### Practice Aptitude Questions (Mensuration) With Solutions

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.

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

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%

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

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

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

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

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

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.

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
CD2  = 52  - 42 = (5 + 4) (5 – 1) = 9
CD = √9 = 3cm

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

10). The length of the sheet 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