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# Mensuration Tricks & Tips

Mensuration is a pure formula-based topic and tricks/shortcuts are seldom applied here. So in this series I will try to solve all the mensuration problems that have appeared in SSC & Banking Exams lately and in the process I will share the important concepts/formulas.

### Quick Method to Solve Mensuration Problems

This quick trick to solve Mensuration problems is totally based on your thinking. It involves next to no calculations. This trick will save you precious time in your IBPS Clerk or SBI PO exams.

The trick will help you get answers in just a few seconds. Solve the questions in detail first to understand this method more clearly. Once you get acquainted with the mechanisms of this trick, you will find it extremely helpful.

Below are given types of questions where this trick can be applied.   ## How to Use this Trick

Check whether π=22/7 has been used in the formula for finding out the Particular Area, Curved Surface Area, Total Area, Volume, etc. If it is so, then

Here are examples to explain the chart given above.

Example 1:

Find the surface area of a sphere whose volume is 4851 cubic meters.

a) 1380 m2
b) 1360 m2
c) 1368 m2
d) 1386 m2

Using the Trick:

We know that surface area of a sphere = 4πr2

It means ‘π’ has been used in finding out the surface area of the sphere.

We can easily see that only ‘1386’ from the given options is divisible by ‘11’

Hence, surface area of the sphere = 1386 m2

Example 2:

The radius and height of a right circular cylinder are 14 cm & 21 cm respectively. Find its volume.

a) 12836 cm3
b) 12736 cm3
c) 12936 cm3
d) 12837 cm3

Using the Trick:

The know that volume of Cylinder = πr2h

We must check divisibility by ‘11’. Here, both ‘12936’ and ‘12837’ are divisible by 11. But you also notice that radius (14 cm) & height (21 cm) are both multiples of 7. So the option divisible by ‘7’ is your answer.

Hence, volume of a right circular cylinder = 12936 cm3 (since this is the only option divisible by 7)

We can test this as follows:

Volume of the given cylinder

= (22/7) × 14 × 14 × 14 × 21 cm3

= 22 × 14 × 14 × 3 cm3

⇒ Volume must be divisible by ‘7’.

Example 3:

The radius and height of a right circular cone are 7 cm & 18 cm respectively. Find its volume.

a) 814 cm3
b) 624 cm3
c) 825 cm3
d) 924 cm3

Using the Trick:

The option should be divisible by ‘11’ because ‘π’ has been used in finding its volume. One of the parameters is a multiple of 7 without being a higher power. So we must go through fundamentals.

Now, volume of a right circular cone = (1/3)πr2h

= (1/3) × (22/7) × 7 × 7 × 18 cm3

= 22 × 7 × 6

Clearly, we need an answer that is a multiple of 11, 7 as well as 3.

Among the given options, 814, 825 and 924 are all multiples of 11. However, we see that only one option is divisible by 7. So this is the correct answer.

Hence, volume of the given cone = 924 cm3

Example 4:

Find the circumference of a circle whose radius is 49 cm.

a) 208 cm
b) 288 cm
c) 308 cm
d) 407 cm

Using the Trick:

The option should be divisible by ‘11’ because ‘π’ has been used in finding its circumference. One of the parameters is a higher power of 7. Thus, we need to find the only option that is a multiple of 7. If, however, we find more than one option that is a multiple of 7, we need to go through fundamentals.

Among the options, 308 and 407 are both multiples of 11. However, only 308 is a multiple of 7. So circumference of the circle = 308 cm.

We can test this as follows:

Circumference of circle = 2πr cm

= 2 × (22/7) × 7 × 49 cm

= 2 × 22 × 7 cm

Remember: If there is only one parameter equal to ‘7’ or multiple of ‘7’ and this parameter is not in a higher power in the formula, the answer will not be divisible by ‘7’

Example 5:

Find the curved surface area of a right circular cylinder whose radius & height are 14 cm & 50 cm respectively.

a) 3300 cm2
b) 3420 cm2
c) 4440 cm2
d) 4400 cm2

Solution:

Curved surface area of a right circular cylinder = 2πrh

Here only one parameter (r = 14 cm) is a multiple of 7 (without being a higher power of 7). This parameter is used only as ‘r’ and not in its higher powers. So we see that our answer will not be a multiple of 7. However, the presence of π means that it will still be a multiple of 11.

Among the given options, 3300 and 4400 are both multiples of 11. We also see that both are not multiples of 7 either. However, we can see that none of our parameters are multiples of 3, so curved surface area cannot be a multiple of 3 either. So our answer cannot be 3300.

Therefore curved surface area of right circular cylinder must be 4400 cm2.

