1
ABOUT DISHA PUBLICATION
One of the leading publishers in India, Disha Publication provides books and study materials for
schools and various competitive exams being continuously held across the country. Disha's sole
purpose is to encourage a student to get the best out of preparation. Disha Publication offers an
online bookstore to help students buy exam books online with ease. We, at Disha provide a wide
array of Bank Exam books to help all those aspirants who wish to crack their respective different
levels of bank exams. At Disha Publication, we strive to bring out the best guidebooks that
students would find to be the most useful for Bank Probationary exams.

MENSURATION
Mensuration is the branch of mathematics which deals with the
study of different geometrical shapes, their areas and volumes in
the broadest sense, it is all about the process of measmement.
These are two types of geometrical shapes (1) 2D (2) 3D
Perimeter : Perimeter is sum of all the sides. It is measured in cm,
m, etc.
Area : The area of any figure is the amount of surface enclosed
within its boundary lines. This is measured in square unit like
cm2, m2, etc.
Volume : If an object is solid, then the space occupied by such an
object is called its volume. This is measured in cubic unit like cm3,
m3, etc.
Basic Conversions :
I.
1 km = 10 hm
1 hm = 10 dam
1 dam = 10 m
1 m = 10 dm
1 dm = 10 cm
1 cm = 10 mm
1 m = 100 cm = 1000 mm
1km = 1000 m
II.

III.
IV.

1 km =

where s =

1
2

Also, A = ´ bh ; where b ® base
(a)

a

a

a

Perimeter = 3a
A=

(b)

3 2
a ; where a ® side
4

Right triangle

h

p

1 mile = 1.6 km
1 inch = 2.54 cm
1 mile = 1760 yd = 5280 ft.
1 nautical mile (knot) = 6080 ft
100 kg = 1 quintal
10 quintal = 1 tonne
1 kg = 2.2 pounds (approx.)
l litre = 1000 cc
1 acre = 100 m2
1 hectare = 10000 m2 (100 acre)

b

1
A = pb and h 2 = p2 + b2
2

(Pythagoras triplet)

where p ® perpendicular
b ® base
h ® hypotenuse
Example 1 :
Find the area of a triangle whose sides are 50 m, 78m, 112m
respectively and also find the perpendicular from the
opposite angle on the side 112 m.
Solution :
Here a = 50 m, b = 78 m, c = 112m

PART I : PLANE FIGURES
B

TRIANGLE
a

h

Area (A) = s ( s – a )( s – b )( s – c )

h ® altitude

Equilateral triangle

5
miles
8

Perimeter (P) = a + b + c

a +b+c
and a, b and c are three sides of the triangle.
2

C

b

c

1
(50 + 78 + 112 ) = 120m
2
s – a = 120 – 50 = 70 m
s – b = 120 – 78 = 42 m
s – c = 120 – 112 = 8 m
s=

A

\

Area = 120 ´ 70 ´ 42 ´ 8 = 1680 sq.m.

2

b

1
Q Area = base ´ perpendicular
2
2Area 1680 ´ 2
=
= 30m.
\ Perpendicular =
Base
112
Example 2 :
The base of a triangular field is 880 m and its height 550 m.
Find the area of the field. Also calculate the charges for
supplying water to the field at the rate of ` 24.25 per sq.
hectometre.
Solution :

NOTE :
•
•

•
•

where a ® breadth
b ® base (or length)
h ® altitude

RHOMBUS

d1

a

2

Cost of supplying water to 1 sq. hm = ` 24.25
\ Cost of supplying water to the whole field
= 24.20 × 24.25 = ` 586.85

a

b

Perimeter = 2 (a + b)
Area = b × h;

Area of the field = Base ´ Height
880 ´ 550
=
= 242000 sq.m. = 24.20 sq.hm
2

h

a

d2
Perimeter = 4 a
1
2

Area = d1 ´ d 2

where a ® side and
d1 and d2 are diagonals.

IRREGULAR QUADRILATERAL

(Perimeter)2
= (diagonal) 2 + 2 ´ Area
4
In an isosceles right angled triangle,

p

In a rectangle,

b
4a 2 - b 2
Area =
4
where a is two equal side and b is different side
In a parallelogram,
Area = Diagonal × length of perpendicular on it
If area of circle is decreased by x%, then the radius of

h1

q
d

h2

s

r

Perimeter = p + q + r + s
1
2

Area = ´ d ´ ( h1 + h 2 )
TRAPEZIUM

a

circle is decreased by (100 - 10 100 - x )%

m

h

n

RECTANGLE
Perimeter = a + b + m + n

b

Area =
Perimeter = 2 ( l + b)
Area = l × b; where l ® length
b ® breadth
SQUARE

1
(a + b ) h;
2

b

where a and b are two parallel sides;

m and n are two non-parallel sides;
h ® perpendicular distance
between two parallel sides.
Area of pathways running across the middle of a rectangle

a

b

a

a

Perimeter = 4 × side = 4a
Area = (side)2 = a2;
where a ® side
PARALLELOGRAM

A = a ( l + b) – a2;
PATHWAYS OUTSIDE

where l ® length
b ® breadth,
a ® width of the pathway.

3
D

b + 2a

m
.5c
27

b

A

a

A = (l + 2a) (b + 2a) – lb;

and AB =

where l ® length
b ® breadth
a ® width of the pathway

Also, AO =
\

l – 2a
b – 2a

O

BO =

55
= 27.5m
2

( 36.5) 2 – ( 27.5) 2
1
2

Now, Area of the rhomhus = AC´ BD
=

l

where l ® length
b ® breadth
a ® width of the pathway

Example 3 :
A 5100 sq.cm trapezium has the perpendicular distance
between the two parallel sides 60 m. If one of the parallel
sides be 40m then find the length of the other parallel side.
Solution :
Since, A =

1
(a + b ) h
2

Þ 5100 =

= 24 cm

Hence, the other diagonal BD = 48 cm

a

A = lb – (l – 2a) (b – 2a);

B

36.5cm

146
= 36.5 cm
4

+ 2a

PATHWAYS INSIDE

b

C

1
´ 55 ´ 48 = 1320 sq.cm.
2

Example 6 :
Find the area of a quadrilateral piece of ground, one of whose
diagonals is 60 m long and the perpendicular from the other
two vertices are 38 and 22m respectively.
Solution :
Area = 1 ´ d ´ ( h1 + h 2 )
2
1
= ´ 60 ( 38 + 22 ) = 1800 sq.m.
2

CIRCLE

1
( 40 + x ) ´ 60
2

Þ 170 = 40 + x
\ other parallel side = 170 – 40 = 130 m
Example 4 :
A rectangular grassy plot is 112m by 78 m. It has a gravel
path 2.5 m wide all round it on the inside. Find the area of the
path and the cost of constructing it at ` 2 per square metre?
Solution :
A = lb – (l – 2a) (b – 2a)
= 112 × 78 – (112 – 5) (78 – 5)
= 112 × 78 – 107 × 73 = 8736 – 7811
= 925 sq.m
\ Cost of construction = rate × area
= 2 × 925 = ` 1850
Example 5 :
The perimeter of a rhombus is 146 cm and one of its
diagonals is 55 cm. Find the other diagonal and the area of
the rhombus.
Solution :
Let ABCD be the rhombus in which AC = 55 cm.

r

Perimeter (Circumference) = 2pr = pd
Area = pr2;
where r ® radius
d ® diameter
and p =

22
or 3.14
7

SEMICIRCLE
Perimeter = pr + 2r
1
2

Area = ´ pr 2
r

SECTOR OFA CIRCLE

O
q r
A

l

B
Segment

4
Area of sector OAB =
Length of an arc (l) =

q
´ pr 2
360

q
´ 2 pr
360

\

2p r

2
Area of sector = p r .

Area of segment = Area of sector – Area of triangle OAB
=

q
1
´ pr 2 – r 2 sin q
360°
2

Perimeter of segment = length of the arc + length of segment
AB =

prq
q
+ 2r sin
180
2

q
= 22cm.
360°
q
q
1
= r.2p r
360° 2
360°

1
1
= r.l = ´ 6 ´ 22 sq.cm = 66 sq.cm.
2
2

Example 10 :
The radius of a circular wheel is 1

3
m. How many revolutions
4

will it make in travelling 11 km ?
Solution :
Distance to be travelled = 11km = 11000 m

RING

Radius of the wheel = 1

R2

R1
\

22 7
´ = 11 m
7 4
In travelling 11 m, wheel makes 1 revolution.

