# RS Aggarwal Quantitative Aptitude PDF Free Download: VOLUME AND SURFACE AREA

Contents

- 1 VOLUME AND SURFACE AREA
- 1.1 IMPORTANT FORMULAE
- 1.2 I. CUBOID
- 1.3 II. CUBE
- 1.4 III. CYLINDER
- 1.5 IV. CONE
- 1.6 V. SPHERE
- 1.7 VI. HEMISPHERE
- 1.7.1 SOLVED EXAMPLES
- 1.7.2
- 1.7.3
- 1.7.4 Ex. 12. If each edge of a cube is increased by 50%, find the percentage increase in
- 1.7.5 Ex.16. 2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm
- 1.7.6 Ex. 22. The heights of two right circular cones are in the ratio 1 : 2 and the
- 1.7.7 Ex. 27. Find the number of lead balls, each 1 cm in diameter that can be

__VOLUME AND SURFACE AREA__

__VOLUME AND SURFACE AREA__

** **

## IMPORTANT FORMULAE

__ __

## I. CUBOID

Let length = 1, breadth = b and height = h units. Then, 1. Volume = (1 x b x h) cubic units.

**Surface area=**2(lb + bh + lh) sq.units.**Diagonal**.=Öl^{2}+b^{2}+h^{2}units

## II. CUBE

Let each edge of a cube be of length a. Then,

**Volume**= a^{3}cubic units.**Surface area**= 6a^{2}sq. units.**Diagonal**= Ö3 a units.

## III. CYLINDER

Let radius of base = r and Height (or length) = h. Then,

**Volume**= (P r^{2}h) cubic units.**Curved surface area**= (2P rh). units.**Total surface area**=2Pr (h+r) sq. units

## IV. CONE

Let radius of base = r and Height = h. Then,

**Slant height, l =****Ö**h^{2}+r^{2}**Volume**= (1/3) Pr^{2}h cubic units.**Curved surface area**= (Prl) sq. units.**Total surface area**= (Prl + Pr^{2 }) sq. units.

## V. SPHERE

Let the radius of the sphere be r. Then,

**Volume**= (4/3)Pr3 cubic units.**Surface area**= (4Pr^{2}) sq. units.

## VI. HEMISPHERE

Let the radius of a hemisphere be r. Then,

**Volume**= (2/3)Pr^{3}cubic units.**Curved surface area**= (2Pr^{2}) sq. units.**Total surface area**= (3Pr^{2}) units.

**Remember**: 1 litre = 1000 cm3.

### SOLVED EXAMPLES

.

**Ex. 1. Find the volume and surface area of a cuboid 16 m long, 14 m broad and**

**7 m high.**

** **

**Sol**. Volume = (16 x 14 x 7) m3 = 1568 m3.

- Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] cm
^{2}= (2 x 434) cm^{2}= 868 cm^{2}.

**Ex. 2. Find the length of the longest pole that can be placed in a room 12 m long**

**8m broad and 9m high.**

i

**Sol**. Length of longest pole = Length of the diagonal of the room

= Ö(12^{2}+8^{2}+9^{2}= .Ö(289)= 17 m.

**Ex. 3. Tbe volume of a wall, 5 times as high as it is broad and 8 times as long as**

**it is high, is 12.8 cu. metres. Find the breadth of the wall.**

**Sol.** Let the breadth of the wall be x metres.

Then, Height = 5x metres and Length = 40x metres.

\x * 5x * 40x = 12.8 Û x^{3}=12.8/200 = 128/2000 = 64/1000

So, x = (4/10) m =((4/10)*100)cm = 40 cm

**Ex. 4. Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar?**

**Sol**. Volume of the wall = (2400 x 800 x 60) cu. cm.

Volume of bricks = 90% of the volume of the wall

=((90/100)*2400 *800 * 60)cu.cm.

Volume of 1 brick = (24 x 12 x 8) cu. cm.

\Number of bricks=(90/100)*(2400*800*60)/(24*12*8)=45000.

**Ex. 5. Water flows into a tank 200 m x 160 m througb a rectangular pipe of**

**1.5m x 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2**

** metres?**

**Sol**. Volume required in the tank = (200 x 150 x 2) m^{3} = 60000 m^{3}.

. .

Length of water column flown in1 min =(20*1000)/60 m =1000/3 m

Volume flown per minute = 1.5 * 1.25 * (1000/3) m^{3} = 625 m^{3}.