We can test this as follows:

Curved surface area of a right circular cylinder = 2πrh

=2 × (22/7) × 14 × 50

=2 × 22 × 2 × 50

= 4400 cm3

NOTE:

1. The whole trick is based on the multiplication & divisibility by 11 & 7. So you are advised to use this trick very carefully using your quick mental mathematics. Please learn divisibility rules for both 11 and 7 for this purpose – they are both pretty simple! Please do not waste time solving these types of questions.
2. If there is a ‘None of these’ among the given options, then don’t use this trick.

## Mensuration (2D) Formulas Tips And Tricks

It is very important to have an understanding of Different Formulas of quadrilaterals and circle for you to comfortably attempt Advanced Maths questions which covers a major portion of Quant Section of Competitive Exams. Here we are providing you formulas and shortcuts on how to solve mensuration questions.

## Important Formulas on Quadrilateral and Circle

Rectangle

A four-sided shape that is made up of two pairs of parallel lines and that has four right angles; especially: a shape in which one pair of lines is longer than the other pair. The diagonals of a rectangle bisect each other and are equal.

Area of rectangle = length x breadth = l x b

OR Area of rectangle = if one sides (l) and diagonal (d) are given.

OR Area of rectangle = if perimeter (P) and diagonal (d) are given.

Perimeter (P) of rectangle = 2 (length + breadth) = 2 (l + b).

OR Perimeter of rectangle = if one sides (l) and diagonal (d) are given.

Square

A four-sided shape that is made up of four straight sides that are the same length and that has four right angles. The diagonals of a square are equal and bisect each other at 900.

(a) Area (a) of a square Perimeter (P) of a square

= 4a, i.e. 4 x side Length (d) of the diagonal of a square Circle

A circle is the path traveled by a point which moves in such a way that its distance from a fixed point remains constant. The fixed point is known as center and the fixed distance is called the radius.

(a) Circumference or perimeter of circle = where r is radius and d is diameter of circle

(b) Area of circle is radius is diameter is circumference circumference x radius

(c) Radius of circle =  Sector :

A sector is  a figure enclosed by two radii and an arc lying between them. here AOB is a sector

length of arc AB= 2πrΘ/360°

Ring or Circular Path: area=π(R2-r2)

Perimeter=2π(R+r)

Rhombus

Rhombus is a quadrilateral whose all sides are equal. The diagonals of a rhombus bisect each other at 900

Area (a) of a rhombus

= a * h, i.e. base * height Product of its diagonals since d2 since d2 Perimeter (P) of a rhombus

= 4a,  i.e. 4 x side Where d1 and d2 are two-diagonals.

Side (a) of a rhombus Parallelogram

A quadrilateral in which opposite sides are equal and parallel is called a parallelogram. The diagonals of a parallelogram bisect each other.

Area (a) of a parallelogram = base × altitude corresponding to the base = b × h Area (a) of parallelogram where a and b are adjacent sides, d is the length of the diagonal connecting the ends of the two sides and  In a parallelogram, the sum of the squares of the diagonals = 2

(the sum of the squares of the two adjacent sides).

i.e., Perimeter (P) of a parallelogram

= 2  (a+b),

Where a and b are adjacent sides of the parallelogram.

Trapezium (Trapezoid)

A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term ‘trapezium,’ from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning ‘little table.’ Area (a) of a trapezium

1/2 x (sum of parallel sides) x perpendicular

Distance between the parallel sides

i.e.,  Where,  l = b – a if b > a = a – b if a > b

And Height (h) of the trapezium Pathways Running across the middle of a rectangle: X is  the width of the path

Area of path= (l+b-x)x

perimeter=  2(l+b-2x)

Outer Pathways: Area=(l+b+2x)2x

Perimeter=4(l+b+2x)

Inner Pathways:

Area=(l+b-2x)2x

Perimeter=4(l+b-2x)

Some useful Short trick:

• If there is a change of X% in defining dimensions of the 2-d figure then its perimeter will also changes by X%
• If all the sides of a quadrilateral is changed by  X% then its diagonal will also changes by X%.
• The area of the largest triangle that can be inscribed in a semi circle of radius r is r2.
• The number of revolution made by a circular wheel of radius r in travelling distance d is given by

number of revolution =d/2πr

• If the length and breadth of rectangle are increased by x% and y% then the area of the rectangle will increased by.

(x+y+xy/100)%

• If the length and breadth of a rectangle is decreased by by x% and y% respectively then the area of the rectangle will  decrease by:

(x+y-xy/100)%

• If the length of a rectangle is increased by x%, then its breadth will have to be decreased by (100x/100+x)% in order to maintain the same area of the rectangle.
• If each of the defining dimensions or sides of any 2-D figure is changed by x% its area changes by

x(2+x/100)%

where x=positive if increase and negative if decreases.