\

In travelling 11000 m the wheel makes

\

Area of ring = p

(

R 22

–

R12

)

Example 7 :
A wire is looped in the form of a circle of radius 28 cm. It is
re-bent into a square form. Determine the length of a side of
the square.
Solution :
(a) Length of the wire = Perimeter of the circle
= 2p × 28
= 176 cm2

7
3
m = m
4
4

Circumference of the wheel = 2 ´

1
´ 11000
11

revolutions, i.e., 1000 revolutions.
Example 11 :
The circumference of a circular garden is 1012m. Find the
area of outsider road of 3.5 m width runs around it. Calculate
the area of this road and find the cost of gravelling the road
at ` 32 per 100 sqm.
Solution :
A = pr2 , C = 2pr = 1012

Side of the square = 176 = 44 cm
4

Example 8 :
The radius of a wheel is 42 cm. How many revolutions will it
make in going 26.4 km ?
Solution :
Distance travelled in one revolution = Circumference of the
22
´ 42cm. = 264 cm
7
\ No. of revolutions required to travel 26.4 km

wheel = 2pr = 2 ´

26.4 ´ 1000 ´ 100
= 10000
264
Example 9 :
Find the area of sector of a circle whose radius is 6 cm when–
(a) the angle at the centre is 35°
(b) when the length of arc is 22 cm
Solution :
(a) Area of sector

=

2
= pr .

q
22
35
=
´6´6´
cm 2 = 11 sq.cm.
360° 7
360

(b) Here length of arc l = 22 cm.

1
2

Þ r = 1012 ´ ´

7
= 161m
22

22
´ 161´161 = 81466 sq. m
7
Area of the road = area of bigger circle – area of the garden

\

Area of garden =

Now, radius of bigger circle = 161 + 3.5 =
\ Area of bigger circle =

329
m
2

1
22 329 329
= 85046 sq.m
´
´
2
7
2
2
1
2

1
2

Thus, area of the road = 85046 – 81466 = 3580 sqm.
Hence, cost = Rs.

7161 32
´
= ` 1145.76
2 100

5
Example 12 :
There is an equilateral triangle of which each side is 2m.
With all the three corners as centres, circles each of radius
1 m are described.
(i) Calculate the area common to all the circles and the
triangle.
(ii) Find the area of the remaining portion of the triangle.
Solution :
A

Total surface area of a cube = 6a2
Volume of longest the cube = a3

a

a
Length of longest diagonal = 3a
a
RIGHT PRISM
A prism is a solid which can have any polygon at both its ends.
Lateral or curved surface area = Perimeter of base × height
Total surface area = Lateral surface area + 2 (area of the end)
Volume = Area of base × height

60°
M
B

N

60°
1m

Area of each sector =
=

60°
C

q
60 22
´ pr 2 =
´
´1´1
360
360 7
1 22 11 2
=
´
m
6 7
21

Area of equilateral triangle =

3 2
a
4

RIGHT CIRCULAR CYLINDER
It is a solid which has both its ends in the form of a circle.
Lateral surface area = 2prh
Total surface area = 2pr (r + h)
Volume = pr2h; where r is radius of the base and h is height

3
´ 2 ´ 2 = 3m 2
4
Common area = 3 × Area of each sector
=

(i)

11 11
= = 1.57m 2
21 7
Area of the remaining portion of the triangle = Ar. of
equilateral triangle – 3(Ar. of each sector)

h

= 3´

(ii)

3 - 1.57 = 1.73 – 1.57 = 0.16m2

PART-II : SOLID FIGURE
CUBOID
A cuboid is a three dimensional box.
Total surface area of a cuboid = 2 (lb + bh + lh)
Volume of the cuboid = lbh
Length of diagonal

l 2 + b 2 + h2

r

PYRAMID
A pyramid is a solid which can have any polygon at its base and
its edges converge to single apex.
Lateral or curved surface area
=

1
1
(perimeter of base) × slant height = pl
2
2

Total surface area = lateral surface area + area of the base
Volume =
(i)

1
(area of the base) × height
3

Triangular Pyramid :

h

l

b

Area of four walls = 2(l + b) × h

l

Rectangular Parallelpiped box. It is same as cuboid. Formally a
polyhedron for which all faces are rectangles.
CUBE
A cube is a cuboid which has all its edges equal.
a

6
(i)

Area of the lateral surface of the pyramid
=

3
1
1
× perimeter × slant height = × 3a ×l = al
2
2
2

(ii) Volume =

3 2
1
1
× h× area of base = ×h×
.a
3
3
4
=

(iii) Total Area of the pyramid =
(ii)

ha 2
4 3

1
3 2
3al +
a
2
4

Square Pyramid :
(i)

r

HEMISPHERE
It is a solid half of the sphere.
Lateral surface area = 2pr2
Total surface area = 3pr2
Volume = 2 pr 3 ;
3

1
1
Volume = h × area of base = h × a2
3
3

r
r

where r is radius

FRUSTUM OFACONE
When a cone cut the left over part is called the frustum of the
cone.
Curved surface area = pl (r1 + r2)
Total surface area = pl ( r1 + r2 ) + pr12 + pr22

h

where l = h 2 + ( r1 – r2 ) 2

l

1
3

(

2
2
Volume = ph r1 + r1r2 + r2

a
1
× perimeter × slant height =
2

(ii) Lateral surface area =
1
× 4a × l = 2al
2

(iii) Total area of the pyramid = 2al + a2 = a (2l + a)
RIGHT CIRCULAR CONE
It is a solid which has a circle as its base and a slanting lateral
surface that converges at the apex.
Lateral surface area = prl
Total surface area = pr (l + r)
1
3

Volume = pr 2 h ;

where r : radius of the base
h : height
l : slant height

h

Where r1 and r2 : radii
r2
h : height
Example 13 :
The sum of length, breadth and height of a room is 19 m.
The length of the diagonal is 11 m. Find the cost of painting
the total surface area of the room at the rate of ` 10 per m2.:
Solution :
Let length, breadth and height of the room be l, b and
h, respectively. Then,
l + b + h = 19
...(i)
l 2 + b2 + h 2 = 11
+ h2 = 121
...(ii)
Þ
Area of the surface to be painted
= 2(lb + bh + hl)
(l + b + h)2 = l2 + b2 + h2 + 2 (lb + bh + hl)
Þ 2(lb + bh + hl) = (19)2 – 121 = 361 – 121 = 240
Surface area of the room =240 m2.
Cost of painting the required area = 10 × 240 = ` 2400
Example 14 :
ABCD is a parallelogram. P, Q, R and S are points on sides
AB, BC, CD and DA, respectively such that AP = DR. If the
area of the rectangle ABCD is 16 cm2, then find the area of
the quadrilateral PQRS.

and

l 2 + b2

l

D

R

where r is radius.

C
Q

SPHERE
It is a solid in the form of a ball with radius r.
Lateral surface area = Total surface area = 4pr2
4
3

h

)

r

3
Volume = pr ;

r1

S
A

Solution :

P

Area of the quadrilateral PQRS
= Area of DSPR + Area of DPQR

B

7
=

1
1
´ PR ´ AP + ´ PR ´ PB
2
2

=

1
1
´ PR (AP + PB ) = ´ AD ´ AB
2
2

(Q PR = AD and AP + PB = AB)
1
1
´ Area of rectangle ABCD = ´16 = 8 cm 2
2
2
Example 15 :
A road roller of diameter 1.75 m and length 1 m has to press
a ground of area 1100 sqm. How many revolutions does it
make ?
Solution :
Area covered in one revolution = curved surface area
=

\

Number of revolutions =
=

Total area to be pressed
Curved surface area
1100
=
2prh

1100
22 1.75
2´ ´
´1
7
2

= 200
Example 16 :
The annual rainfall at a place is 43 cm. Find the weight in
metric tonnes of the annual rain falling there on a hectare of
land, taking the weight of water to be 1 metric tonne to the
cubic metre.
Solution :
Area of land = 10000 sqm
Volume of rainfall =

10000 ´ 43
= 4300 m3
100

Weight of water = 4300 × 1 m tonnes = 4300 m tonnes
Example 17 :
The height of a bucket is 45 cm. The radii of the two circular
ends are 28 cm and 7 cm, respectively. Find the volume of
the bucket.
Solution :
Here r1 = 7 cm, r2 = 28 cm and h = 45 cm
7cm