\ Required time = (60000/625)min = 96min

**Ex. 6. Tbe dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is**

**2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box**.

** Sol**. Volume of the metal used in the box = External Volume – Internal Volume

= [(50 * 40 * 23) – (44 * 34 * 20)]cm^{3}

= 16080 cm^{3}

\ Weight of the metal =((16080*0.5)/1000) kg = 8.04 kg.

**Ex. 7. The diagonal of a cube is 6****Ö****3cm. Find its volume and surface area. **

** **

**Sol. ** Let the edge of the cube be a.

\Ö3a = 6../3 _ a = 6.

So,Volume = a^{3} = (6 x 6 x 6) cm^{3} = 216 cm^{3}.

Surface area = 6a^{2} = (6 x 6 x 6) cm^{2} == 216 cm^{2}.

** **

**Ex. 8. The surface area of a cube is 1734 sq. cm. Find its volume. **

** **

**Sol.** Let the edge of the cube bea. Then,

6a^{2} = 1734 Þ a^{2} = 289 => a = 17 cm.

\ Volume = a^{3} = (17)^{3} cm^{3} = 4913 cm^{3}.

Ex. 9. A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.

** **

** Sol**. Volume of the block = (6 x 12 x 15) cm^{3} = 1080 cm^{3}.

Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm.

Volume of this cube = (3 x 3 x 3) cm^{3} = 27 cm^{3}.

Number of cubes = 1080/27 = 40.

**Ex.l0. A cube of edge 15 cm is immersed completely in a rectangular vessel containing water . If the dimensions of the base of vessel are 20 cm x 15 cm, find the rise in water level. **

** **

**Sol.** Increase in volume = Volume of the cube = (15 x 15 x 15) cm^{3}.

\Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.

**Ex. 11. Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new**

**cube. Find the surface area of the cube so formed.**

** **

**Sol. **Volume of new cube = (1^{3} + 6^{3} + 8^{3}) cm+ = 729 cm^{3}.

Edge of new cube = _{3}Ö729 cm = 9 cm.

\ Surface area of the new cube = (6 x 9 x 9) cm^{2} = 486 cm^{2}.

### Ex. 12. If each edge of a cube is increased by 50%, find the percentage increase in

**Its surface area. “**

** **

**Sol.** Let original length of each edge = a.

Then, original surface area = 6a^{2}.

New edge = (150% of a) = (150a/100) = 3a/2

New surface area = 6x (3a/2)^{2 } = 27a^{2}/2

Increase percent in surface area = ((__15a ^{2}__) x (

__1__) x 100)% = 125%

- 6a
^{2}

^{ }

** **

**Ex. 13. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their**

**surface areas. **

** **

**Sol.** Let their edges be a and b. Then,

a^{3}/b^{3} = 1/27 (or) (a/b)^{3 }= (1/3)^{3 } (or) (a/b) = (1/3).

\Ratio of their surface area = 6a^{2}/6b^{2 }= a^{2}/b^{2 }= (a/b)^{2} = 1/9, i.e. 1:9.

** **

**Ex.14.Find the volume , curved surface area and the total surface area of a cylinder with diameter of base 7 cm and height 40 cm.**

** **

**Sol.** Volume = ∏r^{2} h = ((22/7)x(7/2)x(7/2)x40) = 1540 cm^{3.} .

Curved surface area = 2∏rh = (2x(22/7)x(7/2)x40)= 880 cm^{2} .

Total surface area = 2∏rh + 2∏r^{2} = 2∏r (h + r)

= (2 x (22/7) x (7/2) x (40+3.5)) cm^{2}

^{ }= 957 cm^{2}

^{ }

**Ex.15. If the capacity of a cylindrical tank is 1848 m ^{3} and the diameter of its base**

**is 14 m, then find the depth of the tank.**

** **

**Sol.** Let the depth of the tank be h metres. Then,

∏ x 7^{2} x h = 1848 Þ h = (1848 x (7/22) x (1/49) = 12 m

### Ex.16. 2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm

**diameter. Find the length of the wire in metres.**

** **

**Sol**. Let the length of the wire be h metres. Then,

∏ (0.50/(2 x 100))^{2} x h = 2.2/1000

- h = ( (2.2/1000) x (100 x 100)/(0.25 x 0.25) x (7/22) ) = 112 m.

