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.
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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°
Answers:
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
Answer: b)
2). V = πr2
h
550 = π × 5x × 5x × 7x
550 = (22 / 7) × 25 × 7x3
x3 = 550 / (22 × 25) = 1
x = 1
∴ Area of curved surface
2 × (22 / 7) × 5 × 7 = 220 sq.cm
Answer:
c)
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
Answer:
a)
4). Volume of
earth : Volume of moon
(4 / 3) π r3 :
(4 / 3) π
(r / 4)3
= 64 : 1
Answer:
a)
5).
Answer:
d)
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
Answer:
d)
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
Answer:
b)
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
Answer:
a)
9). In an
equilateral triangle, centroid, incentre etc lie at the same point
Answer:
d)
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°
Answer:
a)