Example 18 :
A hollow cylindrical tube open at both ends is made of iron
2 cm thick. If the external diameter be 50 cm and the length of
the tube be 140 cm, find the number of cubic cm of iron in it.
Solution :
Height = 140 cm
External diameter = 50 cm
\ External radius = 25 cm

A

O

Also, internal radius OA = OB – AB= 25 – 2 = 23 cm
\ Volume of iron = Vexternal – Vinternal
=

(

)

22
´ 140 252 – 232 = 42240 cu. cm.
7

Example 19 :
A cylindrical bath tub of radius 12cm contains water to a
depth of 20 cm. A spherical iron ball is dropped into the tub
and thus the level of water is raised by 6.75 cm. What is the
radius of the ball?
Solution :
Volume of the spherical ball = volume of the water
displaced.
Þ

4 3
pr = p (12)2 × 6.75
3

144 ´ 6.75 ´ 3
= 729
4
or r = 9 cm
Example 20 :
A toy is in the form of a cone mounted on a hemisphere with
the same radius. The diameter of the base of the conical
portion is 6 cm and its height is 4 cm. Determine the surface
area of the toy. (Use p = 3.14).
Solution :

Þ r3 =

The radius of the hemisphere =
45cm

1
´ 6 = 3 cm
2

Now, slant height of cone = 32 + 42 = 5 cm

28cm

5cm

1
2
2
Volume of the bucket = ph(r1 + r2 + r1r2 )
3
Hence, the required volume
1 22
= ´ ´ 45(282 + 7 2 + 28 ´ 7) = 48510 cm3
3 7

B

4cm

3cm

8
The surface area of the toy
= Curved surface of the conical portion
+ Curved surface of the hemisphere
= (p × 3 × 5 + 2p × 32) cm2
= 3.14 × 3 (5 + 6) cm2 = 103.62 cm2.
Example 21 :
A solid is composed of a cylinder with hemispherical ends.
If the whole length of the solid is 104 cm and the radius of
each of the hemispherical ends is 7 cm, find the cost of
polishing its surface at the rate of Re. 1 per dm2.
Solution :

1
\ Volume of the pyramid = ´ 100 ´12 = 400 cu.cm.
3
Example 24 :
Semi-circular lawns are attached to both the edges of a
rectangular field measuring 42 m × 35m. The area of the total
field is :
(a) 3818.5 m2
(b) 8318 m2
2
(c) 5813 m
(d) 1358 m2
Solution :

90cm

7cm

Example 23 :
A right pyramid, 12 cm high, has a square base each side of
which is 10 cm. Find the volume of the pyramid.
Solution :
Area of the base = 10 × 10 = 100 sq.cm.
Height = 12 cm

7cm

Let the height of the cylinder be h cm.
Then h + 7 + 7 = 104
Þ h = 90
Surface area of the solid
= 2 × curved surface area of hemisphere
+ curved surface area of the cylinder
22
22
æ
ö
= ç2´ 2´
´7´7+2´
´ 7 ´ 90÷ cm 2
è
ø
7
7

= 616 + 3960 cm2 = 4576 cm2
Cost of polishing the surface of the solid
4576 ´ 1
=` 45.76
100
Example 22 :
A regular hexagonal prism has perimeter of its base as 600
cm and height equal to 200 cm. How many litres of petrol
can it hold ? Find the weight of petrol if density is 0.8 gm/cc.
Solution :

=`

Side of hexagon =

600
Perimeter
=
= 100 cm
6
Number of sides

3 3
´100 ´100 = 25950 sq.cm.
Area of regular hexagon =
2
Volume = Base area × height
= 25950 × 200 = 5190000 cu.cm. = 5.19 cu.m.
Weight of petrol = Volume × Density
= 5190000 × 0.8 gm/cc
= 4152000 gm = 4152 kg.

(a) Area of the field
= 42 ´ 35 + 2 ´

1 22
1 22
´
´ (21)2 + 2 ´ ´
´ (17.5)2
2 7
2 7

= 1470 + 1386 + 962.5 = 3818.5 m2
Example 25 :
A frustum of a right circular cone has a diameter of base 10
cm, top of 6 cm, and a height of 5 cm; find the area of its
whole surface and volume.
Solution :
Here r1 = 5 cm, r 2 = 3 cm and h = 5 cm.
\

l = h 2 + ( r1 – r2 )
= 52 + ( 5 – 3)

\

2

2

= 29 cm = 5.385 cm

Whole surface of the frustum
= pl ( r1 + r2 ) + pr12 + pr22

=

22
22
22
´ 5.385 ( 5 + 3) + ´ 52 + ´ 32 = 242.25 sq.cm.
7
7
7

Volume =
=

ph 2
(r1 + r1r2 + r22 )
3

22 5 é 2
´ 5 + 5 ´ 3 + 32 ù = 256.67 cu. cm.
û
7 3ë

9
Example 26 :
A cylinder is circumscribed about a hemisphere and a cone
is inscribed in the cylinder so as to have its vertex at the
centre of one end, and the other end as its base. The volume
of the cylinder, hemisphere and the cone are, respectively in
the ratio :
(a) 2 : 3 : 2
(b) 3 : 2 : 1
(c) 3 : 1 : 2
(d) 1 : 2 : 3
Solution :
(b) We have,
radius of the hemisphere = raidus of the cone

= height of the cone
= height of the cylinder = r (say)
Then, ratio of the volumes of cylinder, hemisphere and
cone

2
1
2 1
= pr 3 : pr 3 : pr 3 = 1: : = 3: 2 :1
3
3
3 3

10

EXERCISE
1.

2.

3.

4.

The length and breadth of a rectangle are in the ratio 9 : 5. If
its area is 720 m2, find its perimeter.
(a) 112 metre
(b) 115 metre
(c) 110 metre
(d) 118 metre
A circle and a rectangle have the same perimeter. The sides
of the rectangle are 18 cm and 26 cm. What is the area of the
circle ?
(a) 88 cm2
(b) 154 cm2
2
(c) 1250 cm
(d) 616 cm2
If the perimeter and diagonal of a rectangle are 14 and 5 cms
respectively, find its area.
(a) 12 cm2
(b) 16 cm2
2
(c) 20 cm
(d) 24 cm2
In an isoscele right angled triangle, the perimeter is 20 metre.
Find its area.
(a)

(

)

100 3 - 2 2 m2

(

)

11.

12.

13.

14.

(b) 150 5 - 3 m2

500 m2

5.

6.

(c)
(d) None of these
In a parallelogram, the length of one diagonal and the
perpendicular dropped on that diagonal are 30 and 20 metres
respectively. Find its area.
(a) 600 m2
(b) 540 m2
2
(c) 680 m
(d) 574 m2
The diameter of a garden roller is 1.4 m and it is 2 m long.
æ
How much area will it cover in 5 revolutions ? ç use p =
è

7.

8.

9.

10.

22 ö
÷
7 ø

(a) 40 m2
(b) 44 m2
2
(c) 48 m
(d) 36 m2
A horse is tethered to one corner of a rectangular grassy
field 40 m by 24 m with a rope 14 m long. Over how much
area of the field can it graze?
(a) 154 cm2
(b) 308 m2
(c) 150 m2
(d) None of these
From a square piece of a paper having each side equal to 10
cm, the largest possible circle is being cut out. The ratio of
the area of the circle to the area of the original square is
nearly :
(a)

4
5

(b)

3
5

(c)

5
6

(d)

6
7

15.

16.

17.

18.

