The length of longer pole that can be placed on the floor of a room is 10 m and the length of the longest pole that can be placed in the room is 10? 2m. The height of the room is?

? ?3a = 8?3
? a = 8
? Surface area = 6a²
= (6 x 8 x 8) cm²
= 384 cm²

Let length = 6k, breadth = 5k and height = 4k in cm
? 2(6k x 5k + 5k x 4k + 6k x 4k) = 33300
? 148k² = 33300
? k² = 33300/148 = 225
? k = 15
? Length = 90 cm, Breadth = 75 cm, Height = 60 cm

Number of boxes = (Volumes of wooden box in cm³) / (Volume of 1 small box)
= (800 x 700 x 600) / (8 x 7 x 6) = 1000000

As the box is painted from inside, the dimensions of the box from inside are
Length = (21 - 05 - 05) = 20 cm
Breadth = (11 - 05 - 05) = 10 cm and height = (6 - 05) = 5.5
Total number of faced to be painted = 4 walls + One base
The dimensions of two of the walls = 2 x (10 x 5.5)
The dimensions of the remaining two = 2 x (20 x 5.5) and that of the base = (20 x 10 ) and so the total area to be pained = 2 x (10 x 5.5) + 2 x (20 x 5.5) + (20 x 10 ) = 530cm³
Since, the total expenses of painting this area is ?70, the rate of painting
= 70/530 = 0.13 = ? 0.1 per cm³ (approx)

Volume of the sphere = Volume of the water displaced
Let the required height to which the water rises be h.
Then, ?r₁²h = 4/3 ?r₂³
? 16 h = 4/3 x 27 ? h = 36/16 = 9/4 cm

? ?3 a = 14 x ?3
? a = 14
? Volume of the cube = (14 x 14 x 14)cm³
= 2744 cm³

Length of pencil = ? [(8)² + (6)² + (2)²] cm
= ?104 cm
= 2?26

Speed per min = (3 x 1000)/60 m = 50 m
Volume of water running per min. = (45 x 2 x 50) m²
= 4500 m³

Number of bricks required = Volume of wall in cm³ / Volume of 1 brick
= (200 x 100 x 1.8 x 100 x 37.5)/(25 x 12.5 x 7.5)
= 57, 600

? Required cost = Rs. (750 x 57600)/1000
= Rs. 43, 200

Volume of cube formed = 216 cm³
? Edge of the cube = (6 x 6 x 6)^1/3 = 6 cm
Surface area of original metal sheet = 2(27 x 8 + 8 x 1 + 27 x 1) cm² = 502 cm²

Surface area of the cube formed = [6 x (6)²] cm² = 216 cm²

? Required difference in area of two solids = (502 - 216) cm²
= 286 cm²