Dear Readers, Important Aptitude Questions for Railway / SSC CGL,CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.
1).If x = 3+
2√2, then the value of (√x-1/√x) is
a)
1
b)
2
c)
2√2
d)
3√3
2).If ‘a’ be
a positive number, then the least value of a+1\a is
a)
1
b)
0
c)
2
d)
1\2
3).If a=0, b≠0,
c≠0, then the equation ax + by + c = 0 represents a line parallel to
a)
x
+ y = 0
b)
x
- axis
c)
y
- axis
d)
None
of these
4).If the
inradius of a triangle with perimeter 32 cm is 6 cm, then the area of the
triangle in sq. cm is
a)
48
b)
100
c)
64
d)
96
5).A, B, P
are three points on a circle having centre O. If ∠OAP = 25o and ∠OBP
=35o then the measure of ∠AOB is
a)
120o
b)
60
o
c)
75
o
d)
150
o
6).Side ¯BC
of ∆ABC is produced to D. If ∠ACD = 140o
and ∠ABC = 3 ∠BAC, then find ∠A.
a)
55o
b)
45
o
c)
40
o
d)
35
o
7).The length
of tangent (upto the point of contact) drawn from an external point P to a
circle of radius 5 cm is 12 cm. The distance of P from the centre of the circle
is
a)
11
cm
b)
12
cm
c)
13
cm
d)
14
cm
8).ABCD is a
cyclic quadrilateral, AB is diameter of the circle, if ∠ACD = 50o. then
value of ∠BAD is
a)
30
o
b)
40
o
c)
50
o
d)
60
o
9).Two
circles of equal radii touch externally at a point P. From a point T on the
tangent at P, tangents TQ and TR are drawn to the circles with points of
contact Q and R respectively. The relation of TQ and TR is
a)
TQ
< TR
b)
TQ
> TR
c)
TQ
= 2TR
d)
TQ
= TR
10).when two
circles touch externally the number of common tangents are
a)
4
b)
3
c)
2
d)
1
Answers:
1).b) 2).c) 3). b) 4).d) 5).a) 6).d) 7).c) 8). b) 9).d) 10).b)
Solution:
1).
Answer:
b)
2). Value of
a + 1/a with be minimum if a = 1
Least
value = 1+ 1 = 2
Answer:
c)
3). When a =
0,
ax + by + c = 0
ð by + c = 0
ð y= c/b
Which
is parallel to x-axis
Answer:
b)
4). Area of
triangle = In radius * semi perimeter
=6
× 32/2 = 96 sq. cm.
Answer:
d)
5).
∠OAP = 25o = ∠OPA
[OA = OP = OB = radius]
∠AOP = 180o – 2×35 = 130o
∠AOP = ∠OPB = 35o
∠POB = 180o – 2×35o
= 110o
` ∠AOB = 360o – (130o
+ 110o)
Answer:
a)
6).
∠ACD = 140o
∠ACD = ∠BAC
+ ∠ABC = 140o
ð ∠BAC + 3∠BAC
= 140o
ð 4∠BAC =
140o
ð ∠BAC = 140/4
= 35o
Answer:
d)
7).
∠OAP=90o
OA = 5cm,
AP = 12 cm
OP = √(AP2+ OA2)
=√ (122+52
) = √(144+25)
=√169=13 cm
Answer:
c)
8).
Angle of
semi-circle is a right angle
∠ACB = 90o, ∠ACD = 50o
∠BCD = 90o + 50o =
140 o
∠A + ∠C = 180 o
ð ∠A = 180
o -140 o = 40 o
Answer:
b)
9).
TQ = TR
OQ = O’R
OP= PO’
Answer:
d)
10).
Answer:
B)
