HomeQuantitative TechniquesQuantitative Techniques Practice Paper on Mensuration for CLAT 2022

# Quantitative Techniques Practice Paper on Mensuration for CLAT 2022

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

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

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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