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Aptitude

Volume Surface Area

How To Solve Volume and Surface Area Questions Quickly

Perimeter can be thought of as the length of the outline of a shape.

Surface area is the area of the two-dimensional surface of a three-dimensional object.

Volume is the space that an object occupies.

Type 1: Find the area, perimeter, length, breadth and some other sides of the shapes

1. The sides of a triangle are in the ratio $ \frac{1}{3}: \frac{1}{2} : \frac{1}{4} $ Find the smallest side of the triangle, if the perimeter of the triangle is 78 cm.

Options:

A. 20 cm

B. 24 cm

C. 36 cm

D. 18 cm

Solution:

According to the question, ratio of the side of the triangle are $\frac{1}{3}: \frac{1}{2} : \frac{1}{4} = 6:3:4 $

Perimeter of the triangle = 78 cm

Therefore 6x + 4x + 3x = 78

13x = 78

x = 6

The Sides are : 6 x 6 = 36 cm

6 x 4 = 24 cm

6 x 3 = 18 cm

Correct option: D

2. A rope makes 120 rounds of cylinder with base radius 10 cm. How many times it can go round a cylinder with base radius 20 cm?

Options:

A. 70 cm

B. 60 cm

C. 45 cm

D. 50 cm

Solution:

Le the round be a

If radius is more, then rounds will be less as the length of the ropes remains the same

x = 2*π*10*120…(1)

Similarly,

x = 2 * π * 20 * a…(2)

From (1) and (2)

10 * 120 = 20 * a

=> a = 60

Correct option: B

3. Find the area of a parallelogram with base 20 m and height 4 m.

Options:

A. 120 m²

B. 24 m²

C. 150 m²

D. 80 m²

Solution:

Area of parallelogram = b * h

Area of parallelogram = 20 * 4 = 80 m²

Correct Option: A

Type 2: How To Solve Perimeter, Area and Volume Questions Quickly.By finding the volume and total surface area

1. A cube of 5 cm was cut into as many 1 cm cubes as possible. Find out the ratio of the surface area of the larger cube to that of the surface areas of the smaller cubes?

Options:

A. 1:2

B. 2:3

C. 1:5

D. 1:3

Solution:

Volume of the original cube $ = 53 = 125 cm^3$.

Volume of each smaller cubes $ = 1 cm^3$. It means there are 125 smaller cubes.

Surface area of the cube $ = 6a^2$

Surface area of the larger cube = 6a2 = 6 * 52 = 6 * 25 = 150

Surface area of one smaller cubes = 6 (1²) = 6

Now, surface area of all 125 cubes = 125 * 6 = 750

Therefore,

Required ratio = Surface area of the larger cube: Surface area of smaller cubes = 150: 750

= 1: 5

Correct option: C

2. The curved surface area of a cylindrical pillar is 264 m² and its volume is 924m³. Find the ratio of its diameter to its height.

Options:

A. 7:4

B. 3:4

B. 6:5

B. 7:3

Solution:

Volume of cylinder $ = πr^2h$

Curved Surface area of cylinder = 2 πrh

$\frac{\text{Volume of cylinder}}{\text{Curved Surface area of cylinder}}$

$\frac{πr^2h}{2 πrh} = \frac{924}{264}$

$r = \frac{924}{264} \times 2$

r = 7

Curved Surface area of cylinder = 2πrh = 264

$ 2 \times \frac{22}{7} \times h = 264 $

$ h = 264 \frac{22}{7} \times \frac{1}{2} \times \frac{1}{7} $

h = 6

Now, required ratio $ = 2 \times \frac{r}{h} = 2 \times \frac{7}{6} = \frac{14}{6} = \frac{7}{3} = 7 : 3 $

Correct option: D

Type 3: How To Solve Perimeter, Area and Volume Questions Quickly to find cost

1. What will be the cost of gardening 1 metre broad boundary around a rectangular plot having a perimeter of 340 metres at the rate of Rs.10 per square metre?

Options:

A. Rs. 4356

B. Rs. 5660

C. Rs. 3440

D. Rs. 4040

Solution:

2(l+b) = 400 (given)

Area of the boundary = [(l+2) (b+2) -- lb] = 2 (l + b) + 4 = 404

∴ Cost of gardening = Rs. (404 × 10) = Rs. 4040

Correct option: B

2. What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58 ?

Options:

A. Rs. 3710

B. Rs. 3770

C. Rs. 3944

D. Rs. 3580

Solution:

Let the side of the square plot be 'a' ft.

Given area of the plot (a x a) = 289 => a = 17

Length of the fence = Perimeter of the plot = 4a = 68 ft.

Cost of building the fence = 68 x 58 = Rs. 3944.

Correct option: C

3. A rectangular wall whose length is 10 m more than its breadth. The cost of painting the wall is at Rs 30 per meter is Rs. 1800, what is the length of the wall in meters?

Options:

A. 18.5m

B. 27.5 m

C. 25 m

D. 20 m

Solution:

Breadth of the wall = x

Length of the wall = x + 10

Perimeter of the rectangular wall $ = \frac{1800}{30} = 60m $

2 (l+b) = 60

2 (x + 10 + x) = 60

2 (2x + 10) = 60

2x + 10 = 30

2x = 20

x = 10

Length = x + 10 = 10 + 10 = 20 m

Correct Option: D