{"id":102017,"date":"2024-07-16T19:47:50","date_gmt":"2024-07-16T14:17:50","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=102017"},"modified":"2024-07-16T19:47:51","modified_gmt":"2024-07-16T14:17:51","slug":"mensuration-for-ibps-rrb-exam","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/mensuration-for-ibps-rrb-exam\/","title":{"rendered":"Mensuration For IBPS RRB Exam, Check 10 Questions"},"content":{"rendered":"\n

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Home<\/a><\/span> \u00bb Banking & Insurance<\/a><\/span> \u00bb Mensuration For IBPS RRB Exam<\/span><\/span><\/div>\n\n\n

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Mensuration For IBPS RRB Exam:<\/strong> Mensuration studies the measurement of geometric figures and their parameters like length, area, perimeter, volume, shape, surface area, lateral surface area, etc. It\u2019s an essential topic in the quantitative aptitude section of IBPS RRB Exam. In this blog, we\u2019ll 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.<\/p>\n\n\n\n

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Differences Between 2D and 3D shapes<\/h2>\n\n\n\n

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\u2019s important to learn the difference between these two dimensions.<\/p>\n\n\n\n

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

Mensuration For IBPS RRB Exam, Important Terms<\/h2>\n\n\n\n

Now let\u2019s 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.<\/p>\n\n\n\n

Mensuration For IBPS RRB Exam, Important Terms<\/strong><\/td><\/tr>
Terms<\/strong><\/td>Abbreviation<\/strong><\/td>Unit<\/strong><\/td>Definition<\/strong><\/td><\/tr>
Area<\/td>A<\/td>m\u00b2 or cm\u00b2<\/td>Surface covered by the closed shape.<\/td><\/tr>
Perimeter<\/td>P<\/td>cm or m<\/td>Continuous line along the boundary of the figure.<\/td><\/tr>
Volume<\/td>V<\/td>cm\u00b3 or m\u00b3<\/td>Space occupied by a 3D shape.<\/td><\/tr>
Curved Surface Area<\/td>CSA<\/td>m\u00b2 or cm\u00b2<\/td>Total area of a curved surface (e.g., sphere).<\/td><\/tr>
Lateral Surface Area<\/td>LSA<\/td>m\u00b2 or cm\u00b2<\/td>Total area of all lateral surfaces surrounding the figure.<\/td><\/tr>
Total Surface Area<\/td>TSA<\/td>m\u00b2 or cm\u00b2<\/td>Sum of all curved and lateral surface areas.<\/td><\/tr>
Square Unit<\/td>\u2013<\/td>m\u00b2 or cm\u00b2<\/td>Area covered by a square with one-unit side.<\/td><\/tr>
Cube Unit<\/td>\u2013<\/td>m\u00b3 or cm\u00b3<\/td>Volume occupied by a cube with one-unit side.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Types of Mensuration Questions in IBPS RRB Exam<\/h2>\n\n\n\n

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\u2019s explore the types of questions that appear in the IBPS RRB Exam, along with some tips to master them<\/p>\n\n\n\n

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

    Tips to Master Mensuration For IBPS RRB Exam<\/h2>\n\n\n\n

    Here are five concise tips needed to master the art of solving questions on mensuration for IBPS RRB Exam.<\/p>\n\n\n\n

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

      Mensuration For IBPS RRB Exam, Check 10 Questions<\/h2>\n\n\n\n

      Question 1:<\/strong> The radius of a sphere is equal to the side of a cube having volume 729 cm3<\/sup>. What is the total surface area of the sphere? (Take \u03c0 = 3)<\/p>\n\n\n\n

      A) 680 cm2<\/sup><\/p>\n\n\n\n

      B) 712 cm2<\/sup><\/p>\n\n\n\n

      C) 1080 cm2<\/sup><\/p>\n\n\n\n

      D) 606 cm2<\/sup><\/p>\n\n\n\n

      E) 972 cm2<\/sup><\/p>\n\n\n\n

      Question 2:<\/strong> 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?<\/p>\n\n\n\n