**Ex. 17. How many iron rods, each of length 7 m and diameter 2 cm can be made**

**out of 0.88 cubic metre of iron? **

** **

**Sol.** Volume of 1 rod = (( 22/7) x (1/100) x (1/100) x 7 ) cu.m = 11/5000 cu.m

Volume of iron = 0.88 cu. m.

Number of rods = (0.88 x 5000/11) = 400.

**Ex. 18. The radii of two cylinders are in the ratio 3: 5 and their heights are in tbe**

**ratio of 2 : 3. Find the ratio of their curved surface areas.**

** **

**Sol**. Let the radii of the cylinders be 3x, 5x and their heights be 2y, 3y respectively. Then

Ratio of their curved surface area = __2∏ X 3x X 2y __ = 2/5 = 2.5

2∏ X 5x X 3y

**Ex. 19. If 1 cubic cm of cast iron weighs 21 gms, then find the eight of a cast iron**

** pipe of length 1 metre with a bore of 3 cm and in which thickness of the metal is 1 em.**

** Sol**. Inner radius = (3/2) cm = 1.5 cm, Outer radius = (1.5 + 1) = 2.5 cm.

\Volume of iron = [∏ x (2.5)^{2} x 100 – ∏ x (1.5)^{2} x 100] cm^{3}

= (22/7) x 100 x [(2.5)^{2} – (1.5)^{2}] cm^{3}

= (8800/7) cm^{3}

Weight of the pipe = ((8800/7) x (21/1000))kg = 26.4 kg.

** **

**Ex. 20. Find the slant height, volume, curved surface area and the whole surface**

**area of a cone of radius 21 cm and height 28 cm.**

** **

**Sol**. Here, r = 21 cm and h = 28 cm.

\ Slant height, l = Ör^{2 } + h^{2 } = Ö (21)^{2 } + (28)^{2 = }Ö 1225 = 35cm

**Ex. 21. Find the length of canvas 1.25 m wide required to build a conical tent of base radius 7 metres and height 24 metres**.

**Sol.** Here, r = 7m and h = 24 m.

So,l = Ö(r^{2} + h^{2}) = Ö(7^{2} + 24^{2}) = Ö (625) = 25 m.

Area of canvas = Õrl=((22/7)*7*25)m^{2} = 550 m^{2} .

Length of canvas = (Area/Width) = (550/1.25) m = 440 m.

### Ex. 22. The heights of two right circular cones are in the ratio 1 : 2 and the

**perimeters of their bases are in the ratio 3 : 4. Find the ratio of their volumes.**

** **

**Sol**. Let the radii of their bases be r and R and their heights be h and 2h respectively.

Then,(2Õr/2ÕR)=(3/4) Þ R=(4/3)r.

\ Ratio of volumes = (((1/3)Õ r^{2}h)/((1/3)Õ(4/3r)^{2}(2h)))=9^{ }: 32.

Ex. 23. The radii of the bases of a cylinder and a cone are in the ratio of 3 : 4 and It heights are in the ratio 2 : 3. Find the ratio of their volumes.

**Sol**. Let the radii of the cylinder and the cone be 3r and 4r and their heights be 2h and

3h respectively.

:. __Volume of cylinder __= __Õ____ x (3r) ^{2} * 2h __= 9/8 = 9 : 8.

Volume of cone (1/3)Õr^{2} * 3h

Ex. 24. A conical vessel, whose internal radius is 12 cm and height 50 cm, is full of liquid. The contents are emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the liquid rises in the cylindrical vessel.

** **

**Sol.** Volume of the liquid in the cylindrical vessel

= Volume of the conical vessel

= ((1/3)* (22/7)* 12 * 12 * 50) )cm^{3 }= (22 *4 *12 * 50)/7 cm^{3}.

Let the height of the liquid in the vessel be h.

#### Then (22/7)*10*10*h =(22*4*12*50)/7 or h = (4*12*50)/100 = 24 cm

**Ex. 25. Find the volume and surface area of a sphere of radius 10.5 cm**.

3 3

**Sol.** Volume = (4/3)Õr^{3 }=(4/3)*(22/7)*(21/2)*(21/2)*(21/2) cm^{3} = 4851 cm^{3}.

Surface area = 4Õr^{ 2} =(4*(22/7)*(21/2)*(21/2)) cm^{2} = 1386 cm^{2}

**Ex. 26. If the radius of a sphere is increased by 50%, find the increase percent in**

**volume and the increase percent in the surface area.**

**Sol**. Let original radius = R. Then, new radius = (150/100)R=(3R/2)

Original volume = (4/3)ÕR^{3}, New volume = (4/3)Õ(3R/2)^{3 }=(9ÕR^{3}/2)

Increase % in volume=((19/6)ÕR^{3})*(3/4ÕR^{3})*100))% = 237.5%

Original surface area =4ÕR^{2}. New surface area = 4Õ(3R/2)^{2}=9ÕR^{2}

Increase % in surface area =(5ÕR^{2}/4ÕR^{2}) * 100) % = 125%.

