Mensuration For IBPS RRB Exam, Check 10 Questions

Mensuration For IBPS RRB Exam: Mensuration studies the measurement of geometric figures and their parameters like length, area, perimeter, volume, shape, surface area, lateral surface area, etc. It’s an essential topic in the quantitative aptitude section of IBPS RRB Exam. In this blog, we’ll discuss all differences between 2D and 3 D shapes, important terms and sample questions needed to train your brain to master mensuration for IBPS RRB Exam.

Differences Between 2D and 3D shapes

Mensuration is branch of mathematics that addresses length, volume, or area of different geometric shape. These shapes are either in 2-dimensions (2-D) or 3-dimensions (3-D). IT’s important to learn the difference between these two dimensions.

Differences Between 2D and 3D Shapes
2D Shape	3D Shape
Enclosed by three or more straight lines in a plane.	Enclosed by multiple surfaces or planes.
No depth or height.	Has depth or height; also called solid shapes.
Two dimensions: length and breadth.	Three dimensions: length, breadth, and depth (or height).
Measurable area and perimeter.	Measurable volume, Curved Surface Area (CSA), Lateral Surface Area (LSA), and Total Surface Area (TSA).

Mensuration For IBPS RRB Exam, Important Terms

Now let’s learn all the important mensuration terms. A good understanding of these terms is needed to solve the mensuration problems. Students can also get the sample mensuration questions from this blog. Check out the table below to know the terms, their abbreviations, units and definitions.

Mensuration For IBPS RRB Exam, Important Terms
Terms	Abbreviation	Unit	Definition
Area	A	m² or cm²	Surface covered by the closed shape.
Perimeter	P	cm or m	Continuous line along the boundary of the figure.
Volume	V	cm³ or m³	Space occupied by a 3D shape.
Curved Surface Area	CSA	m² or cm²	Total area of a curved surface (e.g., sphere).
Lateral Surface Area	LSA	m² or cm²	Total area of all lateral surfaces surrounding the figure.
Total Surface Area	TSA	m² or cm²	Sum of all curved and lateral surface areas.
Square Unit	–	m² or cm²	Area covered by a square with one-unit side.
Cube Unit	–	m³ or cm³	Volume occupied by a cube with one-unit side.

Types of Mensuration Questions in IBPS RRB Exam

There are many types of mensuration questions that you can encounter in in the quantitative aptitude section of IBPS RRB Exam. It is essential to understand each type to tackle it successfully in the exam. Let’s explore the types of questions that appear in the IBPS RRB Exam, along with some tips to master them

1. Direct Formulas: Questions where you directly apply the formulas for area, perimeter, volume, or surface area.
2. Composite Figures: Questions involving combinations of shapes (e.g., a rectangle with a semicircle).
3. Missing Dimensions: Given one parameter (e.g., area), find another (e.g., perimeter).
4. Comparisons: Compare areas, volumes, or perimeters of different shapes.
5. Data Interpretation (DI): Sometimes, mensuration-based DI questions appear.

Tips to Master Mensuration For IBPS RRB Exam

Here are five concise tips needed to master the art of solving questions on mensuration for IBPS RRB Exam.

1. Learn Formulas: Memorize all the relevant formulas.
2. Practice Regularly: Solve a variety of mensuration problems daily.
3. Understand Concepts: Understand how formulas are derived.
4. Unit Consistency: Always check that dimensions are in the same unit.
5. Visualize: Draw figures to visualize problems better.

Mensuration For IBPS RRB Exam, Check 10 Questions

Question 1: The radius of a sphere is equal to the side of a cube having volume 729 cm³. What is the total surface area of the sphere? (Take π = 3)

A) 680 cm²

B) 712 cm²

C) 1080 cm²

D) 606 cm²

E) 972 cm²

Question 2: What will be the height of a cylinder of volume equal to the volume of cone of radius 14 cm and height 18 cm, if the radius of cone and cylinder is same?

A) 8 cm

B) 6 cm

C) 7 cm

D) 12 cm

E) 14 cm

Question 3: The ratio of the length and breadth of the rectangular field is 5:4 respectively. If the cost of fencing the rectangular field at the rate of Rs. 4/m is Rs. 720, then find the breadth of the rectangular field.

A) 28 m

B) 32 m

C) 44 m

D) 36 m

E) None of these

Question 4: Find the volume of cube having side equal to the radius of in-circle of equilateral triangle of area 48√3 cm²?

A) 8 cm³

B) 24√3 cm³

C) 27 cm³

D) 8√3 cm³

E) 64 cm³

Question 5: A circular park has been fenced at the rate of Rs. 6 per m. If the total cost of fencing is Rs. 3168, then find the diameter of the circular park.

