The sum of the surface area of a cube and area of a rectangle is 3240 m 2 The length and breadth of ...........

Quantitative Aptitude ( CCS) Arithmatic inequalities

The sum of the surface area of a cube and area of a rectangle is 3240 m^2. The length and breadth of the rectangle is 50% more and 40% more than the side of the

cube.

Quantity I: Find the volume of the sphere that can be fitted inside the given cube such that it touches all the six faces of the cube. (pi = 3)

Quantity II: Find the volume of cuboidal hole of dimension equal to the rectangle and depth 4.5 m.

I =< II

I = II

I < II

I > II

I >=II

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Explanation:

Let l and b be the length and breadth of rectangle and ‘a’ be the side of the cube.
Length of rectangle = 1.5a metre
Breadth of rectangle = 1.4a metre l*b + 6a^2 = 3240
1.5a * 1.4a + 6a^2 = 3240
=> a = 20 m
side of cube = 20 m
length of rectangle = 1.5*20 = 30 m
breadth of rectangle = 1.4*20 = 28 m
From I: Diameter of sphere = side of cube
radius of sphere = 20/2 = 10 m
Volume of sphere = (4/3)*pi*r^3 = 4000 m^3
From II: Volume of cuboidal hole = 30*28*4.5 = 3780 m^3
I > II

Hence, option 4 ist he correct asnwer.