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Aptitude Questions – Mensuration Set 19

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Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample from Mensuration that are important for all the competitive exams. We have included some questions that are repeatedly asked in exams !!

  1. A right circular cone is placed over a cylinder of the same radius. Now the combined structure is painted on all sides. Then they are separated now the ratio of area painted on Cylinder to Cone is 3:1. What is the height of Cylinder if the height of Cone is 4 m and radius is 3 m?
    1. 5 m
    2. 6 m
    3. 8 m
    4. 10 m
    5. Cannot be determined
    Answer – 2. 6 m
    Explanation :
    Cylinder painted area = 2πrh+πr²
    Cone painted area = πrl
    2h+r/√ (r² +h1² ) = 3:1
    h = 6

  2. The diameter of Road Roller is 84 cm and its length is 150 cm. It takes 600 revolutions to level once on a particular road. Then what is the area of that road in m²?
    1. 2376
    2. 2476
    3. 2496
    4. 2516
    5. None
    Answer – 1. 2376
    Explanation :
    Area
    600*2*22/7*42/100*150/100=2376

  3. A smaller triangle is having three sides. Another big triangle is having sides exactly double the sides of the smaller triangle. Then what is the ratio of Area of Smaller triangle to Area of the Bigger triangle?
    1. 1:2
    2. 2:1
    3. 1:4
    4. 4:1
    5. Cannot be determined
    Answer – 3. 1:4
    Explanation :
    Smaller triangle sides = a, b, c
    Area= √s(s-a) (s-b) (s-c);
    s=a+b+c/2
    = √(a+b+c)(b+c-a)(a+c- b)(a+b-c)/4
    Bigger triangle =2a, 2b, 2c
    Area = √(a+b+c)(b+c-a)(a+c- b)(a+b-c)
    Ratio = 1:4

  4. ABCD is a square of 20 m. What is the area of the least-sized square that can be inscribed in it with its vertices on the sides of ABCD?
    1. 120 m²
    2. 100 m²
    3. 200 m²
    4. 250 m²
    5. None
    Answer – 3. 200 m²
    Explanation :
    It touches on midpoints on the sides of the square ABCD
    Side= √ (10² +10²) = √200
    Area= 200 m²

  5. A hemispherical bowl of diameter 16cm is full of ice cream. Each student in a class is served exactly 4 scoops of ice cream. If the hemispherical scoop is having a radius of 2cm, then ice cream is served to how many students?
    1. 16
    2. 32
    3. 64
    4. 128
    5. None
    Answer – 1. 16
    Explanation :
    Solution:
    2/3*π*8³ = n*4*2/3*π*2³
    n = 16

  6. A hollow cylindrical tube is made of plastic is 4 cm thick. If the external diameter is 18 cm and length of the tube is 59cm, then find the volume of the plastic?
    1. 10380 cm³
    2. 10384 cm³
    3. 10440 cm³
    4. 10444 cm³
    5. None
    Answer – 2. 10384 cm³
    Explanation :
    R = 9, r =5
    V =22/7*59(9 2 -5 2) = 10384

  7. What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?
    A. 27m
    B. 28m
    C. 29m
    D. 25m
    E. 15m
    Answer – C. 29m
    Explanation :
    πR² = πr1² + πr2²
    πR² = π(r1² + r2²)
    R² = (400 + 441)
    R = 29

  8. A well with 14 m diameter is dug up to 49 m deep. Now the soil taken out during dug is made into cubical blocks of 3.5m side each. Then how many such blocks were made?
    1. 22
    2. 44
    3. 88
    4. 176
    5. None
    Answer – 4. 176
    Explanation :
    Solution:
    22/7*7²*49 = n*(7/2)³
    n =176

  9. If the ratio of radius two Cylinders A and B are in the ratio of 2:1 and their heights are in the ratio of 2:1 respectively. The ratio of their total surface areas of Cylinder A to B is?
    1. 2:1
    2. 1:2
    3. 1:4
    4. 4:1
    5. Cannot be determined
    Answer – 4. 4:1
    Explanation :
    Explanation :
    Cylinder A: 2πr1 (r1 + h1)
    Cylinder B: 2πr2 (r2 + h2)
    r 1 /r 2 = 2:1; h 1 /h 2 = 2:1
    T A /T B = 2πr1 (r1 + h1)/ 2πr2 (r2 + h2)

  10. The area of the Circular garden is 88704 m². Outside the garden a road of 7m width laid around it. What would be the cost of laying road at Rs. 2/m².
    1. Rs.7546
    2. Rs.10036
    3. Rs.11092
    4. Rs.15092
    5. Rs.16086
    Answer – 4. Rs.15092
    Explanation :
    88704 = 22/7*r
    r =168
    Outer radius = 168+7 = 175
    Outer area = 22/7*175 2 = 96250
    Road area = 96250 – 88704 = 7546
    Cost = 7546*2 = 15092