      A) 8 cm<\/p>\n\n\n\n

      B) 6 cm<\/p>\n\n\n\n

      C) 7 cm<\/p>\n\n\n\n

      D) 12 cm<\/p>\n\n\n\n

      E) 14 cm<\/p>\n\n\n\n

      Question 3:<\/strong> 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.<\/p>\n\n\n\n

      A) 28 m<\/p>\n\n\n\n

      B) 32 m<\/p>\n\n\n\n

      C) 44 m<\/p>\n\n\n\n

      D) 36 m<\/p>\n\n\n\n

      E) None of these<\/p>\n\n\n\n

      Question 4:<\/strong> Find the volume of cube having side equal to the radius of in-circle of equilateral triangle of area 48\u221a3 cm2<\/sup>?<\/p>\n\n\n\n

      A) 8 cm3<\/sup><\/p>\n\n\n\n

      B) 24\u221a3 cm3<\/sup><\/p>\n\n\n\n

      C) 27 cm3<\/sup><\/p>\n\n\n\n

      D) 8\u221a3 cm3<\/sup><\/p>\n\n\n\n

      E) 64 cm3<\/sup><\/p>\n\n\n\n

      Question 5:<\/strong> 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.<\/p>\n\n\n\n

      A) 156 m<\/p>\n\n\n\n

      B) 172 m<\/p>\n\n\n\n

      C) 160 m<\/p>\n\n\n\n

      D) 168 m<\/p>\n\n\n\n

      E) 176 m<\/p>\n\n\n\n

      Question 6:<\/strong> Find the area of the square having perimeter 20% more than 80 cm.<\/p>\n\n\n\n

      A) 484 cm2<\/sup><\/p>\n\n\n\n

      B) 529 cm2<\/sup><\/p>\n\n\n\n

      C) 576 cm2<\/sup><\/p>\n\n\n\n

      D) 625 cm2<\/sup><\/p>\n\n\n\n

      E) None of these<\/p>\n\n\n\n

      Question 7:<\/strong> A cylinder of volume 4335 cm3<\/sup> is to be painted. If the cost of painting the curved surface area of the cylinder at the rate of Rs. 5\/cm2<\/sup> is Rs. 2550, then find the radius of the cylinder. [Use \u03c0 = 3]<\/p>\n\n\n\n

      A) 15 cm<\/p>\n\n\n\n

      B) 17 cm<\/p>\n\n\n\n

      C) 19 cm<\/p>\n\n\n\n

      D) 21 cm<\/p>\n\n\n\n

      E) None of these<\/p>\n\n\n\n

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

      A) 196 cm2<\/sup><\/p>\n\n\n\n

      B) 100 cm2<\/sup><\/p>\n\n\n\n

      C) 144 cm2<\/sup><\/p>\n\n\n\n

      D) 256 cm2<\/sup><\/p>\n\n\n\n

      E) 400 cm2<\/sup><\/p>\n\n\n\n

      Question 9:<\/strong> 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<\/p>\n\n\n\n

      A) 25 cm<\/p>\n\n\n\n

      B) 16 cm<\/p>\n\n\n\n

      C) 14 cm<\/p>\n\n\n\n

      D) 12.5 cm<\/p>\n\n\n\n

      E) None of these<\/p>\n\n\n\n

      Question 10:<\/strong> What is the circumference of a circle of maximum area inscribed in a square having area 196 cm2<\/sup>?<\/p>\n\n\n\n

      A) 88 cm<\/p>\n\n\n\n

      B) 22 cm<\/p>\n\n\n\n

      C) 44 cm<\/p>\n\n\n\n

      D) 66 cm<\/p>\n\n\n\n

      E) None of these<\/p>\n\n\n\n

      Solution 1: E)<\/strong><\/p>\n\n\n\n

      Side of the cube = 7291\/3<\/sup> = 9 cm<\/p>\n\n\n\n

      So, radius of the sphere = 9 cm<\/p>\n\n\n\n

      Total surface area of the sphere = 4 \u00d7 \u03c0 \u00d7 r2<\/sup> = 4 \u00d7 3 \u00d7 92<\/sup> = 972 cm2<\/sup><\/p>\n\n\n\n