Mensuration (3D) Formulas Tips And Tricks

It is very important to have an understanding of Different Formulas of quadrilaterals and circle for you to comfortably attempt Advanced Maths questions which covers a major portion of Quant Section of Competitive Exams. Here we are providing you formulas and shortcuts on how to solve mensuration questions.

## Important Mensuration (3D) Formulas

Cube • s = side
• Volume: V = s^3
• Lateral surface area = 4a2
• Surface Area: S = 6s^2
• Diagonal (d) = s√3

Cuboid • Volume of cuboid: length x breadth x width
• Total surface area = 2 ( lb + bh + hl)

Right  Circular  Cylinder • Volume of Cylinder = π r^2 h
• Lateral Surface Area (LSA or CSA) = 2π r h
• Total Surface Area = TSA = 2 π r (r + h)

Right Circular Cone • l^2 = r^2 + h^2
• Volume of cone = 1/3 π r^2 h
• Curved surface area: CSA=  π r l
• Total surface area = TSA = πr(r + l )

Frustum of a Cone • h = height, s = slant height
• Volume: V = π/ 3 (r^2 + rR + R^2)h
• Surface Area: S = πs(R + r) + πr^2 + πR^2

Sphere • Volume: V = 4/3 πr^3
• Surface Area: S = 4π^2

Hemisphere • Volume-Hemisphere = 2/3 π r^3
• Curved surface area(CSA) = 2 π r^2
• Total surface area = TSA = 3 π r^2

Prism

• Volume = Base area x height • Lateral Surface area = perimeter of the base x height Pyramid • Volume of a right pyramid = (1/3) × area of the base × height.
• Area of the lateral faces of a right pyramid = (1/2) × perimeter of the base x slant height.
• Area of whole surface of a right pyramid = area of the lateral faces + area of the base.

For Prism and Calendar (figures with uniform girth) –
Lateral Surface Area = Height * Perimeter of the Base
Volume = Height * Area of the Base

In this question the Total surface area is being asked
Total Surface Area of a Prism = Lateral Surface Area + Area of the two bases
Height of the prism = 10 cm
Perimeter of the base = 5 + 12 + 13 (Calculate the hypotenuse with Pythagoras Theorem) = 30 cm
So Lateral Surface Area = 10 * 30 = 300 cm
Area of the base = 1/2 * base * height = 1/2 * 5 * 12 = 30 cm
So Total Surface Area = 300 + 2*30 = 360 cm

In such questions, remember one thing SIMILARITY
r/R = h/H     …    (1)
where r = radius of small cone
R = radius of Big cone
h = height of small cone
H = height of big cone
Volume of cone = 1/3 * π* r* h
Given, Volume of big cone = 27 * Volume of small cone
1/3 * π * R* H = 27 * 1/3 * π * r* h

27 * (r/R)^2=H/h
Put the value of r/R from equation (1)

27 * (h/H)^2 = H/h
27 * h3 = H3

Put H = 30 cm

So h = 10 cm
The question asks us the height above the base, which is (30 – h) = 30 – 10 = 20 cm

The base of the prism looks like the figure above.
AD = 12 cm, AB = 9 cm
Hence BD = 15 cm (Pythagoras Theorem)
Area of base = Area of triangle ABD + Area of triangle BDC
Area of triangle ABD = 1/2 * 9 * 12 = 54 cm

Area of triangle BDC = 84 cm (Apply Heron’s formula)
Area of base/quadrilateral = 84 + 54 = 138 cm
Volume = Height * Area of the Base
2070 = Height * 138
So, Height of the prism = 15 cm

Lateral Surface Area = Height * Perimeter of the Base
Perimeter of the base = AB + BC + CD + DA = 48 cm
Lateral Surface Area = 48 * 15 = 720 cm^2

Area of the base = √3/4 * a^2, where a is the side of the equilateral triangle

Perimeter of the base = 3a
Volume of the prism = Area of the base * Height =√3/4 * a^2 * h … (1)
Lateral surface Area of the prism = Perimeter of the base * Height = 3a * h …(2)
Divide equation (1) by (2)
Volume/Area = (1/4√3) * a
40√3/120 = a/4√3 [Since Volume = 40√3 and Lateral surface Area = 120]
a = 160 * 3/120

