﻿ Surface Area and Volume Word problems | Problems & Solutions
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Surface Area and Volume Word problems
• Page 1
1.
An adhesive compound in liquid form is prepared in a container of hemispherical shape having a radius of 180 cm. This compound is to be packed in cylindrical bottles of radius 1 cm and height of 4 cm. How many bottles are needed if the liquid prepared exactly fills the container? [Take $\pi$ = 3.] a. 1.94 b. 388800 c. 972000 d. 16200

#### Solution:

Number of bottles required = total volume of the liquidvolume of one bottle
[Formula.]

Volume of the liquid = Volume of the hemisphere 2 / 3× 3 × 1803 = 11664000 cm3
[Volume of the hemisphere = 2 / 3π r3.]

Volume of one bottle 3 × (1)2 × 4 = 12 cm3
[Volume of the cylinder = π (r)2 h.]

Number of bottles 11664000 / 12 = 972000 (approximately)
[From steps 2 and 3.]

Correct answer : (3)
2.
Find the surface area of the earth assuming the earth to be a sphere of radius 6369 km. a. 509485853 square km b. 509485862 square.km c. 509485863 square km d. 509485861 square km

#### Solution:

The radius of the earth = 6369 km.
[Given.]

The surface area of the earth = 4π r2
[Formula.]

= 4π (6369)2
[Substitute the value of r.]

= 509485862 square km
[Simplify.]

Correct answer : (2)
3.
$\frac{3}{4}$ of the earth's surface area is covered with water. Find the surface area of the land if the earth can be assumed as a sphere of radius 6366 km. a. 6366 square km b. 127251501 square km c. 4774 square km d. 19098 square km

#### Solution:

The radius of the earth = 6366 km.
[Given.]

The surface area of the earth = 4πr2
[Formula.]

= 4π(6366)2

= 509006007

The surface area which is covered with water = 3 / 4 × Total surface area of the earth

The surface area of the land on earth = (1- 3 / 4) × Total surface area of earth

= 1 / 4 × Total surface area of the earth

= 1 / 4 × 509006007

= 127251501 square km

Correct answer : (2)
4.
$\frac{3}{4}$ of the earth's surface area is covered with water. Find the surface area covered by water if the earth can be assumed as a sphere of radius 6365 km. a. 127211526.50 square.km b. 381634579.50 square.km c. 190817289.75 square.km d. 63605763.25 square.km

#### Solution:

The radius of the earth = 6365 km.
[Given.]

The surface area of the earth = 4π r2
[Formula.]

= 4 π(6365)2

= 508846106 square.km

The surface area which is covered with water = 3 / 4 × Total surface area of the earth

= 3 / 4 × 508846106

= 381634579.50 sq.km

Correct answer : (2)
5.
Find the maximum volume of the sphere in cubic cm that can be carved out of a cube of side 20 cm.[Take $\pi$ = 3.14] a. 4186.66 cm3 b. 33493.3 cm3 c. 8000 cm3 d. 1256.8 cm3

#### Solution:

The side of the cube is 20 cm.
[Given.]

The maximum radius of sphere that can be carved out of the cube = 20 / 2 = 10 cm.

The maximum volume of the sphere that can be carved out of the cube = 4 / 3πr3
[Formula.]

= 4 / 3π × 103

= 4186.66 cm3

Correct answer : (1)
6.
Find the sum of the volume of a cone with radius 1 cm and height 5 cm and the volume of a sphere of radius 1 cm. [Take $\pi$ = 3.] a. 9 cm³ b. 14 cm³ c. 13 cm³ d. 12 cm³

#### Solution:

Volume of the cone = 1 / 3× 3 × (12) 5 = 5 cm³
[Volume of the cone = 1 / 3π (r2) h.]

Volume of the sphere = 4 / 3× 3 × (13) = 4 cm3
[Volume of a sphere = 4 / 3π (r)3.]

Sum of the volumes = 5 + 4 = 9 cm3
[Simplify.]

Correct answer : (1)
7.
A cylindrical jar of radius 15 cm is filled with water upto a height of 25 cm. 15 spherical balls of radii 2 cm each are immersed in the jar. Find the new level to which water is filled in the jar. [Take $\pi$ = 3.] a. 25.71 cm b. 32.71 cm c. 27.71 cm d. 30.71 cm

#### Solution:

Volume of one sphere = 4 / 3× 3 × (23) = 32 cm3
[Volume of the sphere = 4 / 3π (r3).]

Volume of 15 spheres = 15 × 32 = 480 cm3
[Multiply.]

Volume of water in the jar = 3 × (152) × 25 = 16875 cm3
[Volume of the cylinder = π (r2) h.]

Total volume of water + balls(V) 480 + 16875 = 17355 cm3
[Simplify.]

Volume of the water in the cylinder when spherical balls are immersed = 17355 cm3
[From step 4.]

3 × (15)2 h = 17355 cm3
[From step 5.]

Height to which water is filled in the jar 173553 × (15)² = 25.71 cm

Correct answer : (1)
8.
Three hemispheres of radius 1, 2 and 1 respectively are melted to form a sphere. What is the radius of the new sphere formed? [Take $\pi$ = 3.] a. $r$ = $\sqrt{15}$ units b. $r$ = $\sqrt{3}$ units c. $r$ = $\sqrt{10}$ units d. $r$ = $\sqrt{5}$ units

#### Solution:

Volume of the sphere = 4 / 3 π r3
[Formula]

Volumes of three hemispheres = 2 / 3 π (1)3 + 2 / 3 π(2)3 + 2 / 3 π (1)3
[Volume of the hemisphere
= 2 / 3π (r3).]

As the sphere is formed by melting three hemispheres, volume of the sphere is equal to the volumes of three hemispheres.

4 / 3 π r3 = 2 / 3 π (1)3 + 2 / 3 π (2)3 + 2 / 3 π (1)3

r = 53 units

Correct answer : (4)
9.
A water tank is in the form of a hemisphere with radius 5 m. It is filled with water at a rate of 8 litres/sec. How much time will it take to fill the tank?

#### Solution:

Radius of the tank = 5 m
[Given.]

Volume of the tank = 2 / 3× 3 × (5)3 × 1000 litres = 250000 litres
[Volume of the tank = 2 / 3π r3.]
[1 m3 = 1000 litres.]

Rate of filling the water = 8 litres/sec,
[Given.]

Time taken to fill the tank = Volume of the tank rate of filling
[Formula.]

Time taken to fill the tank = 250000 / 8
= 31250 seconds = 520.83 min
[Simplify.]
[1 minute = 60 seconds.]

Correct answer : (0)
10.
A rectangular block of lead with dimension 43 cm × 53 cm × 63 cm is melted to mould spherical balls of 3 cm radius. How many balls are made?
(Round your answer to nearest whole number and take $\pi$ = 3. ] a. 1829 b. 1629 c. 1429 d. 1329

#### Solution:

Nuumber of spherical balls made = Total volume of leadvolume of one spherical ball
[Formula.]

Total volume of lead = 43 cm × 53 cm × 63 cm 143577 cm3
[Given.]

Volume of one spherical ball = 4 / 3 × 3 × (3)³ 108 cm³
[Volume of a sphere = 4 / 3π r3.]

Number of spherical balls = 143577 / 108 1329 (approximately)
[Simplify.]

Correct answer : (4)

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