(a)

A square carpet with an area 169 m2 must have 2 metres cutoff one of its edges in order to be a perfect fit for a rectangular
room. What is the area of rectangular room?
(a) 180 m2
(b) 164 m2
2
(c) 152 m
(d) 143 m2
A picture 30" × 20" has a frame 2½" wide. The area of the
picture is approximately how many times the area of the
frame?
(a) 4
(c) 2

1
(b) 2
2
(d) 5

A rectangular plot 15 m ×10 m, has a path of grass outside it.
If the area of grassy pathway is 54 m 2, find the width of the
path.
(a) 4 m
(b) 3 m
(c) 2 m
(d) 1 m
If the area of a circle decreases by 36%, then the radius of a
circle decreases by
(a) 20%
(b) 18%
(c) 36%
(d) 64%
The floor of a rectangular room is 15 m long and 12 m wide.
The room is surrounded by a verandah of width 2 m on all
its sides. The area of the verandah is :
(a) 124 m2
(b) 120 m2
(c) 108 m2
(d) 58 m2
A rectangular lawn 70 m × 30 m has two roads each 5 metres
wide, running in the middle of it, one parallel to the length
and the other parallel to the breadth. Find the cost of
gravelling the road at the rate of ` 4 per square metre.
(a) ` 2,000
(b) ` 1,800
(c) ` 1,900
(d) ` 1,700
A cylindrical bucket of height 36 cm and radius 21 cm is
filled with sand. The bucket is emptied on the ground and a
conical heap of sand is formed, the height of the heap being
12 cm. The radius of the heap at the base is :
(a) 63 cm
(b) 53 cm
(c) 56 cm
(d) 66 cm
The altitude drawn to the base of an isosceles triangle is 8
cm and the perimeter is 32 cm. The area of the triangle is
(a) 72 cm2
(b) 60 cm2
(c) 66 cm2
(d) None of these
The cross section of a canal is a trapezium in shape. If the
canal is 7 metres wide at the top and 9 metres at the bottom
and the area of cross-section is 1280 square metres, find the
length of the canal.
(a) 160 metres
(b) 172 metres
(c) 154 metres
(d) None of these
It is required to fix a pipe such that water flowing through it
at a speed of 7 metres per minute fills a tank of capacity 440
cubic metres in 10 minutes. The inner radius of the pipe
should be :

(c)
19.

20.

2 m
1
m
2

(b) 2 m
(d)

1
2

m

The area of a rectangular field is 144 m 2. If the length had
been 6 metres more, the area would have been 54 m 2 more.
The original length of the field is
(a) 22 metres
(b) 18 metres
(c) 16 metres
(d) 24 metres
A rectangular parking space is marked out by painting three
of its sides. If the length of the unpainted side is 9 feet, and
the sum of the lengths of the painted sides is 37 feet, then
what is the area of the parking space in square feet?

21.

22.

23.

24.

25.

26.

27.

28.

29.

(a) 46
(b) 81
(c) 126
(d) 252
A rectangular paper, when folded into two congruent parts
had a perimeter of 34 cm for each part folded along one set
of sides and the same is 38 cm when folded along the other
set of sides. What is the area of the paper?
(a) 140 cm2
(b) 240 cm2
2
(c) 560 cm
(d) None of these
The length and breadth of the floor of the room are 20 feet
and 10 feet respectively. Square tiles of 2 feet length of
different colours are to be laid on the floor. Black tiles are
laid in the first row on all sides. If white tiles are laid in the
one-third of the remaining and blue tiles in the rest, how
many blue tiles will be there?
(a) 16
(b) 24
(c) 32
(d) 48
Four equal circles are described about the four corners of a
square so that each touches two of the others. If a side of
the square is 14 cm, then the area enclosed between the
circumferences of the circles is :
(a) 24 cm2
(b) 42 cm2
2
(c) 154 cm
(d) 196 cm2
The ratio between the length and the breadth of a rectangular
park is 3 : 2. If a man cycling along the boundary of the park
at the speed of 12km / hr completes one round in 8 minutes,
then the area of the park (in sq. m) is:
(a) 15360
(b) 153600
(c) 30720
(d) 307200
The water in a rectangular reservoir having a base 80 metres
by 60 metres is 6.5 metres deep. In what time can the water
be emptied by a pipe whose cross section is a square of
side 20 cm, if the water runs through the pipe at the rate of
15 km per hour?
(a) 52 hrs
(b) 26 hrs
(c) 65 hrs
(d) 42 hrs
The ratio of height of a room to its semi-perimeter is 2 : 5. It
costs ` 260 to paper the walls of the room with paper 50 cm
wide at ` 2 per metre alllowing an area of 15 sq. m for doors
and windows. The height of the room is:
(a) 2.6 m
(b) 3.9 m
(c) 4 m
(d) 4.2 m
Wheels of diameters 7 cm and 14 cm start rolling
simultaneously from X and Y, which are 1980 cm apart,
towards each other in opposite directions. Both of them
make the same number of revolutions per second. If both of
them meet after 10 seconds, the speed of the smaller wheel
is:
(a) 22 cm / sec
(b) 44 cm / sec
(c) 66 cm / sec
(d) 132 cm / sec
A metal cube of edge 12 cm is melted and formed into three
smaller cubes. If the edges of two smaller cubes are 6 cm
and 8 cm, then find the edge of the third smaller cube.
(a) 10 cm
(b) 14 cm
(c) 12 cm
(d) 16 cm
The length, breadth and height of a cuboid are in the ratio
1 : 2 : 3. The length, breadth and height of the cuboid are
increased by 100%, 200% and 200%, respectively. Then, the
increase in the volume of the cuboid will be :
(a) 5 times
(b) 6 times
(c) 12 times
(d) 17 times

30.

The surface area of a cube is 150
diagonal is
(a)

32.

33.

34.

35.

36.

37.

38.

39.

40.

11
The length of its

(b) 5 m

10
m
(d) 15 m
3
A copper sphere of radius 3 cm is beaten and drawn into a
wire of diametre 0.2 cm. The length of the wire is
(a) 9 m
(b) 12 m
(c) 18 m
(d) 36 m
A plot of land in the form of a rectangle has a dimension 240
m × 180 m. A drainlet 10 m wide is dug all around it (outside)
and the earth drug out is evenly spread over the plot,
increasing its surface level by 25 cm. The depth of the drainlet
is
(a) 1.225 m
(b) 1.229 m
(c) 1.227 m
(d) 1.223 m
The water from a roof, 9 sq metres in area, flows down to a
cylinder container of 900 cm2 base. To what height will the
water rise in cylinder if there is a rainfall of 0.1 mm ?
(a) 1 cm
(b) 0.1 metre
(c) 0.11 cm
(d) 10 cms
The length of a cold storage is double its breadth. Its height
is 3 metres. The area of its four walls (including the doors) is
108 m2. Find its volume.
(a) 215 m3
(b) 216 m3
(c) 217 m3
(d) 218 m3
How many spherical bullets can be made out of a lead
cylinder 28 cm high and with base radius 6 cm, each bullet
being 1.5 cm in diameter?
(a) 1845
(b) 1824
(c) 1792
(d) 1752
A rectangular reservoir is 54 m ×44 m × 10 m. An empty pipe
of circular cross-section is of radius 3 cms, and the water
runs through the pipe at 20 m section. Find the time the
empty pipe will take to empty the reservoir full of water.
(a) 116.67 hours
(b) 110 .42 hours
(c) 120.37 hours
(d) 112 hours
A spherical ball of lead, 3 cm in diameter, is melted and recast
into three spherical balls. The diameter of two of these balls
are 1.5 cm and 2 cm respectively. The diameter of the third
ball is
(a) 2.5 cm
(b) 2.66 cm
(c) 3 cm
(d) 3.5 cm
A cube of 384 cm2 surface area is melt to make x number of
small cubes each of 96 mm2 surface area. The value of x is
(a) 80,000
(b) 8
(c) 8,000
(d) 800
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.
(a) 18 cm
(b) 22 cm
(c) 24 cm
(d) None of these
The trunk of a tree is a right cylinder 1.5 m in radius and 10
m high. The volume of the timber which remains when the
trunk is trimmed just enough to reduce it to a rectangular
parallelopiped on a square base is
(a) 44 m3
(b) 46 m3
3
(c) 45 m
(d) 47 m3
(c)

31.

5 3m

m2.