a = 4 cm

This question is about ‘Pyramid’. So let me just give you some basic understanding of Pyramids. CGL can ask questions about two types of Pyramids – Pyramid with a Triangular Base and Pyramid with a square base. Both these pyramids have different formulas. Look at the below figure and understand the labellings, i.e., Slant edge and Slant Height
In the below image, I have written formulas for both types of Pyramids. The formula for Volume is same for both the Pyramids-
V = 1/3 * A * h
where A = Area of the base (calculation of A will be different for both)
h = Height of the Pyramid
When lateral surface area is asked, you will first calculate the ‘Slant Height’. Then with the help of slant height  you will find the area of one lateral face (let’s call this area M). If the pyramid is having a triangular base then multiply M with 3, to get the lateral surface area of the pyramid. And if the pyramid is having a square base, then multiply M with 4.
The area of the square is 324, hence its side is 18 cm
Volume of the pyramid = 1/3 * Area of the base * Height
1296 = 1/3 * 324 * Height
So Height = 12 cm
Slant Height of the pyramid with square base  = √ (h^2 + a^2/4) = √(12^2 + 18^2/4)
Slant Height = 15cm
Area of the lateral face = 1/2 * Base * Height = 1/2 * 18 * 15 = 135 cm^2
Pyramid with a square base has 4 lateral faces, so lateral surface area of the pyramid = 4 * 135 = 540 cm^2 This is a famous question. Just remember whenever you are forming a circle and then a square,

the side of that square is given by, a = 1.6*r (approx.), where r = radius of the circle

I have written approx. because the actual formula is 1.57*r, but it will make the calculations a bit lengthy. So, just find 1.6*r and the answer will be little less than that. Like here

Side = 1.6*84 = 134.4, so the answer is 132 cm

When the wire is bent in the form of a circle of radius 84cm, that means the circumference (or the length of the wire) of the circle is 2*π*84 = 44*12 cm

Now this wire forms a square of (let’s say) side ‘a’

Then, 4a (perimeter of the square) = 44*12

Hence a = 132 cm   If the given rectangular sheet of paper (length =l, breadth = b) is rolled across its length to form a cylinder, having a height b, then volume of cylinder = (l*l*b)/4π

If rolled across its breadth, then = (b*b*l)/4π

In this question the sheet is rolled along its length, so volume = (l*l*b)/4π = 12*12*5/(4* π)

Volume = 180/ π cm3 In this question, the ratio of surface areas is given and they are asking the ratio of volumes. The word “sphere” is useless here. In such questions, just imagine area as A2 and volume as A3. Now A3 is given and you have to find A3. How will you do it? Simple, first take the square-root of A2 to convert it into A, and then take the cube of A to find A3.

So for solving this question, we just have to take the square-root of 4:9. The ratio will become 2:3. Then take the cube of 2:3. Hence the answer is 8:27 Here again ratio of areas is given, that means A2 is given, and we have to find A. So 4:9 will become 2:3 Diameter and perimeter are directly proportional, P = D*π, where P is the perimeter and D is the diameter.

Hence a 75% increase in diameter means a 75% increase in perimeter The area of base and the volume of a cone are directly proportional V = A * h/3, where V = volume and h = height of the cone

Hence a 100% increase in the area of the base would mean a 100% increase in the volume A is increased by 50% hence A(or surface area) will increase by (1.5*1.5 – 1)*100 % = 125%

Note: Similarly A(or volume) will increase by (1.5*1.5*1.5 – 1)*100 % = 237.5% Where ever the word “melting” is used in mensuration, it means only one thing – equate the volume

The volume of the rectangular block = l*b*h = 21*77*24 cm3

Now this volume will be equal to the volume of the sphere formed after melting the block

Volume of sphere = (4/3) * π * r3 = 21*77*24

Hence, r = 21 cm The water rises by 5.6 cm. Take this 5.6 cm as the height of the cylindrical beaker and find its volume.

Volume of a cylinder = π*r*r*h = π * (7/2) * (7/2) * 5.6

Volume of the marbles (spherical in shape) = (4/3) * π * r3 = (4/3) * π * 0.7 * 0.7 * 0.7

No. of marbles dropped = Volume of beaker/Volume of a marble = 150 Let the radius of the big sphere be R.

Volume of a cone = (1/3) * π * R3 (since radius and volume are same as the radius of the sphere)

Let the radius of the smaller sphere = r

Then volume of cone = volume of smaller sphere

(1/3) * π * R3 = (4/3) * π * r3

r : R = 1 : 22/3

Surface area of smaller sphere(s) = 4 * π * r2

Surface area of larger sphere(S) = 4 * π * R2

S/s = (r/R)2 = 1 : 24/3  When a cone is hollowed out from a cylinder, we get the above figure

The whole surface area of the remaining solid = Area of A + Area of B + Area of C

A = curved surface area of the cone

B = curved surface area of the cylinder

C = area of the cylindrical base

A = π * r * l, where = slant height of the cone, which is  or

Hence A = π*3*5 = 15π

B = 2πrh = 2π*3*4 = 24π

C = πr2 = π*32 = 9π

The whole surface area of the remaining solid = 15π + 24π + 9π = 48π  Given, AB = 3 cm, BC = 6 cm and OF = 1 cm

