### Practice Aptitude Questions (Mixed Problems) With Solutions Set-54

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

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%

2). Volume of the pyramid = 1/3 (Base area × height)
Then, (1/3) × 37 × height = 296
∴ height = (296 × 3) / 37 = 24 cm

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 / (a/ 2) = (2 / 1) = 2 : 1

4). sin2 x + sinx =1
∴ sinx = 1 – sin2 x = cos2 x
Now, cos4x + cos2x – 1
= sin2 x  + cos2x – 1 = 1-1 = 0

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

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 )

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)

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

9).

Now, in ∆ABC we get
tanθ = (AB / AC) = (25√3  / 25) = √3
or, tanθ = tan60°
∴  θ = 60°