12
41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

The cost of the paint is ` 36.50 per kg. If 1 kg of paint covers
16 square feet, how much will it cost to paint outside of a
cube having 8 feet each side?
(a) ` 692
(b) ` 768
(c) ` 876
(d) ` 972
A right circular cone and a right circular cylinder have equal
base and equal height. If the radius of the base and the
height are in the ratio 5 : 12, then the ratio of the total surface
area of the cylinder to that of the cone is
(a) 3 : 1
(b) 13 : 9
(c) 17 : 9
(d) 34 : 9
A reservoir is supplied from a pipe 6 cm in diameter. How
many pipes of 3 cms diameter would discharge the same
quantity, supposing the velocity of water is same ?
(a) 4
(b) 5
(c) 6
(d) 7
A conical cavity is drilled in a circular cylinder of 15 cm
height and 16 cm base diameter. The height and the base
diameter of the cone are same as those of the cylinder.
Determine the total surface area of the remaining solid.
(a) 440 p cm2
(b) 215p cm2
2
(c) 542 p cm
(d) 376 p cm2
An ice-cream company makes a popular brand of ice-cream
in rectangular shaped bar 6 cm long, 5 cm wide and 2 cm
thick. To cut the cost, the company has decided to reduce
the volume of the bar by 20%, the thickness remaining the
same, but the length and width will be decreased by the
same percentage amount. The new length L will satisfy :
(a) 5.5 < L < 6
(b) 5 < L < 5.5
(c) 4.5 < L < 5
(d) 4 < L < 4.5
Water flows through a cylindrical pipe of internal diameter 7
cm at 2 m per second. If the pipe is always half full, then
what is the volume of water (in litres) discharged in 10
minutes?
(a) 2310
(b) 3850
(c) 4620
(d) 9240
If the radius of a sphere is increased by 2 cm, then its surface
area increases by 352 cm2. The radius of the sphere before
the increase was:
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
A semicircular sheet of paper of diameter 28 cm is bent to
cover the exterior surface of an open conical ice-cream cup.
The depth of the ice-cream cup is
(a) 10.12 cm
(b) 8.12 cm
(c) 12.12 cm
(d) 14.12 cm
The cost of painting the walls of a room at the rate of ` 1.35
per square metre is ` 340.20 and the cost of matting the floor
at the rate of ` 0.85 per m2 is ` 91.80. If the length of the
room is 12 m, then the height of the room is :
(a) 6 m
(b) 12 m
(c) 1.2 m
(d) 13.27 m
A hollow sphere of internal and external diameters 4 cm and
8 cm respectively is melted into a cone of base diamater 8
cm. The height of the cone is:
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm

51.

52.

A cone of height 9 cm with diameter of its base 18 cm is
carved out from a wooden solid sphere of radius 9 cm. The
percentage of the wood wasted is:
(a) 25%
(b) 30%
(c) 50%
(d) 75%
A hemispherical bowl is filled to the brim with a beverage.
The contents of the bowl are transfered into a cylindrical
vessel whose radius is 50% more than its height. If the
diameter is same for both the bowl and the cylinder, the
volume of the beverage in the cylindrical vessel is:
2
1
%
(b) 78 %
3
2
(c) 100%
(d) More than 100%
A cylindrical container of radius 6 cm and height 15 cm is
filled with ice-cream. The whole ice-cream has to be
distributed to 10 children in equal cones with hemispherical
tops. If the height of the conical portion is four times the
radius of its base, then find the radius of the ice-cream cone.
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
A cylinder is filled to 4/5th its volume. It is then filled so that
the level of water coincides with one edge of its bottom and
top edge of the opposite side, In the process, 30 cc of the
water is spilled. What is the volume of the cylinder?
(a) 75 cc
(b) 96 cc
(c) Data insufficient
(d) 100 cc
There are two concentric circular tracks of radii 100 m and
102 m, respectively. A runs on the inner track and goes once
round on the inner track in 1 min 30 sec, while B runs on the
outer track in 1 min 32 sec. Who runs faster?
(a) Both A and B are equal (b) A
(c) B
(d) None of these
A monument has 50 cylindrical pillars each of diameter 50
cm and height 4 m. What will be the labour charges for getting
these pillars cleaned at the rate of 50 paise per sq. m?
(use p = 3.14)
(a) ` 237
(b) ` 157
(c) ` 257
(d) ` 353
Four sheets 50 cm × 5 cm are arranged without overlapping
to form a square having side 55 cm. What is the area of inner
square so formed?
(a) 2500 cm2
(b) 2025 cm2
2
(c) 1600 cm
(d) None of these
A conical vessel of base radius 2 cm and height 3 cm is filled
with kerosene. This liquid leaks through a hole in the bottom
and collects in a cylindrical jar of radius 2 cm. The kerosene
level in the jar is
(a) p cm
(b) 1.5 cm
(c) 1 cm
(d) 3 cm
A garden is 24 m long and 14 m wide. There is a path 1 m
wide outside the garden along its sides. If the path is to be
constructed with square marble tiles 20 cm × 20 cm, the
number of tiles required to cover the path is
(a) 1800
(b) 200
(c) 2000
(d) 2150

(a)
53.

54.

55.

56.

57.

58.

59.

66

60.

61.

62.

63.

64.

65.

66.

2 cm of rain has fallen on a sq. km of land. Assuming that
50% of the raindrops could have been collected and
contained in a pool having a 100 m × 10 m base, by what
level would the water level in the pool have increased?
(a) 15 m
(b) 20 m
(c) 10 m
(d) 25 m
In a swimming pool measuring 90 m by 40 m, 150 men take a
dip. If the average displacement of water by a man is 8 cubic
metres, what will be the rise in water level?
(a) 33.33 cm
(b) 30 cm
(c) 20 cm
(d) 25 cm
A square is inscribed in a circle of radius 8 cm. The area of
the square is
(a) 16 cm2
(b) 64 cm2
(c) 128 cm2
(d) 148 cm2
The biggest possible circle is inscribed in a rectangle of
length 16 cm and breadth 6 cm. Then its area is
(a) 3p cm2
(b) 4p cm2
2
(c) 5p cm
(d) 9p cm2
If the diagonal of a square is doubled, then its area will be
(a) three times
(b) four times
(c) same
(d) none of these
A metal pipe of negligible thickness has radius 21 cm and
length 90 cm. The outer curved surface area of the pipe in
square cm is
(a) 11880
(b) 11680
(c) 11480
(d) 10080
The base of a right pyramid is an equilateral triangle of side
4 cm each. Each slant edge is 5 cm long. The volume of the
pyramid is
(a)

4 8
cm3
3

(b)

4 60
cm3
3

(c)

4 59
cm3
3

(d)

4 61 3
cm
3

67.

68.

69.

70.

71.

13
There are two cones. The curved surface are aof one is
twice that of the other. The slant height of the latter is twice
that of the former. The ratio of their radii is
(a) 4 : 1
(b) 4 : 3
(c) 3 : 4
(d) 1 : 4
A wire is bent into the form of a circle, whose area is 154
cm2. If the same wire is bent into the form of an equilateral
triangle, the approximate area of the equilateral triangle is
(a) 93.14 cm2
(b) 90.14 cm2
2
(c) 83.14 cm
(d) 39.14 cm2
The radius of a right circular cone is 3 cm and its height is 4
cm. The total surface area of the cone is
(a) 48.4 sq.cm
(b) 64.4 sq.cm
(c) 96.4 sq.cm
(d) 75.4 sq.cm
A wooden box of dimension 8 metre × 7 metre × 6 metre is to
carry rectangular boxes of dimensions 8 cm × 7 cm × 6 cm.
The maximum number of boxes that can be carried in 1
wooden box is
(a) 7500000
(b) 9800000
(c) 1200000
(d) 1000000
Two circular cylinders of equal volume have their heights in
the ratio 1 : 2; Ratio of their radii is (Take p =
(a)

72.

22
)
7

1: 4

(b) 1: 2
(c)
(d) 1 : 2
2 :1
A rectangular piece of paper of dimensions 22 cm by 12 cm
is rolled along its length to form a cylinder. The volume
(in cm3) of the cylinder so formed is (use p =

22
)
7

(a) 562
(b) 412
(c) 462
(d) 362
73 . A sphere is placed inside a right circular cylinder so as to
touch the top, base and the lateral surface of the cylinder. If
the radius of the sphere is R, the volume of the cylinder is
(a) 2pR3
(b) 4pR3
(c)

8pR3

(d)

8 3
pR
3

ANSWER KEY
1
2
3
4
5
6
7
8

(a)
(d)
(a)
(a)
(a)
(b)
(a)
(a)

9
10
11
12
13
14
15
16

(d)
(a)
(d)
(a)
(a)
(c)
(a)
(b)

17
18
19
20
21
22
23
24

(a)
(a)
(c)
(c)
(a)
(a)
(b)
(b)

25
26
27
28
29
30
31
32

(a)
(c)
(a)
(a)
(d)
(a)
(d)
(c)

33
34
35
36
37
38
39
40

(a)
(b)
(c)
(a)
(a)
(c)
(c)
(c)

41
42
43
44
45
46
47
48

(c)
(c)
(a)
(a)
(b)
(c)
(d)
(d)

49
50
51
52
53
54
55
56

(a)
(b)
(d)
(c)
(b)
(d)
(b)
(b)

57
58
59
60
61
62
63
64

(b)
(c)
(c)
(c)
(a)
(c)
(d)
(b)

65
66
67
68
69
70
71
72

(a)
(c)
(c)
(b)
(d)
(d)
(c)
(c)

73

(a)

14

HINTS & SOLUTION
1.