Height of the cone (AC) = √(6^2 – 3^2 ) = 3√3 cm

Triangles ABC and CFO are similar (RHS similarity)

So, OC/BC = OF/AB

OC = 2 cm, therefore CG = 3 cm (OG = 1 cm)

Now, ABC and CEG are similar

GE/AB = CG/AC

So, GE =

Required volume = Volume of cone (CDE) – Volume of Sphere

= 3π – (4/3)π

= (5/3)π

# Practice Sets On Mensuration

1. Poles are to be fixed along the boundary of a rectangular field in such a way that distance between any two adjacent poles is 2 m.The perimeter of the field is 70m and length and the breadth of the field are in the ratio 4:3 resp. How many poles will be required?
A) 42
B) 40
C) 35
D) 38
E) 45
Option C
Solution:
Required between the two poles = (Perimeter/Dist.between any two adjacent poles) = 70 / 2 = 35
• The circumference of  a circular garden is 1320m.Find the area. Outside the garden , a road of 2m width runs around it .What is the area of this road and calculate the cost of gravelling it at the rate of 50 paise per sq. m .
A) 2500.15 m2, Rs.1500.15
B) 2652.57 m2, Rs.1326.285
C) 2541.14 m2, Rs.1600.47
D) 3245.78 m2,Rs.2000
E) 4157.12 m2,Rs.1452.11

Option B
Solution:
Circumference of the garden = 2*pi*R = 1320

R= 210m
Outer radius = 210 +2= 212 m
Area of the road = pi*(212)^a  – pi*(210)^2
= pi*422*2 = 2652.57 m^2
Therefore ,
cost of gravelling = 2652.57 * 0.5 = Rs.1326.285
• A square shape of park of area 23,104 sq. m is to be enclosed with wire placed at heights 1,2,3,4 m above the ground . Find required length of the wire ,,if its length required for each circuit is 10% greater than the perimeter of the field ?
A) 2675.2m
B) 2145.12m
C) 2750m
D) 2478.11m
E) 2400.5m

Option A
Solution:
Perimeter = √23,104  * 4 = (152 * 4)m

Length of each circuit = 152 * 4 *(110/100)
The wire goes around 4 times ,so the total length of the wire required =  152 * 4 *(110/100) * 4 = 2675.2 m
• Area of a hexagon is 54√3 cm^2. What is its side ?
A) 7cm
B) 5cm
C) 4cm
D) 6cm
E) 8cm

Option C
Solution:
(6√3/4) *a^2 = 54√3

=> a^2 = 36
=> a = 6 cm
• Smallest side of a right angled triangle is 8 cm less than the side of a square of perimeter 64cm . Second largest side of the right angled triangle is 4 cm less than the length of rectangle of area 112 sq. cm and breadth 8 cm .What is the largest side of the right angled triangle?
A) 9.2cm
B) 7.75cm
C) 10.50cm
D) 14cm
E) 12.80cm

Option E
Solution:
Side of a square = (perimeter /4) = 64/4 = 16 cm

smallest side  = 16 – 8   = 8cm
Length of the rectangle = Area/Breadth = 112/8 = 14cm
Second side of triangle = 14 – 4 = 10cm
Hypotenuse of the right angled triangle = √(8)^2+(10)^2 = 12.80 cm
• If the radius of the circular field is equal to the side of a square field .If the difference between the area of the circular field and area of the square field is 5145 sq. m ,then calculate the perimeter of the circular field?
A) 421m
B) 315m
C) 310m
D) 308m
E) 300m

Option D
Solution:
Let the radius of the circular field and the side of the square field be r

Then,
pi*r^2 – r^2 = 5145
=> r^2[(22-7)/7] = 5145
=> r = 49 m
Therefore ,
circumference of the circular field = 2*pi*r = 308m
• A rectangular plot has a concrete path running in the middle of the plot parallel to the parallel to the breadth of the plot. The rest of the plot is used as a lawn ,which has an area of 240sq. m. If the width of the path is 3m and the length of the plot is greater than its breadth by 2m ,what is the area of the rectangular plot(in m )?
A) 410m
B) 288m
C) 250m
D) 300m
E) 320m
Option B
Solution:
Let width be x m

and length be (x+2)m
Area of path = 3x sq. m
x(x+2) – 3x = 240
=> x^2 – x – 240 = 0
=> x(x – 16) +15 (x – 16) = 0
=>(x – 16)(x + 15) = 0
=>x = 16
Length =  16 + 2 = 18m
Therefore ,
Area of  plot = 16 * 18 = 288sq. m
• A solid spherical ball  of radius r is converted into a solid circular cylinder of radius R. If the height of the cylinder is twice the radius of the sphere ,then find the relation between these two with respect to radius.
A) R = r√(3/4)
B) R = r√(3/2)
C) R = r√(1/2)
D) R = r√(2/3)
E) R = r√(1/3)

Option D
Solution:
Since one object is converted into another so the volume will remain the same .

