**IBPS CWE NUMERICAL ABILITY SOLVED EXAMPLES**

**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)] cm2 = (2 x 434) cm2 = 868 cm2.

**2. Find the length of the longest pole that can be placed in a room 12 m long 8m broad and 9m high.**

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

17 m.

**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 x3=12.8/200 = 128/2000 = 64/1000

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

**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.

**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) m3 = 60000 m3.

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

Volume flown per minute = 1.5 * 1.25 * (1000/3) m3 = 625 m3.

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

**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)]cm3

= 16080 cm3

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

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

Sol. Let the edge of the cube be a.

3a = 6../3 _ a = 6.

So,Volume = a3 = (6 x 6 x 6) cm3 = 216 cm3.

Surface area = 6a2 = (6 x 6 x 6) cm2 == 216 cm2.

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

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

6a2 = 1734 a2 = 289 => a = 17 cm.

Volume = a3 = (17)3 cm3 = 4913 cm3.

**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) cm3 = 1080 cm3.

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) cm3 = 27 cm3.

Number of cubes = 1080/27 = 40.

**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) cm3.

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

**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 = (13 + 63 + 83) cm+ = 729 cm3.

Edge of new cube = 3 root 729 cm = 9 cm.

Surface area of the new cube = (6 x 9 x 9) cm2 = 486 cm2.

1

**2. 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 = 6a2.

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

New surface area = 6x (3a/2)2 = 27a2/2

Increase percent in surface area = 125%

**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,

a3/b3 = 1/27 (or) (a/b)3 = (1/3)3 (or) (a/b) = (1/3).

Ratio of their surface area = 6a2/6b2 = a2/b2 = (a/b)2 = 1/9, i.e. 1:9.

**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 = Î r2 h = ((22/7)x(7/2)x(7/2)x40) = 1540 cm3. .

Curved surface area = 2Î rh = (2x(22/7)x(7/2)x40)= 880 cm2 .

Total surface area = 2Î rh + 2Î r2 = 2Î r (h + r)

= (2 x (22/7) x (7/2) x (40+3.5)) cm2

= 957 cm2

**15. If the capacity of a cylindrical tank is 1848 m3 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,

Î x 72 x h = 1848 h = (1848 x (7/22) x (1/49) = 12 m

**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,

Î (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.

**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.

**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 = 2Î X 3x X 2y = 2/5 = 2.5

2Î X 5x X 3y

**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 = [Î x (2.5)2 x 100 - Î x (1.5)2 x 100] cm3

= (22/7) x 100 x [(2.5)2 - (1.5)2] cm3

= (8800/7) cm3

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

**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 = r2 + h2 = (21)2 + (28)2 = 1225 = 35cm

**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 = (r2 + h2) = (72 + 242) = (625) = 25 m.

Area of canvas = rl=((22/7)*7*25)m2 = 550 m2 .

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

**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,(2r/2R)=(3/4) R=(4/3)r.

Ratio of volumes = (((1/3) r2h)/((1/3)(4/3r)2(2h)))=9 : 32.

**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 = x (3r)2 * 2h = 9/8 = 9 : 8.

Volume of cone (1/3)r2 * 3h

**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) )cm3 = (22 *4 *12 * 50)/7 cm3.

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

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

Sol. Volume = (4/3)r3 =(4/3)*(22/7)*(21/2)*(21/2)*(21/2) cm3 = 4851 cm3.

Surface area = 4r 2 =(4*(22/7)*(21/2)*(21/2)) cm2 = 1386 cm2

**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)R3, New volume = (4/3)(3R/2)3 =(9R3/2)

Increase % in volume=((19/6)R3)*(3/4R3)*100))% = 237.5%

Original surface area =4R2. New surface area = 4(3R/2)2=9R2

Increase % in surface area =(5R2/4R2) * 100) % = 125%.

**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)*6*6*6) cm3 = 288 cm3.

Volume of 1 small lead ball = ((4/3)*(1/2)*(1/2)*(1/2)) cm3 = /6 cm3.

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

**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 = (Î x 6 x 6 x 28 ) cm3 = ( 9Î /16) cm3.

Number of bullet = Volume of cylinder/ Volume of each bullet = [(36 x 28)Î x 16] /9Î = 1792.

**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 = ((4Î /3) x 9 x 9 x 9 ) cm3 = 972Î cm3

Volume of sphere = (Î x 0.2 x 0.2 x h ) cm3

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

= 243m

**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

= ( 1/3 Î x (2.102) x 4.1 + 1/3 Î x (2.1)2 x 4.3)

Let the radius of sphere be R

(4/3)Î R3 = (1/3)Î (2.1)3 or R = 2.1cm

Hence , diameter of the sphere = 4.2.cm

**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) Î R3 = (1/3) Î R2H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2

Required ratio = 1:2.

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

Sol. Volume = (2Î r3/3) = ((2/3) x (22/7) x (21/2) x (21/2) x (21/2))cm3

= 2425.5 cm3

Curved surface area = 2Î r3 = (2 x (22/7) x (21/2) x (21/2))cm2

=693 cm2

Total surface area = 3Î r3 = (3 x (22/7) x (21/2) x (21/2))cm2

= 1039.5 cm2.

**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 = ((2Î /3) x 9 x 9 x 9 ) cm3 = 486Î cm3.

Volume of 1 bottle = (Î x (3/2) x (3/2) x 4 ) cm3 = 9Î cm3

Number of bottles = (486Î /9Î ) = 54.

**34.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)Î R2 x R : (2/3)Î R3 : Î R2 x R = 1:2:3

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