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