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Mensuration Practice Set 5

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Mensuration Practice Set 5

shape Introduction

Mensuration is a topic in Geometry which is a branch of mathematics. Mensuration deals with length, area, and volume of different kinds of shape- both 2D and 3D. The article Mensuration Practice Set 5 provides information about Mensuration, an important topic of Mathematics Consists of different types of Mensuration questions with solutions useful for candidates preparing for different competitive examinations like RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL, SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc. Prepare better for all exams with this Mensuration Practice Set 5 for SSC CGL & Railways. This Mensuration Practice Set 5 for SSC, Railways Exams will help you learn concepts of mensuration.

shape Quiz

1. The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.

    A. 4192 \({m}^{2}\)
    B. 4304 \({m}^{2}\)
    C. 4312 \({m}^{2}\)
    D. 4360 \({m}^{2}\)
    E. None of these


Answer: Option C


Explanation:

Let the radii of the smaller and the larger circles be s m and l m respectively.

2πs = 264 and 2πl = 352
s = \(\frac{264}{2π}\)and l = \(\frac{352}{2π}\)
Difference between the areas = π\({l}^{2}\) – π\({s}^{2}\)
= π{\(\frac{{176}^{2}}{{π}^{2}}\) – \(\frac{{132}^{2}}{{π}^{2}}\)}
= \(\frac{{176}^{2}}{π}\) – \(\frac{{132}^{2}}{π}\)
= \(\frac{(176 – 132) (176 + 132)}{π}\)
= (\(\frac{(44) (308)}{\frac{22}{7}}\)) = (2) (308) (7) = 4312 sq m


2. A 25 cm wide path is to be made around a circular garden having a diameter of 4 meters. Approximate area of the path is square meters is

    A. 3.34
    B. 2
    C. 4.5
    D. 5.5
    E. 5


Answer: Option A


Explanation:

Area of the path = Area of the outer circle – Area of the inner circle = π(\({\frac{4}{2}} + {\frac{25}{100}}\))\(^{2}\) – π(\(\frac{4}{2}\))\(^{2}\)

= π[\({2.25}^{2}\) – \({2}^{2}\)] = π(0.25)(4.25)
= (3.14) (\(\frac{1}{4}\)) (\(\frac{17}{4}\)) = \(\frac{53.38}{16}\) = 3.34 sq m


3. The parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm. Find the circumference of a semicircle whose diameter is equal to the side of the square. (Round off your answer to two decimal places)


    A. 77.14 cm
    B. 47.14 cm
    C. 84.92 cm
    D. 94.94 cm
    E. 23.57 cm


Answer: Option E


Explanation:

Let the side of the square be a cm.

Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of the square = 60 cm
i.e. 4a = 60
A = 15
Diameter of the semicircle = 15 cm
Circimference of the semicircle
= \(\frac{1}{2}\) (π) (15)
= \(\frac{1}{2}\) (\(\frac{22}{7}\)) (15) = \(\frac{330}{14}\) = 23.57 cm to two decimal places


4. There are two circles of different radii. The are of a square is 784 sq cm and its side is twice the radius of the larger circle. The radius of the larger circle is seven – third that of the smaller circle. Find the circumference of the smaller circle.

    A. 6π cm
    B. 8π cm
    C. 12π cm
    D. 16π cm
    E. None of these.


Answer: Option C


Explanation:

Let the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.

\({a}^{2}\) = 784 = (4) (196) = (22) (142)
a = (2) (14) = 28
a = 2l, l = \(\frac{a}{2}\) = 14
l = (\(\frac{7}{3}\)) s
Therefore s = (\(\frac{3}{7}\)) (l) = 6 Circumference of the smaller circle = 2πs = 12π cm.


5. A cube of side one meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?

    A. 10
    B. 100
    C. 1000
    D. 10000
    E. None of these


Answer: Option C


Explanation:

Along one edge, the number of small cubes that can be cut

= \(\frac{100}{10}\) = 10
Along each edge 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 × 10 × 10 = 1000

1. The dimensions of a room are 25 feet × 15 feet × 12 feet. What is the cost of white washing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet × 3 feet and three windows of dimensions 4 feet × 3 feet each?

    A. Rs. 4800
    B. Rs. 3600
    C. Rs. 3560
    D. Rs. 4530
    E. None of these


Answer: Option D


Explanation:

Area of the four walls = 2h (l + b)

Since there are doors and windows, area of the walls = 2 × 12 (15 + 25) – (6 × 3) – 3(4 × 3) = 906 sq.ft.
Total cost = 906 × 5 = Rs. 4530


2. The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 500 resolutions.

    A. 252 m
    B. 704 m
    C. 352 m
    D. 808 m
    E. None of these


Answer: Option B


Explanation:
In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions.

= 500 × 2 × \(\frac{22}{7}\) × 22.4 = 70400 cm = 704 m


3. The length of a rectangular floor is twice its breadth. If Rs.624 is required to paint the floor at the rate of Rs.8 psm. What is the breadth of the floor ?

