the cuboidal pot is 4:3 then let us write by calculating the length and breadth of the pot.

If the depth of the cuboidal pot would be 5 dcm, let us calculate whether 1620 L oil can be kept or not in that pot.

Let the ratio be in x m

So, length of cuboidal pot = 4x m


breadth of cuboidal pot = 3x m


Given: Height of cuboidal pot = 7 dcm = 0.7 m


As, 1m3 = 1000 L


Volume of 1st drum = 800 L = 0.8 m3


Volume of 2nd drum = 725L = 0.725 m3


Volume of 3rd drum = 575 L = 0.575 m3


A/Q, Volume from all the 3 drums = volume of cuboidal pot


therefore, 0.8 + 0.725 + 0.575 = 4x × 3x × 0.7


2.1 = 8.4x2


x2 = 0.25


x = 0.25


x = 0.5 m


Thus, length of pot = 4x = 4X0.5 = 2 m


breadth of pot = 3x = 3X0.5 = 1.3 m


2nd part: Given if depth of cuboidal pot = 5dcm = 0.5 m


then volume = length×breadth×height


volume = 2×1.5×0.5


volume = 1.5 m3 = 1.5×1000L


volume = 1500 L


Thus the mentioned volume of 1620 L cannot be kept in that pot .


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