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
Answers:
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
∠B =∠ADE = 65o
∠BDE=180o-85o=115o
Answer:
b)
2).
AB=Tower =125 metre
C= Position of car
From ∆ABC
tan 45o=AB/BC
ð 1= AB/BC
ð AB= BC=125
metre
Answer:
a)
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
Answer:
b)
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)
Answer:
c)
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]
Answer:
d)
6). Sum of
three angles of triangle
= π radian
Least angle =2/2+5+3×π
=π/5 radians
Answer:
d)
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
Answer:A
)
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
Answer:
a)
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.
Answer:
c)
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