      Hence, option e.<\/p>\n\n\n\n

      Solution 2: B)<\/strong><\/p>\n\n\n\n

      Let the height of cylinder be \u2018h\u2019 cm.<\/p>\n\n\n\n

      Volume of cylinder = Volume of cone<\/p>\n\n\n\n

      \u03c0 \u00d7 142 <\/sup>\u00d7 h = (1\/3) \u00d7 \u03c0 \u00d7 142 <\/sup>\u00d7 18<\/p>\n\n\n\n

      h = 18\/3 = 6 cm<\/p>\n\n\n\n

      Hence, option b.<\/p>\n\n\n\n

      Solution 3: E)<\/strong><\/p>\n\n\n\n

      Let the length and breadth of the rectangular field be \u20185x\u2019 m and \u20184x\u2019 m respectively.<\/p>\n\n\n\n

      Perimeter of rectangular field = 720\/4 = 180 m<\/p>\n\n\n\n

      According to question,<\/p>\n\n\n\n

      2(5x + 4x) = 180<\/p>\n\n\n\n

      x = 10 m<\/p>\n\n\n\n

      So, breadth of the rectangular filed = 4x = 40 m<\/p>\n\n\n\n

      Hence, option e.<\/p>\n\n\n\n

      Solution 4: E)<\/strong><\/p>\n\n\n\n

      Let the side of the equilateral triangle be \u2018s\u2019 cm.<\/p>\n\n\n\n

      Area of equilateral triangle = 48\u221a3 cm2<\/sup><\/p>\n\n\n\n

      (\u221a3\/4) \u00d7 s2<\/sup>= 48\u221a3<\/p>\n\n\n\n

      s = \u221a(48 \u00d7 4)<\/p>\n\n\n\n

      s = 8\u221a3 cm<\/p>\n\n\n\n

      Radius of the in-circle = (side \u00d7 \u221a3)\/6 = (8\u221a3 \u00d7\u221a3)\/6 = 4 cm<\/p>\n\n\n\n

      So, side of cube = 4 cm<\/p>\n\n\n\n

      Therefore, volume of cube = 43<\/sup> = 64 cm3<\/sup><\/p>\n\n\n\n

      Hence, option e.<\/p>\n\n\n\n

      Solution 5: D)<\/strong><\/p>\n\n\n\n

      Circumference of circular park = 3168\/6 = 528 m<\/p>\n\n\n\n

      So, 2 \u00d7 (22\/7) \u00d7 radius = 528<\/p>\n\n\n\n

      2 \u00d7 radius = (528 \u00d7 7)\/22 = 168 m<\/p>\n\n\n\n

      So, diameter of the park = 168 m<\/p>\n\n\n\n

      Hence, option d.<\/p>\n\n\n\n

      Solution 6: C)<\/strong><\/p>\n\n\n\n

      Let the side of the square be \u2018x\u2019 cm.<\/p>\n\n\n\n

      Perimeter of the square = 4x = 120% of 80<\/p>\n\n\n\n

      4x = 96<\/p>\n\n\n\n

      x = 24<\/p>\n\n\n\n

      So, the area of the square = 242<\/sup> = 576 cm2<\/sup><\/p>\n\n\n\n

      Hence, option c.<\/p>\n\n\n\n

      Solution 7: B)<\/strong><\/p>\n\n\n\n

      Volume of cylinder = \u03c0 r2<\/sup> h, where \u2018r\u2019 and \u2018h\u2019 are the radius and height of the cylinder.<\/p>\n\n\n\n

      3 \u00d7 r2<\/sup> \u00d7 h = 4335<\/p>\n\n\n\n

      r2<\/sup> h = 1445 —-(1)<\/p>\n\n\n\n

      Curved surface area of the cylinder = 2 \u03c0 r h<\/p>\n\n\n\n

      2 \u00d7 3 \u00d7 r \u00d7 h = 2550\/5 = 510<\/p>\n\n\n\n

      r \u00d7 h = 85 —-(2)<\/p>\n\n\n\n

      Dividing eq. (1) by eq. (2), we get<\/p>\n\n\n\n

      r = 17 cm<\/p>\n\n\n\n

      Hence, option b.<\/p>\n\n\n\n

      Solution 8: B)<\/strong><\/p>\n\n\n\n

      Solution 1: C)<\/strong><\/p>\n\n\n\n

      Let the breadth of rectangle R1<\/sub> and rectangle R2 <\/sub>be \u20185x\u2019 cm and \u20184x\u2019 cm respectively.<\/p>\n\n\n\n

