# Mensuration For IBPS RRB Exam, Check 10 Questions

Jul 16, 2024

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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.

## 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.

## 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 = 7291/3 = 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 × 14× h = (1/3) × π × 14× 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 Rbe ‘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 R= 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.

## Mensuration For IBPS RRB Exam FAQs

What is mensuration in mathematics?

Mensuration involves measuring geometric figures’ length, area, perimeter, volume, and surface area.

What are 2D shapes?

2D shapes are enclosed by straight lines in a plane and have length and breadth but no depth.

What are 3D shapes?

3D shapes, also known as solid shapes, have length, breadth, and depth, and are enclosed by surfaces or planes.

What is the difference between area and perimeter?

Area measures the surface covered by a shape, while perimeter measures the continuous line along a shape’s boundary.

What types of mensuration questions appear in the IBPS RRB Exam?

Mensuration questions include direct formulas, composite figures, missing dimensions, comparisons, and data interpretation.

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