Therefore ,
(4/3)*pi*r^3 = pi*R^2*H
=>R = r√(2/3)
• A rectangular tank of length 37 (1/3) m internally , 12 m in breadth and 8 m in depth is full of water .Find the weight of water in metric tons, given that one cubic metre of water weighs 1000kg.
A) 3584 metric tons
B) 4500 metric tons
C) 4101 metric tons
D) 3870 metric tons
E) 5721 metric tons

Option A
Solution:
Volume of  water = 37(1/3)*12*8 m^3

Weight of water = (112/3)*12*8*1000 = 3584metric tons.
• An equilateral triangle and a regular hexagon have equal perimeters. The ratio of the area of the triangle and that of the hexagon is :
A) 3:4
B) 4:9
C) 1:2
D) 2:3
E) 4:5

Option D
Solution:
Let side of triangle be x  and the side of regular hexagon be y .

3x = 6y
=>x = 2y
Area of triangle = (√3/4)x2
Area of hexagon = 6*(√3/4) * y2 = (3√3/8)*x2
Required ratio = 2 : 3
• A room 10mtr long 4mtr broad and 4mtr high has two windows of 2*1mtr and 3*2mtr. Find the cost of papering the walls with paper 50cm wide at 25paisa per meter?
A) Rs48
B) Rs50
C) Rs52
D) Rs54
E) Rs46
Option C
Solution:
Area of walls = 2(10+4)*4 =112
Area of windows = 2+6 = 8
Area to be covered = 112−8= 104mtr
Length of paper = 104/50*100 =208m
Cost = 208*25/100 = 52
• A cubical block of 8m*12m*16m is cut into exact number of equal cubes. The least possible number of cubes will be?
A) 9
B) 24
C) 18
D) 30
E) 12

Option B
Solution:
H.C.F of 8,12,16 =4
Least number of cubes = 8*12*16/4*4*4 = 24
• Find the volume, curved surface area and the total surface area of a hemisphere of radius 21cm?
A) 19404cm³, 2772cm², 4158cm²
B) 4158cm³, 5000cm², 4000cm²
C) 20000cm³, 40000cm²,1000cm²
D) 30000cm³, 2000cm²,5000cm²
E) 40302cm³, 3320cm²,5650cm²
Option A
Solution:
The option which gets divided by 11, will be the answer
Method to check – 19404 = add alternate number = 1+4+4 =9
0+9= 9
Find difference = 9-9=0
If difference is either 0 or divisible of 11 then number is divisible of 11.
Ans ¬– A
• A right circular cone is exactly fitted inside a cube in such a way that the edges of the base of the cone are touching the edges of one of the faces of the cube and the vertex is on the opposite face of the cube. If the volume of cube is 2744 cubic cm, what is the approximate volume of the cone?
A) 715
B) 719
C) 729
D) 725
E) 710

Option B
Solution:
side of cone 3√2744 = 14

Height = 14
Volume = 1/3 ∏r²h
1/3*22/7*7*7*14 = 718.66= 719
• A hollow cylindrical tube is open at both ends is made of iron 4cm thick. If the external diameter be 52cm and the length of the tube be 120cm, find the number of cubic cm of iron in it?approx
A) 72419
B) 72425
C) 72405
D) 72411
E) 72534

Option D
Solution:
H = 120 external diameter – 52
Volume of iron = external volume – internal volume
22/7  * 26 * 26 * 120 – 22/7 * 22 * 22 * 120 = 72411
• A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2cm and the diameter of the base is 4cm. If a right circular cylinder circumscribe the solid, find how much more space will it cover?
A) 4π cm³
B) 2π cm³
C) 16π cm³
D) 8π cm³
E) 8π cm³

Option D
Solution: R of hemisphere =4/2 = 2cm
H of cylinder = 4cm
R of cone = 2cm
V of cylinder – volume of solid =
=π*2²*4 –(2/3 π*2³+1/3*π*2³)
= 16π − 8π
= 8π
• The ratio between volumes of a hemisphere and a cone is 1:1. If the cone’s height is equal to its diameter, then find the ratio of diameter of hemisphere and cone ?
A) 2:1
B) 1:1
C) 3:2
D) 2:3
Option B
Solution:
let the radius of hemisphere and cone are r1 and r2
H’s volume/c’s volume = 1/1
So [2/3 πr13]/[1/3 πr22*2r2] = 1/1
So r1 : r2 = 1 : 2 or D1 : D2 = 1: 1
• If the height of a pyramid is 12cm and its base is a square which perimeter is 40cm, then find the volume of pyramid?
A) 300 cm³
B) 200 cm³
C) 400 cm³
D) 500 cm³

