Question 1

If each edge of a cube is increased in increased by 50%, the percentage increase in surface area is?

Accepted Answer

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 %

Question 2

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?

Accepted Answer

Let the edges of the two cubes be x and y metres Then, x3 - y3 = 152 and x2 - y2 = 20 Also, x + y = 10 So, x - y = (x2 - y2) / (x + y) = 20/10 = 2 Now, (x3 - y3) / (x - y) = 152/2 ⇒ x2 + y2 + xy = 76 ⇒ (x + y)2 - xy = 76 ⇒ xy = (x + y)2 - 76 = (10)2 - 76 = 24

Question 3

The sun of length, breadth and depth of a cuboid is 19 cm and its diagonal is 5 √ 5 cm. Its surface area is?

Accepted Answer

∵ 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

Question 4

A tank 30 m long, 20 m wide and 12 m deep is dug in a field 500 m long and 30 m wide. By how much will the level of the field rise, if the Earth dug of the tank is evenly spread over the field?

Accepted Answer

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

Question 5

A right circular cone of height h is cut by a plane parallel to the base at a distance h/3 from the base, then the volumes of the resulting come and the frustum are in the ratio?

Accepted Answer

The volume of the original cone is V = (πR2h)/3 The height and the radius of the smaller cone are 2h/3 and 2R/3, respectively ⇒ 1/3π (2R/3)2 x 2h/3 = 8V/27 ∴ Volume of the frustum = (V - 8V/27) = 19V/27 So, the required ratio is 8 : 19 .

Question 6

If height of a cylinder is increased by 17.5 % while its radius remains unchanged, by what percent does its volume increase?

Accepted Answer

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

Question 7

The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is 5√ 5 cm, its surface area is?

Accepted Answer

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

Question 8

The sum of length, breadth and depth of a cuboid is 19 cm and its diagonal is 5√ 5 cm. Its surface area would be?

Accepted Answer

L + B + H = 19 cm, where L is the length, B is the breadth and H is the height of the cuboid. ∴ (√L2 + B2 + H2 = (5√5) On squaring both sides, we get L2 + B2 + H2 + = 125 (L + B + H)2 = L2 + B2 + H2 + 2LB + 2LH + 2BH ⇒ (19)2 = 125 + 2(LB + LH + BH) ∴ Surface area = 2(LB + LH + BH) = 192 = 125 + 2(LB + LH + BH) = 192 - 125 = 361 - 125 = 192 - 125 = 361 - 125 = 236 cm2

Question 9

If the number representing volume and surface area of a cube are equal, then the length of edge of a cube in terms of the unit of measurement will be?

Accepted Answer

Let the side of cube be x . Then, Volume of cube = Surface area of cube ⇒ x3 = 6x2 ∴ x = 6

Question 10

A rectangular tank has 2.6 cu m of water. If the area of the base of the tank is 6500 sq cm, then the depth of water (in m) is?

Accepted Answer

Volume of the tank = 2.6 cu m Base area of the tank = 6500 cm = 0.65 sq m ∴ Depth of the tank = 26/0.65 = 4m

Question 11

The length, breadth and height of a room are l, b and h, respectively. The perimeter of the ceiling expressed as a percentage of the total area of the four walls, is?

Accepted Answer

Total area of four walls = 2h(l + b) Perimeter of the celling = 2l + 2b = 2(l + b) Perimeter of the ceiling in percentage = 2(l +b ) x 100/2h(l + b) % = 100/h %

Question 12

A rectangular sump of dimension 6 m x 5 m x 4 m is to be built by using bricks to make the outer dimension 6.2 m x 5.2 m x 4.2 m. Approximately how many bricks of size 20 cm x 10 cm x 5 cm are required to build of size 20 cm x 10 cm 5 cm are require…

Accepted Answer

The volume that is needed to built by the bricks = Volume of outer dimensions of sump - Volume of sump = 6.2 x 5.2 x 4.2 - 6 x 5 x 4 = 135.408 - 120 = 15.405 m3 Volume of one brick = 20/100 x 10/100 x 5/100 = 0.01 m3 ∴ Total number of bricks required to build the sump = 15.408 / 0.001 = 15408

Question 13

A sphere and a cube have the same surface area. The ratio of the volume of the sphere to that of the cube is?

Accepted Answer

Let r be the radius of the sphere and a be the side of the cube ∴ 4π r2 = 6a2 r = √6 a / 2√π ⇒ r3 = 6√6a3 / 8π√π ⇒ Ratio of volume = (4/3 πr3)/a3 ∴ Required ratio = √6 / √π

Question 14

A rectangular tank measuring 5 m x 4.5 m x 2.1 m is dug in the centre of the field measuring 13.5 m x 2.5 m . The Earth dug out is spread evenly over the remaining portion of the field. How much is the level of the field raised?

Accepted Answer

Volume of Earth dug out = 5 x 4.5 x 2.1 = 47.25 m2 Area over which Each is spread = 13.5 x 25 - 5 x 4.5 = 33.75 - 220 = 11.75 m Rise in level = 47.25/11.75 = 4.02 m

Question 15

The base radius and height of a cone are doubled. The ratio of the new volume to original volume is?

Accepted Answer

Let original radius = r and height = h Original volume = (1/3)πr2h New radius = 2r and new height = 2h ∴ New volume = 1/3 x π(2r)2 x 2h = 8/3πr2h Required ratio = 8/3πr2h : 1/3πr2h = 8 : 1

Question 16

If radius of sphere is decreased by 24%. By what per cent does the surface area decrease?

Accepted Answer

The decrease in volume is given [ x + y + xy/100 ] % Here , x = y = -24 = [(-24) + (-24) + (-24) x (-24)/100] % = [-48 + 576/100] % = -48 + 5.76 = - 42.24 % (-ve sign denotes decrease in volume)

Question 17

If each side of a cube is decreased by 19 %, then decrease in surface area of cube is?

Accepted Answer

The net decrease in the surface area of the cube is given by [x + y + xy/100] % Here, x = y = 19% = [-19 + (-19) + (-19) x (-19)/100] % = [-38 + 361/100] % = [-38 + 3.61] %= -34.39% (-ve sign implies for decrease)

Question 18

If the height of the right circular cone is increased by 200% and the radius of the base is reduced by 50%. Then, the change in the volume of cone is?

Accepted Answer

From the formula the net change in volume is given as [x + y + z + (xy + yz + zx)/100 + xyz/(100)2 ] % Here, x = 200, y = z = -50 (-ve sign for decrease) = [ 200 + (-50) + (-50) + (200) x (-50) + (-50) x (-50) + (-50) x (200)/100 + (200) x (-50) x (-50)/(100)2 ] % = 100 + (-10000 + 2500 - 10000)/100 + 500000/100 x 100 ] % = [ 100 + (-17500)/100 + 50]% = (150 - 175)% = -25% (-ve sign implies for a decrease ) So, the volume is decreased by 25%.

Question 19

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Accepted Answer

Volume of the spherical ball = Volume of the water displaced ⇒ 4/3 πr3 = π(12)2 x 6.75 ⇒ r3 = 144 x 6.75 x 3/4 = 729 ∴ r = 9 cm

Question 20

A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm lowered into the water until it is completely immersed. The water level in the vessel will rise by?

Accepted Answer

Volume of the sphere = Volume of the water displaced Let the required height to which the water rises be h. Then, πr12h = 4/3 πr23 ⇒ 16 h = 4/3 x 27 ⇒ h = 36/16 = 9/4 cm

Volume and Surface Area Questions