Practice Questions: Mensuration | Quantitative Techniques for CLAT PDF Download

Q1: 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
(a) 1500cm² 
(b) 1518cm²
(c) 1650cm²
(d) 1620cm²
Ans: b
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

Q2: 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 volume of the metal of the pipe is
(a) 1138.5cm³
(b) 1100cm³
(c) 1239cm³
(d) 1139.5cm³
Ans: (a)
Volume of hollow cylinder = (pi) (R² - r²) * h = 22*(6.5² -5²) * 21/7
= 1138.5cm²

Q3: 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.
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.
(a) 2788.5cm³
(b)2888.5cm³
(c) 2898.5cm³
(d) 2988cm³
Ans: (a)
Volume of the cylinder if not hollow = (pi) R²h
= 22*6.5*6.5*21/7 = 2788.5cm²

Q4: 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.
What would be the volume of a cone with the same height and inner radius as the radius of the cone?
(a) 540cm³
(b) 560cm³
(c) 550 cm³
(d) 530cm³
Ans: c
Volume of a cone with the same height and inner r = 1/3 * (pi) r²h
= 1/3 * 22/7 *5*5*21
= 550cm²

Q5: 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.
If the hollow pipe is melted and remoulded in a sphere, then what is the radius of the new sphere?
(a) 12cm
(b) 4.9cm
(c) 5.5cm
(d) 6.47cm
Ans: d
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

Q6: 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.
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
Ans: b
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

Q7: 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.
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
Ans: c
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

Q8: 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.
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
Ans:  a
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] 

Q9: 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.
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
Ans: d
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

Q10: 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.
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
Ans: 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