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

Practice Aptitude Questions With Solutions Railway / SSC CGL, CHSL Exams Set-16:
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).D and E are mid-points of AB and AC of ∆ABC, if ∠A=80o, ∠C=35 o, then ∠EDB is equal to
a)    100 o
b)    115 o
c)    120 o
d)    125 o

2).From 125 metre high tower, the angle of depression of a car is 45 o.  Then how far the car is from the tower?
a)    125 metre
b)    60 metre
c)    75 metre
d)    95 metre

3).The value of (sin 53 o/cos 37 o)+(cot 65 o/tan25 o) is
a)    2
b)    1
c)    3
d)    0

4).The value of (cos 60 o+sin 60 o)/(cos 60 o-sin 60 o) is
a)    -1
b)    √3+2
c)    –(2+√3)
d)    √3-2

5).The value of cot 5o. cot 10o.cot 15 o. cot 60 o. cot 75 o. cot 80 o. cot 85 o / { (cos220o+cos270o)+2 } is
a)    9/√3
b)    1/9
c)    1/√3
d)    √3/9

6).In a triangle the angles are in the ratio 2: 5: 3. What is the value of the least angle in the radian?
a)    π /20
b)    π /10
c)    2π/5
d)    π/5

7).If x= a cos θ – b sin θ, y = b cos θ + a sin θ, then find the value of x2 + y2
a)    a2
b)    b2
c)    a2/b2
d)    a2+b2

8).If tan ɑ + cot ɑ =2, then the value of tan7 ɑ + cot7 ɑ is
a)    2
b)    16
c)    64
d)    128

9).12 men construct 1. 5 km of road in 7 days.28 men will construct 12 km of roads in
a)    20 days
b)    24 days
c)    28 days
d)    38 days

10).A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in
a)    15 days
b)    20 days
c)    25 days
d)    30 days

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

Solution:
1).

∠B=180o-80o-35o=65o=65o
Points D and E are the midpoints of sides AB and AC respectively
DE ІІ BC
∠BDE=180o-85o=115o
2).

AB=Tower =125 metre
C= Position of car
From ∆ABC
tan 45o=AB/BC
ð  1= AB/BC
ð  AB= BC=125 metre
3). Sin 53o/cos 37o ÷ cot 65o/tan25o
= sin 53o/cos (90o-53o) ÷ cot(90o-25o)/tan 25o
= sin 53o /sin 53o ÷ tan 25o/tan 25o+
=1 ÷ 1=1
4). [Cos 60o + sin60o] / [cos 60o - sin 60o] = { 1/2+√3/2 } / { 1/2-√3/2 }
= (1+√3) / (1-√3) =(1+√3)(1+√3) / (1-√3)(1+√3)
= (1+3+2√3) / -2 = (4+2√3) / -2 = -(2+√3)
5). Cos 85o= cot (90o-5o) = tan 5o
Cot 80o= cot (90o-10o) = tan 10o
Cot 80o= cot (90o-15o) = tan 15o
Cot 70o = cot (90o-20o) = tan 20o
Expression
(Cot 5o. cos 10o. cos 15o. cos 60o. tan 15o . tan 10o. tan 5o) / {(cos2200+sin220o)+2}
{(Cot 5o. tan 5o)(Cot 10o. tan 10o)(Cot 15o. tan 15o) . cot 60o } / (1+2)
1/(√3(2+1)) = 1/3√3 = 1×√3 / (3√3×√3) =√3/9
[tan θ cot θ=1, sin2 θ+cot2 θ=1]
6). Sum of three angles of triangle
Least angle =2/2+5+3×π
7). x= a cos θ – b sin θ
x= b cos θ + sin θ
x2+ y2=(a cos θ – b sin θ)2+(b cos θ + a sin θ)2
=a2cos2 θ + b2 sin2 θ- 2ab sin θ. Cos θ + b2cos2+a2sin2 θ+2ab sin θ cos θ
=a2cos2 θ+b2sin θ+a2sin2 θ+b2cos2 θ
`           =a2cos2 θ + a2sin2 θ+b2sin2 θ+b2cos2 θ
=a2(cos2 θ +sin2 θ)+b2(sin2 θ+cos2 θ)
=a2+b2
8). tan ɑ + cot ɑ = 2
=> tan ɑ + 1/tan ɑ =2
=>tan2 ɑ +1 = 2 tan ɑ
=> tan2 ɑ-2 tan ɑ+1 = 0
=>(tan ɑ-1)2=0
=>tan ɑ-1=0
=>tan ɑ=1
Cot ɑ=1/tan ɑ=1
tan2 ɑ+cot2=1+1=2

10). (A+B)’s 1 day’s work = 1/10
C’s 1 day’s work = 1/50
(A+B+C)’s 1 day’s work
=1/10+1/50= 5+1 /50=3/25
As 1 day’s work
=1/x = (B+C)’s 1 day’s work
ð  2/x = 3/25
ð  x=50/3
B’s 1 day’s work = 1/10 - 1/x
ð  1/10 - 3/50 = 5-3 /50 =1/25
B alone will complete the work in 25 days.