Option C
Solution:
perimeter of base =40
Side of base = 10
Area of base = 100
Volume = 1/3 * area of base * height
= 1/3 * 100 * 12 = 400cm³
• If the perimeter of square, circle, rectangle, are equal. Then whose area is largest?
A) Circle
B) Square
C) Rectangle
D) All are equal

Option A
Solution:
when perimeter of these are equal then descending order of area is
Circle >square> rectangle.
So option A is Ans
• A rectangular plot of grass is 50m long and 40m broad. From the center of each side a path of 3m wide goes across the center of the opposite side. Find the area of path?
A) 270
B) 280
C) 251
D) 261
Option D
Solution: area of road = 3*50 + 3*40 −3²
= 270 −9=261

• The height of the cone is 24 cm and the curved surface area of cone is 550 cm2. Find its volume.
A) 1200 cm2
B) 1232 cm2
C) 1240 cm2
D) 1260 cm2
E) 1262 cm2
Option B
Solution
:

Volume= 1/3 π*r2 * h
Answer will be divisible by 11, as in pie we have 2*11. As only 1232 is divisible by 11, it is the answer
• The side of a square base of a pyramid increases by 20% and its slant height increases by 10%. Find the per cent change in Curved Surface Area.
A) 28%
B) 58.4%
C) 32%
D) 45.20%
E) 48%
Option C
Solution
:

C.S.A=1/2*(perimeter of base)*l
20+10+(20*10)/100=32%
• If a copper wire is bend to make a square whose area is 324 cm2. If the same wire is bent to form a semicircle, then find the radius of semicircle.
A) 7 cm
B) 14 cm
C) 11 cm
D) 21 cm
E) 12 cm

Option B
Solution
:

Area of square= 324, hence side =18
Perimeter = 4a =4*18=72
Circumference of semicircle= 2r+Pie *r
r(2+pie)=72
r=14 cm
• A man wants to make small sphere of size 1 cm of radius from a large sphere of size of 6 cm of radius. Find out how many such sphere can be made?
A) 216
B) 125
C) 36
D) 200
E) 64

Option A
Solution
:

Volume of sphere1/volume of sphere 2= required number of sphere
=6*6*6/1*1*1=216
• A sphere of radius 9 cm is dip into a cylinder who is filled with water upto 20 cm. If the radius of cylinder is 6 cm find the percentage change in height.
A) 50%
B) 40%
C) 55%
D) 45%
E) 57%

Option D
Solution
: Volume of sphere= volume of cylinder from height 20 cm to upwards.
4/3 * π * 9*9*9 = π * 6*6*h
h=9
new height=20+9=29
%change= 9/20*100=45%
• The length of the perpendicular drawn from any point in the interior of an equilateral triangle to the respective sides are P1, P2 and P3. Find the length of each side of the triangle.
A) 2/√3 *(P1 + P2 + P3)
B) 1/3 * (P1 + P2 + P3)
C) 1/√3 *(P1 + P2 + P3)
D) 4/√3 *(P1 + P2 + P3)
E) 5/√3 *(P1 + P2 + P3)

Option A
• A conical cup is filled with ice cream. The ice cream forms a hemispherical shape on its top. The height of the hemispherical part is 7 cm. The radius of the hemispherical part equals the height of cone then the volume of ice cream is?
A) 1078 cm3
B) 1708 cm3
C) 7108 cm3
D) 7180 cm3
E) 1808 cm3

Option A
Solution
:

Volume = volume of hemisphere + volume of cone= 2/3* π *r3 + 1/3 π * r2 *h
=1078
• Assume that a drop of water is spherical and its diameter is one tenth of a cm. A conical glass has equal height to its diameter of rim. If 2048000 drops of water fill the glass completely then find the height of the glass.
A) 12 cm
B) 16 cm
C) 20 cm
D) 8 cm
E) 10 cm

Option B
Solution
:

diameter of drop of water=1/10 => radius=1/20
volume of 204800 drop of water=204800*4/3* π*1/20 *1/20*1/20 = 1024 π/3
Volume of cone=1024 π/3 = 1/3 * π *r2 * h (r=h/2)
h=16
• If the radius of a sphere increase by 4 cm then the surface area increase by 704 cm2 . The radius of the sphere initially was?
A) 5
B) 4
C) 6
D) 8
E) 10

Option A
Solution
:

4 π(r+4)2 – 4 * π*r2 = 704
solve and get r=5
• By melting two solid metallic spheres of radii 1 cm and 6 cm, a hollow sphere of thickness 1 cm is made. The external radius of the hollow sphere will be.
A) 8 cm
B) 9 cm
C) 6 cm
D) 7 cm
E) 10 cm