      Length of rectangle = side of square = \u221a256 = 16 cm<\/p>\n\n\n\n

      According to question,<\/p>\n\n\n\n

      Sum of areas of two rectangles (R1<\/sub> and R2<\/sub>) = 432 cm2<\/sup><\/p>\n\n\n\n

      16 \u00d7 5x + 16 \u00d7 4x = 432<\/p>\n\n\n\n

      16(5x + 4x) = 432<\/p>\n\n\n\n

      9x = 432\/16<\/p>\n\n\n\n

      9x = 27<\/p>\n\n\n\n

      x = 3 cm<\/p>\n\n\n\n

      So, the breadth of rectangle R2 <\/sub>= 4 \u00d7 3 = 12 cm<\/p>\n\n\n\n

      Therefore, required area of square = 12 \u00d7 12 = 144 cm2<\/sup><\/p>\n\n\n\n

      Hence, option b.<\/p>\n\n\n\n

      Solution 9: A)<\/strong><\/p>\n\n\n\n

      Let the breadth of the rectangular page be \u2018x\u2019 cm.<\/p>\n\n\n\n

      Then, length of rectangular page = 1.25x cm<\/p>\n\n\n\n

      According to question,<\/p>\n\n\n\n

      2 \u00d7 (x + 1.25x) = 90<\/p>\n\n\n\n

      4.5x = 90<\/p>\n\n\n\n

      x = 20 cm<\/p>\n\n\n\n

      So, breadth = 20 cm and length = 25 cm<\/p>\n\n\n\n

      Hence, option a.<\/p>\n\n\n\n

      Solution 10: C)<\/strong><\/p>\n\n\n\n

      Side of square = \u221a196 = 14 cm<\/p>\n\n\n\n

      So, diameter of the circle having maximum area inscribed in square = 14 cm<\/p>\n\n\n\n

      Therefore, circumference of the circle = 2 \u00d7 (22\/7) \u00d7 7 = 44 cm<\/p>\n\n\n\n

      Hence, option c.<\/p>\n\n\n\n

      <\/p>\n\n\n\n

      \n
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      Mensuration For IBPS RRB Exam FAQs<\/h2>\n\n\n\n
      What is mensuration in mathematics?<\/strong>

      Mensuration involves measuring geometric figures’ length, area, perimeter, volume, and surface area.<\/p> <\/div>

      What are 2D shapes?<\/strong>

      2D shapes are enclosed by straight lines in a plane and have length and breadth but no depth.<\/p> <\/div>

      What are 3D shapes?<\/strong>

      3D shapes, also known as solid shapes, have length, breadth, and depth, and are enclosed by surfaces or planes.<\/p> <\/div>

      What is the difference between area and perimeter?<\/strong>

      Area measures the surface covered by a shape, while perimeter measures the continuous line along a shape’s boundary.<\/p> <\/div>

      What types of mensuration questions appear in the IBPS RRB Exam?<\/strong>

      Mensuration questions include direct formulas, composite figures, missing dimensions, comparisons, and data interpretation.<\/p> <\/div> <\/div>\n","protected":false},"excerpt":{"rendered":"

      Master the art of answering questions on Mensuration for IBPS RRB Exam. Read to check out 10 sample Questions.<\/p>\n","protected":false},"author":6,"featured_media":102061,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[22],"tags":[],"class_list":["post-102017","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-banking-insurance"],"yoast_head":"\nMensuration For IBPS RRB Exam, Check 10 Questions<\/title>\n<meta name=\"description\" content=\"Master the art of answering questions on Mensuration for IBPS RRB Exam. 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