## Volume and capacity Questions

1.  The figure below shows a bucket of depth 30cm used to fill a cylindrical tank of radius 1.2m and height 1.35m which is initially three-fifth full of water.

1. Calculate, in terms of Π;
1. The capacity of the bucket in litres  (5mks)
2. The volume of water required to fill the tank in litres (2mks)
3. Calculate the number of buckets that must be drawn to fill the tank    (3mks)

2.   A bucket is in the shape of a frustum with base radius 12cm and top radius 16cm. The slant height of the bucket is 30cm as shown below. The bucket is full of water.

(a) Calculate the volume of the water. (Take  = 3.142)  (6 marks)

16

30

12

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(b)  All the water is poured into a cylindrical container of circular radius 12cm. If the cylinder has height 45cm, calculate the surface area of the cylinder which is not in contact with water.  (4 marks)

3.  The British government hired two planes to airlift football fans to South Africa for the World cup tournament. Each plane took 10 ½ hours to reach the destination.

Boeng 747 has carrying capacity of 300 people and consumes fuel at 120 litres per minute. It makes 5 trips at full capacity. Boeng 740 has carrying capacity of 140 people and consumes fuel at 200 litres per minute. It makes 8 trips at full capacity. If the government sponsored the fans one way at the cost of 800 dollars per fan, calculate:

(a) The total number of fans airlifted to South Africa. (2mks)

(b) The total cost of fuel used if one litre costs 0.3 dollars. (4mks)

(c) The total collection in dollars made by each plane. (2mks)

(d) The net profit made by each plane.  (2mks)

4.  The figure below represents a part in form of a frustum of a right circular cover. The upper and the lower radii are 50cm and 15cm respectively. The slant height is 70cm.

a.  Calculate the height of the pail. (5 cm)

b.  Find the capacity of the pail to the nearest a litre. (5 mks)

5.  Consider the vessel below

a) Calculate the volume of water in the vessel.

b) When a metallic hemisphere is completely submerged in the water, the level of the water rose by 6cm. Calculate:

i) the radius of the new water surface.

ii) the volume of the metallic hemisphere (to 4 s.f)

iii) the diameter of the hemisphere (10 mks)

6. A village water tank is in the form of a frustrum of a cone of height 3.2m.

The top and bottom radii are 18m and 24m respectively

(a) Calculate:

(i) The surface area of the tank excluding the bottom

(ii) The capacity of the water tank

(b) 15 families each having 15 members use the water tank and each person uses

65 litres of water daily. How long will it take for the full tank to be emptied

7.  The diagram below shows a bucket with a top diameter 30cm and bottom diameter 20cm.

The height of the bucket is 28cm

(a) Calculate the capacity of the bucket in litres

(b) Find the area of the metal sheet required to make 100 such buckets taking 10% extra for

overlapping and wastage

8.  A rectangular water tank measures 2.6m by 4.8m at the base and has water to a height

of 3.2m. Find the volume of water in litres that is in the tank

9.  The figure alongside shows a cone from which a frustum is made. A plane parallel to the base

cuts the cone two thirds way up the vertical height of the cone to form frustum ABCD.

The top surface radius of the frustum is labeled r and the bottom radius is R

a) Find the ratio r:R

b) Given that r = 7cm, find R

c) If the height VY of the original cone is 45cm, calculate to the nearest whole number the volume

of the frustum

d) The frustum represents a bucket which is used to fill a rectangular tank measuring 1.5m

long, 1.2m wide and 80cm high with water. How many full buckets of water are required

to fill the tank

10.  Three litres of water (density1g/cm3) is added to twelve litres of alcohol (density 0.8g/cm3.

What is the density of the mixture?

11.  A rectangular tank whose internal dimensions are 2.2m by 1.4m by 1.7m is three fifth full

of milk.

(a) Calculate the volume of milk in litres

(b) The milk is packed in small packets in the shape of a right pyramid with an equilateral

base triangle of sides 10cm. The vertical height of each packet is 13.6cm. Full packets

obtained are sold at shs.30 per packet. Calculate:

(i) The volume in cm3 of each packet to the nearest whole number

(ii) The number of full packets of milk

(iii) The amount of money realized from the sale of milk

12.  An 890kg culvert is made of a hollow cylindrical material with outer radius of 76cm and

an inner radius of 64cm. It crosses a road of width 3m, determine the density of the material

used in its construction in Kg/m3 correct to 1 decimal place.

 1. M1M1M1 A1 4 2. LSF 1 cm rep 50000cm1cm rep 500mASF 1cm2 rep 250000m2Area = = 154ha B1 M1  A1 ASF given 03 3. Area = 4 x 4sin 420 – = 10.71 – 5.867= 4.796 M1 M1A1 area of rhombus & sectordifference in area 03 4. (i) h = 90 Π × 1600 × 120 − ⅓ Π× 900 × 90 (64 000 Π − 27 000) ÷ 1000 37 Π litres (ii) Volume of water = =777.6 Π litres = 22 M1 A1M1M1A1M1A1M1M1A1 Divide by 1000 Mult by 1000