    A. 6.24m
    B. 5.10m
    C. 7.8m
    D. 4.6m
    E. None of these


Answer: Option A


Explanation:
Area = \(\frac{624}{8}\) = 78 \({m}^{2}\)
2x × x = 78
2\({x}^{2}\) = 78
\({x}^{2}\) =78/2 = 39
x = 6.24

Only radius (r) and height (h) are varying.
Hence, ( \(\frac{1}{3}\))π may be ignored.
\(\frac{V1}{V2}\) =\(\frac{{r1}^{2}h1}{{r2}^{2}h2}\) =>\(\frac{1}{10} \) = \(\frac{(1) h1}{{(2)}^{2}h2}\)
=>\(\frac{h1}{h2}\) =
i.e. h1 : h2 = 2 : 5


4. A metallic sphere of radius 12 cm is melted and drawn into a wire, whose radius of cross section is 16 cm. What is the length of the wire?

    A. 45 cm
    B. 18 cm
    C. 90 cm
    D. 180 cm
    E. None of these


Answer: Option E


Explanation:

Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.

π\({16}^{2}\) × h = (\(\frac{4}{3}\))π \({(12)}^{3}\) => h = 9 cm


5. The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas?

    A. 81 : 121
    B. 9 : 11
    C. 729 : 1331
    D. 27 : 121
    E. None of these


Answer: Option A


Explanation:

Ratio of the sides = ³√729 : ³√1331 = 9 : 11

Ratio of surface areas = \({9}^{2}\) : \({11}^{2}\) = 81 : 121

1. The length of a rectangle is two – fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?

    A. 140
    B. 156
    C. 175
    D. 214
    E. None of these


Answer: Option A


Explanation:

Given that the area of the square = 1225 sq.units

=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = \(\frac{2}{5}\) × 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 × 10 = 140 sq.units


2. The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter?

    A. 91.5 cm
    B. 93.5 cm
    C. 94.5 cm
    D. 92.5 cm
    E. None of these


Answer: Option A


Explanation:

Perimeter of the sector = length of the arc + 2(radius)

= (\(\frac{135}{360}\) × 2 × \(\frac{22}{7}\) × 21) + 2(21)
= 49.5 + 42 = 91.5 cm


3. An order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2. Subsequently, the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter. Find the ratio of the areas of the carpets in both the cases.

    A. 4 : 3
    B. 8 : 7
    C. 4 : 1
    D. 6 : 5
    E. None of these


Answer: Option B


Explanation:

Let the length and breadth of the carpet in the first case be 3x units and 2x units respectively.

Let the dimensions of the carpet in the second case be 7y, 3y units respectively.
From the data,.
2(3x + 2x) = 2(7y + 3y)
=> 5x = 10y
=> x = 2y
Required ratio of the areas of the carpet in both the cases
= 3x × 2x : 7y × 3y
= 6\({x}^{2}\) : 21\({y}^{2}\)
= 6 × \({(2y)}^{2}\) : 21y2
= 6 × 4\({y}^{2}\) : 21\({y}^{2}\)
= 8 : 7

4. A wire when bent in the form of a square encloses an area of 484 sq.cm. What will be the enclosed area when the same wire is bent into the form of a circle? (Take π=227)

    A. 462 \({cm}^{2}\)
    B. 539 \({cm}^{2}\)
    C. 616 \({cm}^{2}\)
    D. 693 \({cm}^{2}\)


Answer: Option C


Explanation:

Area of a square = side × side

The perimeter of a square = 4 × side

Given, wire, when bent in the form of a square, encloses an area of 484 sq.cm

Side = √484 = 22 cm

Perimeter = 88 cm

Same wire is used to make a circle.

The perimeter of a circle of radius r = 2πr

⇒ 2πr = 88

⇒ \(\frac{44r}{7}\) = 88

⇒ r = 14 cm

Area of a circle of radius r = π\({r}^{2}\)
Area enclosed by the circle = 227 × 142

⇒ Area enclosed by the circle = 616 \({cm}^{2}\)


5. A square of side 22 cm is made from a single wire. The same wire is then bent to form a circle. The area of the circle formed is (π = 22/7)

    A. 416 \({cm}^{2}\)
    B. 343 \({cm}^{2}\)
    C. 528 \({cm}^{2}\)
    D. 616 \({cm}^{2}\)


Answer: Option D


Explanation:
Perimeter of a square = 4 × side

Given, side of the square = 22 cm

∴ The perimeter of the square = 88 cm

Now, the same wire is used to form a circle.

∴ The circumference of the circle formed = perimeter of the square

Circumference of a circle = 2πr

⇒ 2πr = 88

⇒ r = \(\frac{44}{π}\) = 14 cm

Area of a circle = π\({r}^{2}\)

⇒ Area of the circle formed =227 × 142 = 616 \({cm}^{2}\)


Mensuration – Related Information
Mensuration Practice Set 1
Mensuration Practice Set 2
Mensuration Practice Set 3