A) 156 m

B) 172 m

C) 160 m

D) 168 m

E) 176 m

Question 6: Find the area of the square having perimeter 20% more than 80 cm.

A) 484 cm²

B) 529 cm²

C) 576 cm²

D) 625 cm²

E) None of these

Question 7: A cylinder of volume 4335 cm³ is to be painted. If the cost of painting the curved surface area of the cylinder at the rate of Rs. 5/cm² is Rs. 2550, then find the radius of the cylinder. [Use π = 3]

A) 15 cm

B) 17 cm

C) 19 cm

D) 21 cm

E) None of these

Question 8: The sum of areas of two rectangles (R₁ and R₂) of same length is 432 cm², and the ratio of breadth of rectangle R₁ and breadth of rectangle R₂ is 5:4, respectively. If the area of square having length of side equal to the length of rectangle is 256 cm², then find the area of square having length of side equal to the breadth of rectangle R₂.

A) 196 cm²

B) 100 cm²

C) 144 cm²

D) 256 cm²

E) 400 cm²

Question 9: The length of the rectangular page is 25% more than the breadth of the rectangular page. If the perimeter of the rectangular page is 90 cm, then find the length of the page

A) 25 cm

B) 16 cm

C) 14 cm

D) 12.5 cm

E) None of these

Question 10: What is the circumference of a circle of maximum area inscribed in a square having area 196 cm²?

A) 88 cm

B) 22 cm

C) 44 cm

D) 66 cm

E) None of these

Solution 1: E)

Side of the cube = 729^1/3 = 9 cm

So, radius of the sphere = 9 cm

Total surface area of the sphere = 4 × π × r² = 4 × 3 × 9² = 972 cm²

Hence, option e.

Solution 2: B)

Let the height of cylinder be ‘h’ cm.

Volume of cylinder = Volume of cone

π × 14² × h = (1/3) × π × 14² × 18

h = 18/3 = 6 cm

Hence, option b.

Solution 3: E)

Let the length and breadth of the rectangular field be ‘5x’ m and ‘4x’ m respectively.

Perimeter of rectangular field = 720/4 = 180 m

According to question,

2(5x + 4x) = 180

x = 10 m

So, breadth of the rectangular filed = 4x = 40 m

Hence, option e.

Solution 4: E)

Let the side of the equilateral triangle be ‘s’ cm.

Area of equilateral triangle = 48√3 cm²

(√3/4) × s²= 48√3

s = √(48 × 4)

s = 8√3 cm

Radius of the in-circle = (side × √3)/6 = (8√3 ×√3)/6 = 4 cm

So, side of cube = 4 cm

Therefore, volume of cube = 4³ = 64 cm³

Hence, option e.

Solution 5: D)

Circumference of circular park = 3168/6 = 528 m

So, 2 × (22/7) × radius = 528

2 × radius = (528 × 7)/22 = 168 m

So, diameter of the park = 168 m

Hence, option d.

Solution 6: C)

Let the side of the square be ‘x’ cm.

Perimeter of the square = 4x = 120% of 80

4x = 96

x = 24

So, the area of the square = 24² = 576 cm²

Hence, option c.

Solution 7: B)

Volume of cylinder = π r² h, where ‘r’ and ‘h’ are the radius and height of the cylinder.

3 × r² × h = 4335

r² h = 1445 —-(1)

Curved surface area of the cylinder = 2 π r h

2 × 3 × r × h = 2550/5 = 510

r × h = 85 —-(2)

Dividing eq. (1) by eq. (2), we get

r = 17 cm

Hence, option b.

Solution 8: B)

Solution 1: C)

Let the breadth of rectangle R₁ and rectangle R₂ be ‘5x’ cm and ‘4x’ cm respectively.

Length of rectangle = side of square = √256 = 16 cm

According to question,

Sum of areas of two rectangles (R₁ and R₂) = 432 cm²

16 × 5x + 16 × 4x = 432

16(5x + 4x) = 432

9x = 432/16

9x = 27

x = 3 cm

So, the breadth of rectangle R₂ = 4 × 3 = 12 cm

Therefore, required area of square = 12 × 12 = 144 cm²

Hence, option b.

Solution 9: A)

Let the breadth of the rectangular page be ‘x’ cm.

Then, length of rectangular page = 1.25x cm

According to question,

2 × (x + 1.25x) = 90

4.5x = 90

x = 20 cm

So, breadth = 20 cm and length = 25 cm

Hence, option a.

Solution 10: C)

Side of square = √196 = 14 cm

So, diameter of the circle having maximum area inscribed in square = 14 cm

Therefore, circumference of the circle = 2 × (22/7) × 7 = 44 cm

Hence, option c.

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