5. 16 12

12 = L

16 30 + L

L = 90 B1 for 90

h =  902 – 122

= 89.2

H = 1202 – 162

= 118.9  B1 for both 89.2

1189

Vol. big core = 1/3 x 3.142 x 162 x 118.9 M1

31879.151

Small cone = 1/3 x 3.142 x 122 x 89.2  M1

= 13452.789

 Volume of water

31879.151 – 13452.789  M1

= 18426.3645 A1

(b)  4.5 12

3.142 x 122 x h=

18426.364 M1

h = 40.73 A1

S.A = 2 x 3.142 x 12 (45 – 40.73) M1

= 321.99cm2 A1

10

 6. (a) (300×5) + (140×8)= 1500 + 1120= 2620fans(b) Cost of fuelBoeng 747= 120×10.5x60x5x2x0.3=226800 dollarsBoeng 740= 200×10.5x60x8x2x0.3=604,800 dollars(c) Total collectionBoeng 747= 300x5x800= 1,200,000 dollarsBoeng 740= 140x8x800= 896,000 dollars(d) Net profitBoeng 747= 1200000 – 226800= 973,200 dollarsBoeng 740= 896,000 – 604,800= 291,200 dollars M1 A1  M1A1 M1A1  B1   B1  B1  B1 10 7.a.)                                   b.) 50   h 70      y x   50/15 = 70 + x/x 50x = 15(70 + x)50x = 1050 + 15x35x = 1050 = 30 cm Total height = 1002 – 502  = 7500  = 86.60am 50 = 86.60 Y = 86.6 X 1515 Y 50  = 25.98  Height = 86.60 – 25.98 = 60.62  Volume= (1/3 x 22/7 x 502 x 86.60) – (1/3 x 22/7 x 152 x 25.98)  = 1/3 x 22/7 (216500 – 5845.5) = 220685.67am3 = 221litres M1    A1   B1 M1    A1   A1    M1M1 M1A110

 8. a)b)ii) New volume = 1/3 x 3.142 x 25.2 x 25.2 x 36 = 23943.55cm3Volume change = 23943.55 – 13856.22 = 10087.33cm3iii) 2/3r3 = 10087.33r3 = 10087.3 x 3/2 x 1/r3 = 4815.72r = r = 16.89cmdiameter = 16.89 x 2 = 33.78cm M1 A1        M1 A1 M1 A1 B1 M1 A1 B1 Attempt                vol of hemisphere 10

9.  L.s.f. = 18 = 3

24 4

A.s.f = 9

16

v.s.f = 27

64

h = 3
 4h = 3h + (3 x 3.2)

3.2 4

h = 9.6

(i) surface area of small cone:

L = 92 + 9.62 = 13.16m

S.A = (3.142 x 9 x 13.6) = 384.581

Curved area of frustrum

= 7 x 3.142 x 9 x 13.16

1 9

= 289.4

Top area = (3.142 x 92) = 254.5cm

Total area = 543.9m2

(ii)Volume of smaller cone = 3.142 x 92 x 9.6

3

= 814.41

Volume of frustrum = (37 x 814.41)

27

= 1116.043m3

= 1116043L

Litres used per day= (15 x 15 x 40) + (116 x 65)= 16540L

No. of days = 1116043

16540 = 67.5days

10.  L.S.F = 3 = 28 + h

2 h

56 + 2h = 3h

h = 56cm

Volume = 1/3 r2H – 1/3r2h

=1/3 x 22/7 x 15x 15×56 –1/3 x 22/7 x 10 x 10 x 28

= 13200 – 29331/3

= 10.2667litres

(b) Slant height = 152 + 562  = 3361

= 57.97cm

Curved surface – RL – rl

11.  2.6 x 4.8 x 3.2 = 39.936m3

1m3 = 1000litres

39.936m3 = 39.936 x 1000

= 39936 litres

12.  The top surface of the frustrum is-2/3 way up the vertical height of the original one.

VX: XY = 1/3h: h = 1:3

Using similar triangle we have

R = VX = 1

R VY 3

R:R = 1:3

r = 1
 R = 3r

R 3

R = 3×7 = 21cm

(c)height of removed cone is 1/3 height of original cone

h = 1/3 x 45 = 15cm

volume of removed cone = 1/3 r2h

= 1 x 22 x 7 x7 x15

7

= 770cm3

Now L. S. F = 1/3

V. S. F= (1/3)3 = 1/27

Hence ratio of volumes = 1:27

Volume of original cone = 27x Vol. of small cone

= 770 x 27= 20790cm3

Capacity of frustrum

= vol. of original cone – vol. of removed cone

= 20790 – 770= 20020cm3

20200

1000 = 20 l

(d) capacity of tank = 150 x 120 x 80

1000 = 1440l

No. of buckets = 1440

20 = 72buckets

13.  Mass of water = 1 x 3000 cm3 = 3000 g

Mass of alcohol = 0.8 x 1200 = 9600g

Mass of mixture = 12,600g

Volume of mixture = 15,000 cm3

Density of mixture = 12600

15000

= 0.84g/cm3

14.  (a)  Vol. of tank = 22 x 144 x 1.7 = 5.236

Vol. of milk = 3/5 x 5.236 = 3.146m3

Vol. in liters = 3.1416 x 1000 = 3141.6litres

(b) (i) Vol. of packet (1/3 x 10 sin 60) x 13.6

= 26.97 x 13.6

= 3.66.75cm3

= 367cm3

(ii) No. packets = (3141.6 x 1000)

367

(iii) Amount = 8560.2 x 20

= 171204.3597

= Shs.171,204.40

15.  Volume of culvert

= 22/7 (762 – 642) x 300 x 10-6

= 22/7 x 1680 x 300

10000000000

= 1.584m3

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