Practice Aptitude Questions (Mixed Problems) With Solutions Set-54:
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 edge of a cube increases by
80% then the surface area of the cube increase by
a)
175%
b)
235%
c)
224%
d)
214%
2). If the
volume and the base area of a pyramid are 296 cm3 and 39 cm2 respectively,
then the height of the pyramid is
a)
30
cm
b)
24
cm
c)
16
cm
d)
32cm
3). A square
and a rhombus have the same base. If the rhombus is inclined at 30°, find the
ratio of the area of the square to the area of the rhombus.
a)
2
: 1
b)
2
: √3
c)
1
: 2
d)
√3
: 2
4). If sin2x
+ sinx =1, then the value of cos4x + cos2x – 1 is
a)
1
b)
1
/ 2
c)
0
d)
√3
/ 2
5). If cot
(A-B) = √3 and tan (A+B) = √3 when A > B >0 and A + B is an acute angle,
then the value of B is
a)
π
/ 12
b)
π
/ 6
c)
3π
/ 8
d)
π
/ 4
6). If cosecθ
+ sinθ = 2, then the value of cot2 θ – cos2θ is
a)
2
b)
5
c)
1
d)
0
7). If xtanθ
= 2 and xsinθ = √3, then the value of x is
a)
2√3
b)
√3
/ 2
c)
2√3
d)
√3
8). The value
of √(sec2θ + cosec2θ) is
a)
1
+ secθ
b)
tanθ
- cotθ
c)
secθ
. cosecθ
d)
secθ
+ tanθ
9). Find the
angular elevation of the sun when the shadow of a 25√3 m – long pole is 25m.
a)
30°
b)
45°
c)
55°
d)
60°
10). The value
of [(cos3 θ – sin3 θ) / (cosθ – sinθ)] – sinθ.cosθ is
a)
1
b)
2
c)
0
d)
1/2
Answers:
1). c) 2).b ) 3)a. ) 4). c) 5).a
) 6).d ) 7).a
) 8).c ) 9).d
) 10).a )
Solution:
1). Let each edge of the smaller cube be 1m.
Now,
increase = 80%
Side
of the new cube =1.8m
Surface
area of the cube = 6 × 12 = 6m2
Area
of the new cube = 6 × (1.8)2 = 19.44 m2
Reqd
% increase in the surface area
=
[(19.44 – 6) / 6] ×100 =224%
Answer:
c)
2). Volume of
the pyramid = 1/3 (Base area × height)
Then,
(1/3) × 37 × height = 296
∴
height = (296 × 3) / 37 = 24 cm
Answer:
b)
3).
Now,
in ∆AEH we get
Or,
sin 30° = (h / a)
Or,
(1 / 2) = h / a
∴
h = (a / 2)
∴
Area of the rhombus = a × h
=
a × (a / 2) = (a2 / 2)
Area
of the square = a2
∴
Required ratio = a2 / (a2
/ 2) = (2 / 1) = 2 : 1
Answer:
a)
4). sin2
x + sinx =1
∴
sinx = 1 – sin2 x = cos2 x
Now,
cos4x + cos2x – 1
=
sin2 x + cos2x – 1
= 1-1 = 0
Answer:
c)
5). cot (A- B)
= √3 = cot ( / 6)
Then,
A – B = π / 6
⋯(i)
Again,
tan (A + B) = √3
Or,
A + B = π / 3 ⋯(ii)
Now,
2A= (π / 6 ) + (π / 3 ) = (π / 2π) / 6 = (3π / 6) = (π / 2)
∴
A = π / 4
So,
B = (π / 4 ) - (π / 6 ) = (3π - 2π) / 12 = π / 12
Answer:
a)
6). cosecθ +
sinθ = 2
Squaring
both sides, we get
cosec2θ
+ sin2θ + 2 cosecθ . sinθ = 4
or,
cosec2θ + sin2θ = 2
⋯(i)
now,
cot2θ - cos2θ =
cosec2θ – 1 – (1 - sin2θ)
=
cosec2θ – 1 – 1 + sin2θ = 2 – 2 =0
(∴cosecθ
+ sinθ = 2 )
Answer:
d)
7). Given
xtanθ = 2 ⋯(i)
xsinθ
= √3 ⋯(ii)
or,
(xtanθ / xsinθ) = (2 / √3)
or,
sinθ / (cosθ . sinθ) = (2 / √3)
or,
cosθ = (√3 / 2) = cos 30°
∴θ
= 30°
Now,
x.tan30° 2
x
× (1/ √3) =2
∴
x = (2√3)
Answer:
a)
8). √(sec2θ
+ cosec2θ)
=
√[ ( 1 + tan2θ) + (1 + cot2θ)]
=
√( tan2θ + cot2θ) + 2
= √(tanθ + cotθ)2
= tanθ + cotθ
=
(sinθ / cosθ) + (cosθ / sinθ) = (sin2θ + cos2θ) / (sinθ +
cosθ)
=
1 / (sinθ . cosθ) = secθ . cosecθ
Answer:
c)
9).
Now,
in ∆ABC we get
tanθ
= (AB / AC) = (25√3 / 25) = √3
or,
tanθ = tan60°
∴ θ = 60°
Answer:
d)
10). [(cos3θ
- sin3θ) / (cosθ - sinθ)] - sinθ . cosθ
=
{[(cosθ - sinθ)(cos2θ + sin2θ + sinθ
. cosθ)] / (cosθ – sinθ)} - sinθ . cosθ
=
1 + sinθ . cosθ - sinθ . cosθ = 1
Answer:
a)
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