### Ex. 27. Find the number of lead balls, each 1 cm in diameter that can be

**a sphere of diameter 12 cm.**

** **

**Sol.** Volume of larger sphere = (4/3)Õ*6*6*6) cm^{3} = 288Õ cm^{3}.

Volume of 1 small lead ball = ((4/3)Õ*(1/2)*(1/2)*(1/2)) cm^{3} = Õ/6 cm^{3}.

\ Number of lead balls = (288Õ*(6/Õ)) = 1728.

** **

**Ex.28.How many spherical bullets can be made out of a lead cylinder 28cm high and with radius 6 cm, each bullet being 1.5 cm in diameter ? **

** **

**Sol.** Volume of cylinder = (∏ x 6 x 6 x 28 ) cm^{3} = ( 9∏/16) cm^{3.}

^{ }Number of bullet = __Volume of cylinder__ = [(36 x 28)∏ x 16] /9∏ = 1792.

##### Volume of each bullet

** **

**Ex.29.A copper sphere of diameter 18cm is drawn into a wire of diameter 4 mm Find the length of the wire.**

** **

**Sol. **Volume of sphere = ((4∏/3) x 9 x 9 x 9 ) cm^{3} = 972∏ cm^{3}

Volume of sphere = (∏ x 0.2 x 0.2 x h ) cm^{3}

\ 972∏ = ∏ x (2/10) x (2/10) x h Þ h = (972 x 5 x 5 )cm = [( 972 x 5 x5 )/100 ] m

= 243m

**Ex.30.Two metallic right circular cones having their heights 4.1 cm and 4.3 cm and the radii of their bases 2.1 cm each, have been melted together and recast into a sphere. Find the diameter of the sphere.**

** **

**Sol.** Volume of sphere = Volume of 2 cones

= ( __1__ ∏ x (2.10^{2}) x 4.1 + __1 __∏ x (2.1)^{2} x 4.3)

- 3

Let the radius of sphere be R

\(4/3)∏R^{3} = (1/3)∏(2.1)^{3} or R = 2.1cm

Hence , diameter of the sphere = 4.2.cm

**Ex.31.A Cone and a sphere have equal radii and equal volumes. Find the ratio of the sphere of the diameter of the sphere to the height of the cone.**

** **

**Sol**. Let radius of each be R and height of the cone be H.

Then, (4/3) ∏ R^{3 }= (1/3) ∏ R^{2}H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2

\Required ratio = 1:2.

**Ex.32.Find the volume , curved surface area and the total surface area of a hemisphere of radius 10.5 cm.**

** **

**Sol.** Volume = (2∏r^{3}/3) = ((2/3) x (22/7) x (21/2) x (21/2) x (21/2))cm^{3 }

^{ } = 2425.5 cm^{3}

Curved surface area = 2∏r^{3} = (2 x (22/7) x (21/2) x (21/2))cm^{2}

=693 cm^{2}

Total surface area = 3∏r^{3} = (3 x (22/7) x (21/2) x (21/2))cm^{2}

= 1039.5 cm^{2.}

** **

Ex.33.Hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl ?

**Sol.** Volume of bowl = ((2∏/3) x 9 x 9 x 9 ) cm^{3} = 486∏ cm^{3}.

Volume of 1 bottle = (∏ x (3/2) x (3/2) x 4 ) cm^{3} = 9∏ cm^{3}

^{ }Number of bottles = (486∏/9∏) = 54.

**Ex34.A Cone,a hemisphere and a cylinder stand on equal bases and have the same height.Find ratio of their volumes.**

**Sol.** Let R be the radius of each

Height of the hemisphere = Its radius = R.

\Height of each = R.

Ratio of volumes = (1/3)∏ R^{2 }x R : (2/3)∏R^{3} : ∏ R^{2 }x R = 1:2:3

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