Option B
Solution
:

4/3* π (R3 + r3)= 4/3* π * ((x+1)3 – x3)
R=6 cm; r=1 cm; x= radius of hollow sphere inner; (x+1)=outer radius
solve and get x=8
outer=x+1=9 cm

• A right circular cone is exactly fitted inside a cube in such a way that the edges of the base of the cone are touching the edge of one of the faces of the cube and the vertex is on the opposite face of the cube. If the volumes of cube is 216 cm3 , what is the volume of the cone (approximately)?
A) 56 cm3
B) 60 cm3
C) 46 cm3
D) 50 cm3
E) None of these

Option A
Solution
: volume(a3)=216 , hence a=6
r= 3 cm; height of the cone= 6cm (as it is fitted in this cube of side 6 cm, hence its height will also be 6 cm)
Volume of cone= 1/3 π*r2 * h
=56
• The diagram shows a section of a rocket firework. If this section can be completely filled with gunpowder what is the volume of gunpowder required?
A) 1882 cm3
B) 1782 cm3
C) 1982 cm3
D) 1682 cm3
E) None of these

Option B
Solution
: sin 60 = P/H=r/6=√3/2
=> r=3√3 cm
In the cone; 62 = h2 + r
h=3 cm
Volume of Gunpowder= Volume of Cone+ Volume of Cylinder=1/3 πr2h + πr2h = πr(1/3 h+h)
=22/7*27*21=1782
• If a square, circle and rectangle has same perimeter then which one of them has the maximum area?
A) Square
B) Circle
C) Rectangle
D) All have equal area
E) Cannot be determined

Option B
Solution
: In such case the area in descending order is: Circle> Square> Rectangle
• A cylinder has some water at height 20 cm. If a sphere of radius 6 cm is poured into it then find the rise in height of water if the radius of cylinder is 4 cm.
A) 3 cm
B) 9 cm
C) 18 cm
D) 15 cm
E) None of these

Option C
Solution
: Volume of ball= volume of rising water in the cylinder
4/3 * π*r= π*r*h
4/3*6*6*6=4*4*h
h=18 cm
• If the base of a pyramid is square and its side is 4√2 cm and slant height of pyramid is 5 cm, find the volume of pyramid.
A) 48 cm3
B) 16 cm3
C) 24 cm3
D) 32 cm3
E) None of these

Option D Solution:
l=slant height=5 cm ; h=height; side=4√2 cm
l= h2 + [(side*√2)/2]2
Note: The content inside bracket is the calculation for half of the diagonal of the square.
h= 3 cm
volume= 1/3 * Area of base * h
=1/3 * 32 * 3= 32
• A sphere of 5 cm radius is melted and small sphere of radius 1 cm is made from it. Find the number of sphere that can be made from it.
A) 25
B) 125
C) 50
D) 100
E) None of these

Option B
Solution
: Number of sphere=Volume of large sphere/volume of small sphere

[4/3* π *r1]/ [4/3* π *r23 ]=5*5*5/1*1*1=125
• A person wants to make a cylindrical box which is open from the top. If the height of that box is 10 cm and radius is 7 cm find the area of sheet which is required to make it.
A) 880 cm2
B) 1188 cm2
C) 594 cm2
D) 440 cm2
E) None of these

Option C
Solution
: Area required=Curved surface area + Area of base= 2 π r h + π r2 = 594
• A square park has a 2 m wide cross road in middle of it. If the side of park is 100 m then find the remaining area of the park.
A) 9650 m2
B) 9596 m2
C) 9600 m2
D) 9604 m2
E) None of these

Option D
Solution
: Total area= 10000
road area= 2*100 + 2*100- 2*2=396
remaining area=10000-396=9604
• In a right circular cone the radius of its base is 6 cm and its height is 14 cm. A cross section is made through the mid-point of the height parallel to the base. The volume of the lower portion is?
A) 528 cm3
B) 366 cm3
C) 498 cm3
D) 462 cm3
E) None of these

Option D Solution: Volume of cone= 1/3 π*r2 * h
Volume of lower part=volume of full cone-volume of upper cone
for full cone take r=6, h=14
for upper cone take r1=r/2=3 and h=7
volume of lower part=528-66=462
• If radius of cone decrease by 50% and height increase by 20%. Then find the percentage change in the volume.A) 70% decrease
B) 70% increase
C) 40% decrease
D) 40% increase
E) 20% increase

Option A
Solution
:

Volume of cone= 1/3 π*r2 * h
r=50% dec =1/2 =>2————1
2———–1(dec)
h=20% inc =1/5 =>5————-6 (inc)
2*2*5:1*1*6=10:3
(3-10)/10*100=70% dec