2.

(a) Let the length and breadth of a rectangle are 9 xm and
5 xm respectively.
In a rectangle, area = length × breadth
\ 720 = 9x × 5x
or x2 = 16 Þ x = 4
Thus, length = 9 × 4 = 36 m
and breadth = 5 × 4 = 20 m
Therefore, perimeter of rectangle = 2(36 + 20) = 112 m
(d) Perimeter of the circle = perimeter of rectangle
2pr = 2(18 + 26)

Þ
\

22
´ r = 88 Þ r = 14
7
Area of the circle

7.

14 m

D

14 m

40 m

A

B

Area of the shaded portion
=

2´

8.

22
´ 14 ´14 = 616 cm 2 .
7
(a) In a rectangle,

1
´ p(14 )2 = 154 m2
4

(a) Area of the square = (10)2 = 100 cm2
The largest possible circle would be as shown in the
figure below :
R

S

(perimeter) 2
= (diagonal) 2 + 2 ´ area
4

5 cm

(14)2
= 52 + 2 ´ area
Þ
4
49 = 25 + 2 × area

P

10 cm

Q
10 cm

49 - 25 24
\ Area =
=
= 12cm 2
2
2

Area of the circle =
Required ratio =

4.

C

24 m

2
= pr =

3.

(a)

(a)

22
22 ´ 25
2
´ ( 5) =
7
7

22 ´ 25 22 11
=
=
7 ´ 100 28 14

= 0.785 » 0.8 =
Perimeter of triangle = a + a +

2 a = 20m

9.

(d) Side of square carpet = Area = 169 = 13m
After cutting of one side,
Measure of one side = 13 – 2 = 11 m
and other side = 13 m (remain same)
\ Area of rectangular room = 13 × 11 = 143 m2

10.

(a)

a(2 + 2) = 20
a=

20
2+ 2

´

(2 - 2)
( 2 - 2 ) = 10 ( 2 - 2 ) m

Area of triangle =
=

5.
6.

(

2.5''

1
´a´a
2

)

(

1
´ 10 2 - 2 ´ 10 2 - 2
2

2.5''

)

= 50 (4 + 2 – 4 2 )
= 100 (3 – 2 2 )m2
(a) In a parallelogram.
Area = Diagonal × length of perpendicular on it.
= 30 × 20 = 600 m2
(b) Required area covered in 5 revolutions
= 5 × 2prh = 5 × 2 ×

20''

22
× 0.7 × 2 = 44 m2
7

30''

Length of frame = 30 + 2.5 × 2 = 35 inch
Breadth of frame = 20 + 2.5 × 2 = 25 inch
Now, area of picture = 30 × 20 = 600 sq. inch
Area of frame = (35 × 2.5) + (25 × 2.5) = 150

x=

600
= 4times
150

4
5

15
11.

15.

(d)
10 W

(a) Volume of the bucket = volume of the sand emptied
Volume of sand = p (21)2 × 36
Let r be the radius of the conical heap.

15

Then,

Let the width of the path = W m
then, length of plot with path = (15 + 2W) m
and breadth of plot with path = (10 + 2 W) m
Therefore, Area of rectangular plot (wihout path)
= 15 × 10 = 150 m2
and Area of rectangular plot (with path)
= 150 + 54 = 204 m2
Hence, (15 + 2W) × (10 + 2W) = 204
Þ 4W2 + 50 W – 54 = 0
Þ 2W2 + 25 W – 27 = 0
Þ (W – 1) (2W + 27) = 0

or r2 = (21)2 × 9 or r = 21 × 3 = 63
(b) Let ABC be the isosceles triangle and AD be the altitude.
Let AB = AC = x. Then, BC = (32 – 2x).
A

x
B

or

-

(100 - 10 100 - x )% = (100 - 10 100 - 36)%

= 100 - 80 = 20%
(a) Area of the outer rectangle = 19 × 16 = 304 m2

17.

12

7m

9m
Let the length of canal = h m.
Then,

2m

Area of the inner rectangle = 15 × 12 = 180 m2
Required area = (304 – 180) = 124 m2
(c)

(a)

2m

15

14.

1
area of canal = ´ h(9 + 7)
2

5m

or 1280 =

5m

C

æ1
ö
= ç ´12 ´10 ÷ cm 2 = 60 cm 2 .
è2
ø

2m

2m

D

æ1
ö
Hence, required area = ç ´ BC ´ AD ÷
è2
ø

= (100 - 10 64)%
13.

x

Since, in an isosceles triangle, the altitude bisects the
base. So, BD = DC = (16 – x).
In DADC, AC2 = AD2 + DC2
Þ x2 = (8)2 + (16 – x)2
Þ 32x = 320 Þ x = 10.
\ BC = (32 – 2x) = (32 – 20) cm = 12 cm.

27
2
\ with of the path = 1 m
(a) If area of a circle decreased by x % then the radius of a
circle decreases by

Thus W = 1

12.

16.

1 2
pr ´ 12 = p(21)2 ´ 36
3

30 m

1280 ´ 2
= 160 m
16
(a) Let inner radius of the pipe be r.
\h =

18.

Then, 440 = 22 ´ r 2 ´ 7 ´ 10

70 m
Total area of road
= Area of road which parallel to length + Area of road
which parallel to breadth – overlapped road
= 70 × 5 + 30 × 5 – 5 × 5
= 350 + 150 – 25
= 500 – 25 = 475 m2
\ Cost of gravelling the road
= 475 × 4 = ` 1900

1
h(16)
2

7

or

440
=2
22 ´10

r= 2m
(c) Let the length and breadth of the original rectangular
field be x m and y m respectively.
Area of the original field = x × y = 144 m2
or

19.

r2 =

16
\x=

144
y

… (i)

If the length had been 6 m more, then area will be
(x + 6) y = 144 + 54
Þ (x + 6) y = 198
… (ii)
Putting the value of x from eq (i) in eq (ii), we get

21.

26.

Þ 144 + 6y = 198
Þ 6y = 54 Þ y = 9 m
Putting the value of y in eq (i) we get x = 16 m
(c) Clearly, we have : l = 9 and l + 2b = 37 or b = 14.
\ Area = (l × b) = (9 × 14) sq. ft. = 126 sq. ft.
(a) When folded along breadth, we have :

æl
ö
2 ç + b ÷ = 34 or l + 2b = 34
2
è
ø
When folded along length, we have :

22.

=

...(i)

æ bö
....(ii)
2 ç l + ÷ = 38 or 2l + b = 38
è 2ø
Solving (i) and (ii), we get :
l = 14 and b = 10.
\ Area of the paper = (14 × 10) cm2 = 140 cm2.
(a) Area left after laying black tiles
= [(20 – 4) × (10 – 4)] sq. ft. = 96 sq. ft.

Number of blue tiles =

27.

æ 22
ö æ 22
ö
Þ ç ´ 7 ´ x ÷ + ç 2 ´ ´ 7 ´ x ÷ = 198
7
è 7
ø è
ø
Þ 66x = 198 Þ x = 3.
Distance moved by smaller wheel in 3 revolutions
æ 22 7 ö
= ç 2 ´ ´ ´ 3 ÷ cm = 66 cm.
7 2 ø
è

64
= 16.
(2 ´ 2)

7

7
7

29.
7

The shaded area gives the required region.
Area of the shaded region = Area of the square – area
of four quadrants of the circles
= (14)2 – 4´

1
p (7)2
4

30.

22
´ 49 = 196 – 154 = 42 cm 2
7
(b) Perimeter = Distance covered in 8 min.
= 196 –

24.

æ 12000 ö
=ç
´ 8 ÷ m = 1600 m.
è 60
ø
Let length = 3x metres and breadth = 2x metres.

66
cm/s = 22 cm/s.
3
(a) Let the edge of the third cube be x cm.
Then, x3 + 63 + 83 = 123
Þ x3 + 216 + 512 = 1728
Þ x3 = 1000 Þ x = 10.
Thus the edge of third cube = 10 cm.
(d) Let the length, breadth and height of the cuboid be x,
2x and 3x, respectively.
Therefore, volume = x × 2x × 3x = 6x3
New length, breadth and height = 2x, 6x and 9x,
respectively.
New volume = 108x3
Thus, increase in volume = (108 – 6)x3 = 102 x3

\ Speed of smaller wheel =

28.

