Quantitative Techniques Practice Paper on Mensuration
Quantitative Techniques Practice Paper on Mensuration
Question 1: A hollow metal pipe with the length 21cm and inner radius 5cm and outer radius 6.5cm. Answer the following questions based on given information only.
The Lateral surface area of the pipe is approximately
1500cm2 (b) 1518cm2 (c) 1650cm2 (d) 1620cm2
The volume of the metal of the pipe is
1138.5cm3 (b)1100cm3 (c) 1239cm3 (d) 1139.5cm3
If the pipe wasn’t hollow and the metal was inside then what is the volume of the cylinder? Consider the outer radius for the whole cylinder.
2788.5cm3 (b)2888.5cm3 (c) 2898.5cm3 (d) 2988cm3
What would be the volume of a cone with the same height and inner radius as the radius of the cone?
540cm3 (b) 560cm3 (c) 550 cm3 (d) 530cm3
If the hollow pipe is melted and remoulded in a sphere, then what is the radius of the new sphere?
12cm (b) 4.9cm (c) 5.5cm (d) 6.47cm
Question 2: A chocolate box is 50cm long, 30cm wide and 15cm high. If 1 cubic cm of a cardboard weights 0.5gms then answer the questions based on given data.
2.1 How much chocolate that box can hold in terms of volume not weight(assume 1cm thickness in calculation)?
(a) 225m3 (b) 18816cm3 (c) 22.5m3 (d)22.5cm3
2.2 If the thickness of the box is 1cm then, What is the total weight of the box?
(a) 1300 gms (b) 1270gms (c) 1.84kg (d) 11.7kgs
2.3 What is the outer surface area of the box if it is open from the top(exclude thickness)?
(a) 0.39m2 (b) 390cm2 (c) 39m2 (d) 3900m2
2.4 Suppose the box is filled with chocolate in cube shape with side 2cm. How many choclates the box will hold?
(a) 2500 (b) 2540 (c) 2400 (d) 2352
2.5 If the length of the box is reduced by 10%, then what is the total surface area of the box?
(a) 6938cm2 (b) 6900cm2 (c) 6950cm2 (d) 6900cm2
Answer and Solutions:
Ans 1.1 (b) 1625cm2
Solution: Inner Radius= r = 5cm Outer Radius = R = 6.5cm
Lateral surface area of Pipe = External Surface area + Internal surface area
= 2(pi) * ( R+r) *h
= 2*22*(6.5+5) 21/7 = 1518cm2
Ans 1.2 (a) 1138.5cm2
Solution: Volume of hollow cylinder = (pi) (R2 - r2) * h = 22*(6.52 -52) * 21/7
= 1138.5cm2
Ans 1.3 (a) 2788.5cm2
Solution: Volume of the cylinder if not hollow = (pi) R2h
= 22*6.5*6.5*21/7 = 2788.5cm2
Ans 1.4 (c) 550cm2
Solution: Volume of a cone with the same height and inner r = 1/3 * (pi) r2h
= 1/3 * 22/7 *5*5*21
= 550cm2
Ans 1.5 (d) 6.47cm2
Solution: When one solid is moulded into another one, the volume remains the same. Hence, Volume of sphere = Volume of a hollow cylinder
4/3*(pi)*radius3 = (pi) (R2 – r2) h
radius3 = (6.52-52)*21*3/4
radius = 6.47cm
Ans 2.1 (b) 20160cm3
Solution: As the thickness is 1 cm we will deduce it with actual length and breadth
Volume of inner box = (50-1)*(30-1)*(15-1) = 48*28*14= 18816 cm3
Ans 2.2 (c) 1.17kgs
Solution: Volume of whole Cuboid = l*b*h = 50*30*15 = 22500cm3
Volume of material = Volumeof whole box – Volume of inner box
= 22500 –18816 cm3 = 3684 cm3
1cubic cm = 0.5gm, Hence Weight = 3684*0.5 gms = 1840 gms = 1.840 kgs
Ans 2.3 (a) 0.39m2
Solution: As the upper surface of the box is open hence we only have 5 surfaces whose area we need to calculate.
Surface Area of outer box = base area(l*b) + side wall(2*b*h) + side wall(2*l*b)
= 50*30 + 2*15*30 + 2*50*15
= 3900 cm2
= 0.39m2 [ since 100cm = 1m]
Ans 2.4 (d) 2352
Solution: The number of chocolates in a box can be equal to total volume of box by volume of one chocolate as the volume as to be equated.
N*volume of chocolate = Volume of the box
N = 18816/(23) = 2352
Ans 2.5 (a) 6938cm2
Solution: If the length is reduced by 10% then new length = 50*0.9 = 45cm
Total surface Area = Outer area+ Inner surface area + Area of top surface(thickness)
= [(45*30)+2(45*15)+2(30*15)] + [(43*28)+2(43*14)+2(28*14)] +[2*1*45+2*1*28]
= [1350+1350+900] + [1204+1204*784] + 146
= 3600+3192+146 = 6938 cm2
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First published on June 14, 2020.Â