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Mensuration Practice Set 1

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Mensuration Practice Set 1

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Mensuration Practice set 1 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 Quiz 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.

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Q1. 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?
    A. 5 m B. 6 m C. 8 m D. 10 m
Answer: B
Explanation: Cylinder painted area = 2 [latex] \pi rh + \pi r² [/latex]
Cone painted area = [latex] \pi rl [/latex]
[latex] \frac {2h + r}{\sqrt {(r² + h1² )}} [/latex] = 3 : 1
h = 6
Q2. 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?
    A. 2376 B. 2476 C. 2496 D. 2516
Answer: A
Explanation: Area
[latex]600 \times 2 \times \frac {22}{7} \times \frac {42}{100} \times \frac {150}{100} [/latex] = 2376
Q3. 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 a Smaller triangle to Area of the Bigger triangle?
    A. 1 : 2 B. 2 : 1 C. 1 : 4 D. 4 : 1
Answer: C
Explanation: Smaller triangle sides = a, b, c
Area = [latex] \sqrt {s (s - a) (s - b) (s - c)} [/latex]
s = [latex] \frac {a + b + c}{2} [/latex]
= [latex] \sqrt \frac {(a + b + c)(b + c - a) (a + c - b) (a + b - c)}{4} [/latex]
Bigger triangle = 2a, 2b, 2c
Area = [latex] \sqrt {(a + b + c)(b + c - a)(a + c - b)(a + b - c)} [/latex]
Ratio = 1 : 4
Q4. 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?
    A. 120 m² B. 100 m² C. 200 m² D. 250 m²
Answer: C
Explanation: It touches on midpoints on the sides of the square ABCD
Side = [latex] \sqrt {(10² +10²)} = \sqrt {200} [/latex]
Area = 200 m²
Q5. 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?
    A. 16 B. 32 C. 64 D. 128
Answer: A
Explanation: [latex] \frac {2}{3} \times \pi \times 8³ = n \times 4 \times \frac {2}{3} \pi 2³ [/latex]
n = 16
Q1. 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?
    A. 10380 cm³ B. 10384 cm³ C. 10440 cm³ D. 10444 cm³
Answer: B
Explanation: R = 9, r =5
V = [latex] \frac {22}{7} \times 59(9 2 -5 2) [/latex] = 10384
Q2. 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. 27 m B. 28 m C. 29 m D. 25 m
Answer: C
Explanation: [latex] \pi R² = \pi r1² + \pi r2² [/latex] [latex] \pi R² = \pi (r1² + r2²) [/latex] R² = (400 + 441) R = 29
Q3. 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?
    A. 22 B. 44 C. 88 D. 176
Answer: D
Explanation: [latex] \frac {22}{7} \times 7² \times 49 = n \times (\frac {7}{2})³ [/latex]
n = 176
Q4. 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?
    A. 2:1 B. 1:2 C. 1:4 D. 4:1
Answer:
Explanation: Cylinder A: [latex] 2 \pi r1 (r1 + h1) [/latex] Cylinder B: 2 [latex] \pi r2 (r2 + h2) [/latex] r [latex] \frac {1}{r} 2 = 2 :1 ; h \frac {1} {h 2} = 2:1 [/latex] [latex] {T A }{T B} = 2 \pi r1 \frac {(r1 + h1)}{2} \pi r2 (r2 + h2) [/latex]
Q5. 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.
    A. Rs.7546 B. Rs.10036 C. Rs.11092 D. Rs.15092
Answer: D
Explanation: 88704 = [latex] \frac {22}{7} \times r [/latex]
r = 168
Outer radius = 168 + 7 = 175
Outer area = [latex] \frac {22}{7} \times 175 2 [/latex] = 96250
Road area = 96250 – 88704 = 7546
Cost = [latex] 7546 \times 2 [/latex]= 15092
Q1. A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a rectangular block of iron 22 cm in length, 18 cm in breadth and 7 cm in thickness is wholly immersed in water, by how many cm will the water level rise?
    A. 10 cm B. 18 cm C. 12 cm D. 16 cm
Answer: B
Explanation: Volume of the Rectangular Block = [latex] 22 \times 18 \times 7 [/latex]
Radius of the Cistern = 7 cm
Area of the Cylinder = [latex] \pi \times r² \times h = \frac {22}{7} \times 7 \times 7 \times h [/latex]
[latex] \frac {22}{7} \times 7 \times 7 \times h = 22 \times 18 \times 7; h [/latex]= 18 cm
Q2. If the radius of the cylinder is doubled, but height is reduced by 50%. What is the percentage change in volume?
    A. 100% B. 150% C. 125% D. 175%
Answer: A
Explanation: [latex] \frac {R1}{R2} = \frac {R}{2R} and \frac {H1}{H2} = \frac {H}{\frac {H}{2}} [/latex]
Original Volume = [latex] \pi r²h [/latex]
New Volume = [latex] \pi (2r)² \frac {h}{2} [/latex]
Change in Volume = [latex] \frac {(2 – 1)}{1} \times 100 [/latex] = 100
Q3. The perimeter of a rectangle and a square is 80 cm each. If the difference between their areas is 100 cm. Find the sides of the rectangle.
    A. 30cm, 10cm B. 40cm, 15cm C. 25cm, 10cm D. 20cm, 30cm
Answer: A
Explanation: 2(l + b) = 4a = 80
l + b = 40; a = 20 [latex] \Rightarrow [/latex] a² = 400
a² – lb = 100; 400 – lb = 100; lb = 300
(l – b)² = (l + b)² – 4lb
(l – b)² = 1600 – 1200; l – b = 20;
Q4. The length of a wall is 5/4 times of its height. If the area of the wall will be 500m². What is the sum of the length and height of the wall?
    A. 55m B. 45m C. 40m D. 50m
Answer: B
Explanation: [latex] l = 5x; h = 4x ; l \times h [/latex]= 500
[latex] 20 x² = 500 ; x = 5 [/latex]
l + h = 25 + 20 = 45 m
Q5. A circular path runs around a circular garden. If the difference between the circumference of the outer circle and the inner circle is 88m. Find the width of Path?
    A. 14 m B. 15 m C. 18 m D. 13 m
Answer: A
Explanation: Width of the Road = R – r
2[latex] \pi R – 2 \pi r [/latex]= 88
R – r = 14 m

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