7

50 ö 2
æ
2
Area of the paper = ç 130 ´
÷ m = 65 m .
100 ø
è
Total area of 4 walls = (65 + 15) m2 = 80 m2.
\ 2(l + b) × h = 80 Þ 2 × 5x × 2x = 80
Þ x2 = 4 Þ x = 2.
\ Height of the room = 4 m.
(a) Let each wheel make x revolutions per sec. Then,
7
éæ
ù
ö
êç 2p´ 2 ´ x ÷ + (2p´ 7 ´ x) ú ´ 10 = 1980
ø
ëè
û

(b)
7

60 ´ 6.5 ´ 80
= 52 hours
600
(c) Let h = 2x metres and (l + b) = 5x metres.

Total cost 260
Length of the paper = Rate per m = 2 m = 130 m.

æ1
ö
Area under white tiles = ç ´ 96 ÷ sq. ft = 32 sq. ft.
è3
ø
Area under blue tiles = (96 – 32) sq. ft = 64 sq. ft.

23.

20 20
´
´ 15000 = 600 cubic metre
100 100

Required time =

æ 144
ö
+ 6 ÷ y = 198
ç
è y
ø

20.

25.

Then, 2 (3x + 2x) = 1600 or x = 160.
\ Length = 480 m and Breadth = 320 m.
\ Area = (480 × 320) m2 = 153600 m2.
(a) Volume of the water running through pipe per hour

Increase in volume 102x 3
=
= 17
Original volume
6x 3
(a) In a cube,
Area = 6 (side)2
or 150 = 6 (side)2
\ side =

25 = 5 m

3 ´ side = 5 3 m
(d) Let the length of the wire be h cm.
and radius of sphere and wire are R and r respectively.
Length of diagonal =

31.

17
Then, volume of sphere = volume of wire (cylinder)
or

4
p R 3 = p r 2h
3

or

4 3 2
R =r h
3

or

4 3
(3) = (0.1)2 h
3

32.

i.e.,

(a) Let radius of the 3rd spherical ball be R,
\

4 ´ (3)3

=

38.

1
4

60 ´ 180 27
=
8800
22
h
=
1.227
m.
\
(a) Let height will be h cm.
Volume of water in roof = Volume of water in cylinder

35.

9 ´ 10000 ´ 0.1
Þ
=h
900 ´ 10
\ h = 1 cm
(b) Let l be the length and b be the breadth of cold
storage.
L = 2B, H = 3 metres
Area of four walls = 2[L × H + B × H] = 108
Þ 6BH = 108 Þ B = 6
\ L = 12, B = 6, H = 3
Volume = 12 × 6 × 3 = 216 m3
(c) Volume of cylinder = (p × 6 × 6 ×28)cm3 = (36 × 28)p cm3.

So, Number of small cube =
=

39.

9p 3
cm .
16
Volume of cylinder
Number of bullets =
Volume of each bullet

36.

16 ù
é
= ê (36 ´ 28)p´ ú = 1792.
9p û
ë
(a) Volume of water in the reservoir
= area of empty pipe × Empty rate × time to empty
2

1 ö
æ
or 54 × 44 × 10 = p ´ ç 3 ´
÷ ´ 20 ´ empty time
è 100 ø
54×44×10×100×100×7
or Empty time =
sec.
22×20×9

54 ´ 44 ´ 10 ´ 100 ´ 100 ´ 7
=
hrs = 116.67 hours.
22 ´ 20 ´ 9 ´ 3600

Volume of bigger cube
Volume of smaller cube

(8)3
(0.4)

3

=

512
= 8,000
0.064

(c) Volume of the liquid in the cylindrical vessel
= Volume of the conical vessel

æ 1 22
ö 3
= ç ´ ´12 ´12 ´ 50 ÷ cm
è3 7
ø
æ 22 ´ 4 ´ 12 ´ 50 ö 3
=ç
÷ cm .
7
è
ø
Let the height of the liquid in the vessel be h.
Then,

3 3 3ö 3
æ4
Volume of each bullet = ç p´ ´ ´ ÷ cm
3
4 4 4ø
è
=

3

27 27
125 æ 5 ö
5
-1 =
= ç ÷ Þ R = = 1.25
8 64
64 è 4 ø
4

Diameter of the third spherical ball
= 1.25 × 2 = 2.5 cm.
(c) Let 'A' be the side of bigger cube and 'a' be the side of
smaller cube
Surface area of bigger cube = 6 A2
or 384 = 6A2
\ A = 8 cm.
Surface area of smaller cube = 6 a 2
96 = 6a2
\ a = 4 mm = 0.4 cm

Þ h=

34.

3

\

1
m.
4

\ 8800 h = 240 × 180 ×

33.

3

4 æ3ö
4 æ3ö
4
4
p ç ÷ = p ç ÷ + p(1)3 + pR 3
3 è2ø
3 è4ø
3
3

3
é 3 3
3 ù
Þ R 3 = êæç ö÷ - æç ö÷ ú - 13
êëè 2 ø è 4 ø úû

108
=
= 3600 cm = 36 m
2
0.03
3 ´ (0.1)
(c) Let the depth of the drainlet be h metres.
Volume of the earth dug from the drainlet 10 m wide
= h[260 × 200 – 240 × 180] = 8800 h cu. m.
Now this is spread over the plot raising its height by
25 cm,

\ h=

37.

22
22 ´ 4 ´12 ´ 50
´ 10 ´ 10 ´ h =
7
7

æ 4 ´ 12 ´ 50 ö
or h = ç
÷ = 24 cm.
è 10 ´10 ø
40.

(c)

B
1.5 m
A

1.5 m

From DAOB ,

O

1.5 m

C

D

AB = 1.52 + 1.52 = 2.25 + 2.25 = 4.50
\ Area of the square base of the trunk of the tree
= 4.50 ´ 4.50 = 4.50 m2
\ Volume of the timber = Area of base × height
= 4.50 × 10 = 45 m3

18
41.

(c) Surface area of the cube = (6 ×82) sq. ft. = 384 sq. ft.

42.

æ 384 ö
Quantity of paint required = ç
÷ kg = 24 kg.
è 16 ø
\ Cost of painting = ` (36.50 × 24) = ` 876.
(c) Let the radius of the base are 5k and 12k respectively
Total surface area of the cylinder
Total surface area of the cone

\

=

49.

2pr ´ h + 2pr

340.20
= 252m 2
1.35
Þ (l + b)h = 126
...(i)
= 2 ( l + b) h =

2

pr r 2 + h 2 + pr 2
2h + 2r

=

r2 + h 2 + r

+

91.8
= 108
0.85
12 × b = 108
(Q l = 12 m)
Þ b= 9m

And l ´ b =

24k + 10k
25k 2 + 144k 2 + 5k

34k
34k
17
=
=
13k + 5k 18k
9
(a) Number of discharge pipe

=

43.

Volume of water supply pipe
=
Volume of water in each discharge pipe

44.

45.

46.

47.

50.

Then,

14
cm = 4.45 cm
p

}

1
æ4
ö
p´ 4 ´ 4 ´ h = ç p´ 56 ÷
3
3
è
ø

æ 4 ´ 56 ö
Þ h =ç
÷ = 14 cm.
è 4´ 4 ø
51.

æ4
ö 3
(d) Volume of sphere = ç p´ 9 ´ 9 ´ 9 ÷ cm .
è3
ø
æ1
ö 3
Volume of cone = ç p´ 9 ´ 9 ´ 9 ÷ cm .
è3
ø
Volume of wood wasted
éæ 4
ö æ1
öù 3
= êç p ´ 9 ´ 9 ´ 9 ÷ - ç p ´ 9 ´ 9 ´ 9 ÷ ú cm .
ø è3
øû
ëè 3
3
= (p × 9 × 9 × 9) cm

æ
ç
\ Required percentage = ç
çç
è

æ 7700 ´ 60 ´10 ö
=ç
÷ litres
1000
è
ø
= 4620 litres.
(d) 4p (r + 2)2 – 4pr2 = 352

r=

(b) Volume of material in the sphere

{

æ 22 7 7
ö 3
(c) Volume of one coin = ç ´ ´ ´ 200 ÷ cm =7700 cm3.
è 7 2 2
ø
Volume of water flown in 10 min. = (7700 × 60 ×10) cm3

28
Þ 2r + 2 = 14 Þ r = 6 cm
Þ 2r + 2 =
2
(d) Circumference of the base of ice-cream cup
= Diameter of the sheet = 28 cm
2pr = 28

126
=6m
21

é4
ö 3
3
3 ù
3 æ4
= ê p´ (4) - (2) ú cm = ç p´ 56 ÷ cm .
3
3
ë
û
è
ø
Let the height of the cone be h cm.

p´ (3) ´1

7 1ö
æ
Þ (r + 2)2 – r2 = ç 352 ´ ´ ÷ = 28.
22
4ø
è
Þ (r + 2 + r)(r + 2 – r) = 28

48.

