Practice Aptitude Questions (Mensuration) With Solutions Set-60:
Dear Readers, Important Aptitude Questions for SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.
Height
of the cylindrical tube is 44 mts
1).If
the sides of a square be increased by 50% the present increase in the area will
be
a)
100%
b)
25%
c)
125%
d)
60%
2).In
a rhombus whose area is 144 cm2 one of its diagonals twise as long
as the other. What are the length of the diagonals.
a)
6, 12 cm
b)
8, 16 cm
c)
10, 20 cm
d)
12, 24 cm
3).The
length of a right triangle sides are in the ratio 1 : 2 and its area is 36.
Find the hypotenuse of the triangle
a)
6√5 units
b)
7√5
c)
5√5
d)
4√5
4).If
the sides of a triangle are doubled then the area becomes
a)
3 times
b)
4 times
c)
2 times
d)
same
5).A
circular wire o radius 42 cm is cut and bent in the form of a rectangle whose
sides are in the ratio 6 : 5. Then the smaller side of the rectangle is
a)
48 cm
b)
24 cm
c)
60 cm
d)
21 cm
6).Find
the circumference of a circle whose area is 314 cm2 ( π= 3.14)
a)
62.8 cm
b)
32.4 cm
c)
6.2 cm
d)
12.8 cm
7).How
much distance a wheel having a diameter of 1.26 m will cover in 500 rounds
a)
2120 mts
b)
1981 mts
c)
1680 mts
d)
1980 mts
8).The
length of a chord of a circle is 8 cm and radius is 5 cms. Find the distance
between the centre and the chord.
a)
4 cms
b)
3 cms
c)
5 cms
d)
8 cms
9).The
radius and height of a cylinder are 12 cms, 21 cm respectively. How many
spheres with radius 3 cm can be made if the cylinder is melted.
a)
48
b)
88
c)
84
d)
86
10).The
length and breadth of a iron sheet are 44 mts. and 14 mts. respy. If iron sheet
is made to be a tube by length wise. Find the volume of the cylindrical tube.
a)
1656m3
b)
2126 m3
c)
1456 m3
d)
6776 m3
Answers:
1).
c) 2). d) 3).a
) 4). b) 5).
c) 6). a) 7).
d) 8).b ) 9).
c) 10). d)
Solution:
1).
Let the side of the square be 100 mts.
Area
of the square = 1002
If the side is increased by 50% then the side
= 150 mts
∴
Area of the square = 1502
∴ Increase in area = 1502 - 1002
= (150 + 100)(150 – 100)
= 250 × 50 sq ms.
∴ Increase in area = (250 ×
50) / 100 = 125%
Answer: c )
2).
Let d1 and 2d1 be the length of the diagonals.
∴
area = (d1 × 2d1) = 144
d12
= 144; d1 = √144 = 12
∴
one diagonal is 12 cm.
The
length of other diagonal is 24 cm
Answer: d)
3).
The legs of a right triangle are in the ratio 1 : 2
∴
Let x be the base and 2x be the height
∴
Area = (1 / 2) × x × 2x = 36
x2
= 36; x = 6
∴
Base of the triangle is 6 units
Height
of the triangle is 2 × 6 = 12 units
∴
Hypotenuse = √(122 + 62)
=
√(144 + 36) = √180 = √(36 × 5) = 6√5 units
Answer: a)
4).
Let a, b, c be the sides of a triangle
∴
Area = √[s(s – a) (s – b) (s – c)]
Where
s = (1 / 2) (a + b + c)
For
the new triangle , let the sides be 2a, 2b, and 2c.
s
= 2s
∴
Area of the new triangle = √[2s(2s - 2a) (2s – 2b)( (2s – 2c)]
=
√[16s (s – a) (s – b) (s – c)]
Hence
the area becomes 4 times
=
4 √[s(s – a) (s – b) (s – c)] = 4 × area of the original triangle
Answer: b)
5).
Let 6x and 5x be the sides of a rectangle.
Perimeter
of the rectangle = 2(6x + 5x) = 22x cm
The
radius of the circle = 42 cm
∴
The perimeter of the circle = 2πr = 2 × (22 / 7) × 42 cm
Perimeter
of the circle = the perimeter of the
∴ rectangle = 2 × (22 / 7) × 42 = 22x
x = 2 × (22 / 7) ×
42 × (1 / 22) = 12 cm
The sides of the
rectangle are 72 cm and 60 cm.
The rectangle of
the smallest side = 60 cm
Answer: c)
6).
Area of the circle = 314 cm2
∴πr2
= 314
r2
= 314 × (1 / π)
=
314 × (1 / 3.14) = 100
∴
r = √100 = 10 cm
The
perimeter of the circle = 2 πr
=
2 × 3.14 × 10 = 62.8 cm
Answer: a)
7).
The diameter of the wheel = 1.26 m.
∴
The perimeter of the wheel = 2 π r = 1.26 × 3.14 = 3.96 m
In
one round, the wheel covers a distance of 3.96 mts.
∴
In 500 rounds, the distance covered will be 3.96 × 500 = 1980 mts.
Answer: d)
8).
Let
C be the centre of the circle
AC
is the radius and AB is the chord
AC
= 5 cm (given)
AB = 8 cm
(given)
∴
AD = BD = 4 cm.
∴
In the ∆ADC, AC2 = AC2 + CD2
AC2
- AD2 = CD2
CD2
= 52 - 42 = (5 + 4) (5 – 1) = 9
CD
= √9 = 3cm
Answer: b)
9).
The radius of the cylinder = 12 cm
Height
of the cylinder = 21 cm
∴
volume of the cylinder = πr2h
π
× 12 × 12 × 21 cm3
The
radius of a sphere = 3 cm.
∴
Volume of the sphere = 4πr3 = 4 π ×33
=
(4 / 3) × π × 3 × 3 × 3cm3
Volume
of the cylinder = volume of the sphere
No
of spheres = Volume of the cylinder / volume of the sphere
=
(π × 12 × 12 × 21) / ((4 / 3) × π × 3 × 3 × 3)
=
(12 × 12 × 21) / (4 × 3 × 3) = 3 × 4 × 7 = 84
Answer: c)
10).
The length of the sheet is 44 mts.
Height
of the cylindrical tube is 44 mts
The
breadth of the sheet is 14 mts.
∴
Perimeter of the circular portion = 14 mts.
∴
2πr = 14
r = 14 × (7 / 22) × 2 = 7 cm
∴
Radius of the cylindrical tube = 7 cm.
Volume
of the cylindrical tube = πr2h = π × 7 × 7 × 44
=
(22 / 7) × 7 × 7 × 44
=
154 × 44 = 6776 m3
Answer: d)
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