### Practice Aptitude Questions With Solutions for SSC CGL,CHSL / Railway Exam

Practice Aptitude Questions With Solutions Railway / SSC CGL, CHSLExams Set-15:
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

1).b)  2).c)  3). b) 4).d)  5).a)  6).d)  7).c)  8). b) 9).d)  10).b)

Solution:
1).

2). Value of a + 1/a with be minimum if a = 1
Least value = 1+ 1 = 2
3). When a = 0,
ax + by + c = 0
ð  by + c = 0
ð  y= c/b
Which is parallel to x-axis
4). Area of triangle = In radius * semi perimeter
=6 × 32/2 = 96 sq. cm.
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)
6).

∠ACD = 140o
∠ACD = ∠BAC + ∠ABC = 140o
ð  ∠BAC + 3∠BAC = 140o
ð  4∠BAC = 140o
ð  ∠BAC = 140/4 = 35o
7).

∠OAP=90o
OA = 5cm, AP = 12 cm
OP = √(AP2+ OA2)
=√ (122+52 ) = √(144+25)
=√169=13 cm
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