Using (i), we get, h =

2

= 4 [Since the velocity of water is same]
2
æ3ö
p´ ç ÷ ´1
è2ø
(a) Total surface area of the remaining solid = Curved
surface area of the cylinder + Area of the base + Curved
surface area of the cone
= 2prh + pr2 + pr l
= 2p × 8 × 15 + p × (8)2 + p × 8 × 17
= 240p + 64p + 136p
= 440 p cm2
(b) L × B × 2 = 48
Þ L × B = 24
Now, 6 – 6 × 10% = 5.4,
5 – 5 × 10% = 4.5 and
Therefore, 5.4 × 4.5 = 24.3
Clearly, 5 < L < 5.5
=

Slant height of cone = radius of the sheet = 14 cm
\ 142 = (4.45)2 + h2
or h2 = 196 – 19.80 = 176.20
\ h = 13.27 cm
(a) Let length, breadth and height of the room be l , b and
h, respectively.
Then, area of four walls of the room

52.

ö
÷
´ 100 ÷ %
4
÷÷
´ p´ 9 ´ 9 ´ 9
3
ø
p´ 9 ´ 9 ´ 9

æ3
ö
= ç ´100 ÷ % = 75%.
è4
ø
(c) Let the height of the vessel be x.
Then, radius of the bowl = radius of the vessel = x/2.
3

2 æxö
1 3
Volume of the bowl, V1 = p ç ÷ = px .
3 è 2 ø 12
2

1 3
æxö
Volume of the vessel, V2 = p ç ÷ x = px .
4
è 2ø
Since V2 > V1, so the vessel can contain 100% of the
beverage filled in the bowl.

19
53.

(b) Volume of the cylinder container
= p × 62 × 15 cu. cm
…(1)
Let the radius of the base of the cone be r cm,
then, height of the cone = 4r cm
\ Volume of the 10 cylindrical cones of ice-cream
with hemispherical tops

Now, labour charges at the rate of 50 paise
per sq. m = 314 × 0.5 = 157.0
º ` 157

57.

(b)

58.

Side of the inner square = 55 – 10 = 45
\ Area of inner square = 45 × 45 = 2025 sq. m.
(c) Let the kerosene level of cylindrical jar be h.

2
é1
ù
= 10 ´ ê ´ p ´ r 2 ´ 4 r ú + 10 ´ pr 3
3
ë3
û

40 3 20 3
pr +
pr = 20 pr3 cu. cm …(2)
3
3
Since the whole ice-cream in the cylindrical container
is distributed among 10 children in cones with
hemispherical tops,
\ (1) and (2), gives
Þ p × 62 × 15 = 20pr3
=

1 2
pr h
3
Since, radius (r) = 2 cm and height(h) = 3cm of conical
vessel.

Now, Volume of conical vessel =

36 ´ 15
= 27 Þ r = 3 cm
20
(d) Let the original volume of cylinder be V .

Þ r3 =

54.

4
4
, then it’s volume = V
5
5
When cylinder is filled, the level of water coincides
with opposite sides of bottom and top edges then

When it is filled

1
V
2
Since, in this process 30 cc of the water is spilled,
therefore

Volume become =

59.

4
1
V , 30 < V
5
2
4
1
V , V < 30
5
2
V
(3/10)
= 30
Þ
Þ V = 100 cc
(b) Radius of the inner track = 100 m
and time = 1 min 30 sec º 90 sec.
Also, Radius of the outer track = 102 m
and time = 1 min 32 sec º 92 sec.
Now, speed of A who runs on the inner track

Þ

55.

2 p(100) 20 p
=
= 6.98 m/s
90
9
And speed of B who runs on the outer track

=

2p(102) 51p
=
= 6.96 m/s
90
23
Since, speed of A > speed of B
\ A runs faster than B.
(b) Curved surface area of cylinder = 2prh
\ Surface area of 50 cylindrical pillars = 50 × 2prh
Now, Diameter of each cylindrical pillar = 50 cm

1
p ´ 4 ´ 3 = 4p
3
Now, Volume of cylinderical jar = pr2h
= p (2)2h
= 4ph
Now, Volume of conical vessel = Volume of cylindrical
Jar
Þ 4 p = 4 ph
h = 1cm
Hence, kerosene level in Jar is 1 cm.
(c) Given, length of garden = 24 m and
breadth of garden = 14 m
\ Area of the garden = 24 × 14 m2 = 336 m2.
Since, there is 1 m wide path outside the garden
\ Area of Garden (including path)
= ( 24 + 2) × ( 14 + 2) = 26 × 16 m2 = 416 m2.
Now, Area of Path = Area of garden( inculding path)
– Area of Garden
= 416 – 336 = 80 m2.
Now , Area of Marbles = 20 × 20 = 400 cm 2

\ Volume =

\

80, 0000
= 2000
400
(c) Volume of rain that is to be collected

=

60.

10
in a pool = 2 ´ 1 ´ 10 ´

50
= 25 cm = 0.25 m
2
Also, height = 4m
\ Surface area = 50 × 2 × 3.14 × 0.25 × 4
= 314 × 1 sq m.
= 314 sq. m.

\ Radius =

1
2

= 1010 cm3 = 10 4 meter3
Volume of Pool = L × B × h
104 = 100 × 10 × h

=

56.

Area of Path
Marbles required = Area of Marbles

104
= 10 m .
100 ´ 10
(a) Let the rise in water level = x m
Now, volume of pool = 40 × 90 × x = 3600 x
When 150 men take a dip, then displacement of
water = 8m3

h=

61.

3600 x
900
=8 Þ
x = 2 Þ x = 0.33m
150
150
Þ x = 33.33 cm

\

20
68.
62. (c)

D

(b)

C
O

154
´ 7 = 49
22
r = 7 cm
length of wire = circumference of circle

m
8c

r2 =

m
8c

A

B

22
´ 7 = 44 cm
7
Now, Perimeter of equilateral triangle = 44 cm

= 2´

Diagonal of square = Diameter of circle
2 ´ side of square = 16 cm
Squaring on both sides

(

side =

2 ´ sides of square ) = 16 2

63. (d)

16 ´16
2
Þ Area of square = 128 sq. cm
The area of circle is 9p cm2.

64. (b)

Diagonal of a square (d) =
d=

2a Þ a =

Area of equilateral triangle =

69.

(d)

2
2

2
æ d2 ö
æ 2d ö
=
4
Area of square = ç
ç
÷
÷
ç 2 ÷
è 2ø
è
ø

22
´ 21´ 90 = 11880 sq.cm
7
C1 = 2C2
pr1l1 = 2pr2l2
also, l2 = 2l1
pr1l1 = 2 × 26 pr2l1

(a)

r1 4
=
r2 1

(

)

32 + 4 2 + 3

22
528
´3´8 =
7
7
S = 75.4 sq.cm

70.

(d)

Maximum number of boxes =

(c)

pr12 h1

=

r1
=
r2

h2
=
h1

800 ´ 700 ´ 600 cm3
8 ´ 7 ´ 6 cm3

= 1000000
71.

d2
is area of square
2
Therefore, Area will be four temes.
(a) Curved surface area of cylinder = 2prh
= 2´

22
´3´
7

=

(2d )
2

2

484 3
= 91.42 cm2
9
Area of equilateral triangle is nearly equal to
90.14 cm2
Hence, option (b) is correct.
Total surface area of cone = p r(l + r)
S=

d
2
Now, diagonal gests doubled
a=

3 æ 44 ö
´ç ÷
4 è 3ø

=

2 × side of square (a).

d

Area of square Þ a2 =

67.

44
cm
3

2

Þ (side of square)2 =

65.

Let r be the radius of circle.
pr2 = 154 cm2

72.

(c)

pr22 h 2
2
1

r1 : r2 = 2 :1
2pr = 22 cm
22 ´ 7 7
= cm
2 ´ 22 2
Height, h = 12 cm
r=

22 7 7
´ ´ ´ 12 = 462 cm3
7 2 2
Radius of cylinder = Radius of sphere = R
Height of cylinder = 2 R
Volume of cylinder = pR2 × (2R) = 2pR3

Volume of cylinder =

73.

(a)