### Practice Aptitude Questions With Solutions for RRB / SSC Exam

Practice Aptitude Questions With Solutions Set-2:
Dear Readers, Important Aptitude Questions for RRB/SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).The base of a right circular cone has the same radius ‘a’ as that of a sphere. Both the sphere and the cone have the same volume. Height of the cone is
a)    3a
b)    4a
c)    7 / 4a
d)    7 / 3a

2).The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm3. Calculate its curved surface area is sq.cm
a)    110
b)    444
c)    220
d)    616

3).If the length of each side of an equilateral triangle is increased by  2 unit, the area is found to be increased by 3 + √3 square unit. The length of each side of the triangle is
a)    √3 unit
b)    3 unit
c)    3√3 unit
d)    1 + 3√3 unit

4).The diameter of the moon is assumed to be one fourth of the diameter of the earth. Then the ratio of the volume of the earth to that of the moon is
a)    64 : 1
b)    1 : 64
c)    60 : 7
d)    7 : 60

5).If a linear equation is of the form x = k where k is a constant, then graph of the equation will be
a)    a line parallel to x - axis
b)    a line cutting both the axes
c)    a line marking positive acute angle with x - axis
d)    a line parallel to y - axis

6).If a = 34, b = c, c = 33, then the value of a3 + b3 + c3 is
a)    0
b)    111
c)    50
d)    100

7).If  (a2 + b2 )3 = (a3 + b3 )2 , then (a / b) + (b / a) is
a)    1 / 3
b)    2 / 3
c)    – 1 / 3
d)    – 2 / 3

8).If x + (1 / x) =5, then the value of (x4 + 3x3 + 5x2 + 3x + 1) / (x4 + 1)
a)    43 / 23
b)    47 / 21
c)    41 / 23
d)    45 / 21

9).If in a triangle, the circumcentre, incentre, centroid and orthocenter coincide, then the triangle is
a)    Acute angled
b)    Isosceles
c)    Right angled
d)    Equilateral

10).The internal bisectors of ∠ABC and ∠ACB of ∆ ABC meet each other at O. If ∠ BOC = 110°, then ∠BAC is equal to
a)    40°
b)    55°
c)    90°
d)    110°
1). b) 2). c) 3). a) 4). a) 5). d) 6). d) 7). b) 8). a) 9). d) 10). a)

Solution:
1). (1 / 3) π a2 h =    (4 / 3) π a3
h = 4a
2). V = πr2 h
550 = π × 5x × 5x × 7x
550 = (22 / 7) × 25 × 7x3
x= 550 / (22 × 25) = 1
x  = 1
∴ Area of curved surface
2 × (22 / 7) × 5 × 7 = 220 sq.cm
3). Side of equilateral triangle = x units.
∴ (√3  / 4) [ (x + 2)2 – x2 ] = 3 + 3√3
(√3  / 4) [ 4x + 4 ] = 3 + 3√3
√3 x + √3   = 3 + √3   = √3x  = 3
x = √3  units
4). Volume of earth : Volume of moon
(4 / 3) π r:  (4 / 3) π (r / 4)3
= 64 : 1
5).

6). a3 + b3 + c3 – 3abc
= (1 / 2) (a + b + c ) [ ( a - b)2 + (b – c)2 + (c – a)2 ]
= (1 / 2) × 100 (1 + 0 + 1) = 100
7). (a2 + b2 )3 = (a3 + b3 )2
a6 + b6 + 3a2 b2 (a2 + b2)
a6 + b6 + 2a3 b3
3 (a2 + b2)  = 2ab
(a2 + b2)  / ab = 2 / 3
(a / b) + (b / a) = 2 / 3
8). x + (1 / x) = 5
On squaring both sides,
x2 + (1 / x2) + 2 = 25
x2 + (1 / x2)  = 25  -  2 = 23               … (i)
Expression
= (x4 + 3x3 + 3x + 5x2) / x4 + 1
= (x4 + 1 + 3x3 + 3x + 5x2) / x4 + 1
= [x2 (x2 + (1 / x2)  ) + 3 x2 (x + (1 / x )  ) + 5x2 ]  /  x2 (x2 + (1 / x2)  )
= [ (x2 + (1 / x2)  ) + 3 (x + (1 / x )  ) + 5 ] / x2 (x2 + (1 / x2)  )
= (23 + 3 × 5 + 5) /  23  = 43 / 23
9). In an equilateral triangle, centroid, incentre etc lie at the same point
10).

In ∆ ABC,
∠A + ∠B + ∠C = 180°                        ..(i)
In ∆OBC,
∠OBC + ∠BOC + ∠OCB = 180°
(∠B / 2) + 110° + (∠C / 2)= 180°
(∠B + ∠C) / 2  = 180° - 110° = 70°
∠B + ∠C = 140°
∴∠A = 180° - 140°  =  40°