The difference in volumes of two cubes is 152 m ³ and the difference in their one face areas is 20 m ². If the sum of their edge is 10 m, the product of their edge is?

Let original length of cube = L
Then, its surface area = 6L²
New edge = (150 x L)/10 = 3L/2
New surface area = 6 x (3L/2)² = (6 x 9 x L²)/4 = (27/2) x L²
Increase in surface area = (27/2 - 6) x L² = 15 x L²/2
? Increase percent = (15 x L²/2 ) / (6L²) x 100 % = 125 %

Let the volume be x³ and 27x³
? Their edges are x and 3x

Now Ratio of their surface area = 6x² : 54x² = 1 : 9

Earth dug out = (3 x 2 x 1.5 ) m³ = 9 m³
Area on which earth has been spread = (22 x 14 - 3 x 2 ) m² = 302 m²

? Rise in level = Volume/Area = (9/302) m = (9 x 100/302) cm = 2.98 cm

Given that,
Volume of sphere = Volume of cone
? (4/3)?r³ = (1/3)?r²h
? 4r = h

Total surface area of the pyramid
= [1/2(perimeter of the base) (Slant height) ] + Area of the base
= 1/2 (4) (4) (8) + 4² = 80 cm²

? l + b + h = 19
and l² + b² + h² = (5?5)² = 125
? (l + b + h)² = (19)²
? (l² + b²+ h²) + 2(lb + bh + lh) = 361
? 2 (lb + bh + lh) = (361 - 125) = 236
? Surface area = 236 cm²

Volume of the Earth taken out = 30 x 20 x 12
= 7200 m³

Area of the remaining portion (leaving the area of dug out portion) = 470 x 30 + 30 x 10
= 14100 + 300 = 14400 m³

Let h be height to which the field is raised by spreading the Earth dug out.
Then, 14400 x h = 7200
? h = 7200/14400 = 0.5 m

The volume of the original cone is V = (?R²h)/3
The height and the radius of the smaller cone are 2h/3 and 2R/3, respectively
? 1/3? (2R/3)² x 2h/3 = 8V/27
? Volume of the frustum = (V - 8V/27) = 19V/27
So, the required ratio is 8 : 19 .

If the radius remains unchanged
Then, x = y = 0 and z = 17.5 %
? Net increase in volume = 17.5%

Let x, y and z be the length breadth and deapth of a cuboid.
? x + y + z = 19
x² + y² + z² = (5?5)² = 125
Surface area of the cuboid = 2(xy + yz + zx)
= (x + y + z)²) - (x² + y² + z²)
= 361 - 125 = 236 cm³