Quantitative Aptitude Practice Questions (With Solutions) Set-29:
List of Practice Questions for Upcoming SSC CGL Tier II Exam was
given here with solutions, candidates those who are preparing for those exams
can use this material
1).In a regular polygon if one of its internal
angle is greater than the external angle by 132°, then the number of sides of
the polygon is
a)
14
b)
12
c)
15
d)
16
2). If V1 V2 and V3
be the volumes of a right circular cone, a sphere and a right circular
cylinder having the same radius and same height, then
a)
V1 = V2/4 = V3/3
b)
V1/2 = V2/3 =
V3
c)
V1/3 = V2/2 =
V3
d)
V1/3 = V2 = V3/2
3). Successive discounts of 15% and 20% amount
to a single discount of
a)
35%
b)
32%
c)
30%
d)
28%
4). The marked price of an article is 10%
higher than the cost price. A discount of 10% is given on the marked price. In
this kind of sale, the seller bears
a)
No loss, no gain
b)
a loss of 5%
c)
a gain of 1%
d)
a loss of 1%
5). The ratio of the length of a school ground
to its width is 5:2.If the width is 40m, then the length is
a)
200 m
b)
100 m
c)
50 m
d)
80 m
6). If 7 men working 7 hours a day for each of
7 days produce 7 units of work, then the units of work produced by 5 men
working 5 hrs a day for each of 5 days is
a)
25/343
b)
125/49
c)
49/125
d)
343/25
7). If cos2∝-sin2∝=tan2
β,
then the value of cos2 β - sin2
β
is
a)
Cot2∝
b)
Cos2 β
c)
Tan2∝
d)
Tan2 β
8). If √3 tanθ=3sinθ, then the value of (sin2
θ
-cos2 θ) is
a)
1
b)
3
c)
1/3
d)
None
9). If sin (A+B) =sinA cos B+cosA sin B, then
the value of sin75° is
a)
√3+1/√2
b)
√2+1/2√2
c)
√3+1/2√2
d)
√3+1/2
10). ABC is a right angled triangle, right
angled at B and ∠A=60
and AB=20 cm, then the ratio of sides BC and
CA is
and AB=20 cm, then the ratio of sides BC and
CA is
a)
√3:1
b)
1:√3
c)
√3:√2
d)
√3:2
Answers:
1).
c) 2). a) 3).
b) 4). d) 5).
b) 6). b) 7).
c) 8). c) 9).
c) 10). d)
SOLUTIONS:-
1). Internal angle –External angle =132
[180(n-2)/n]
- 360/n =132
180 -
720/n =132
720/n
=48
∴n=720/48
=15
Answer:
c)
2). V1=Volume of the cone =1/3πr2h=1/3
πr3(r=h)
V2=Volume of a sphere =4/3 πr3 V3=Volume of the cylinder=r2h
r3(r=b)
r2h
r3(r=b)
V2=4V1; V =3V1
∴V1=V2/4
=V3/3
Answer:
a)
3). Formula:
Single discount equal to two successive discounts
X% and Y% is (x+y - xy/100)%
X=15; y=20
Required discount = 15+20 - 15×20/100
=15+20-3 =32%
Answer:
b)
4). Let the cost price be Rs.x.Then
marked price =110% of x
=110/100x =11x/10
After discount, selling price = 90% of 11x/10
= 90/100 × 11x/10 = 99x/100
Loss
=x - 99x/100 =x/100
Loss %= (x/100)/x × 100 =1%
Answer:
d)
5). Let the length and width be 5x and 2x
respectively
Width 2x=40
x=20
∴Length 5x = 5×20 = 100m
Answer:
b)
6). M1 =7; H1=7; D1=7
W1=7 units
M2 =5; H2=5; D2=5
W2=?
Formula:
M1H1D1W2=M2H2D2W1
7×7×7×W2 = 5×5×5×7
∴W2=(5×5×5×7)/(7×7×7)
=125/49
Answer:
c)
7). Cos2 α -Sin2 α =tan2 α
Cos2 α
= tan2 β
Now, Cos2 β -Sin2 β
Cos2
β =1-tan2 β
/ 1+ tan2β = 1- Cos2 α / 1+ Cos2 α
=2
Sin2 α
/2 Cos2 α
= tan2 α
Answer:
c)
8). √3tanθ =3sinθ
Tanθ =3/√3sinθ=√3sinθ
→ sinθ / cosθ=√3sinθ
∴cos
=1/√3
=1/√3
Now, sin2θ-cos2θ
=1-2 cos2θ
= 1-2(1/√3)2 =1-2/3 =1/3
Answer:
c)
9). sin(A+B)=sinA cosB+cosA sinB
Now sin75°=sin(45°+30°)
=sin45°cos30°+cos45°sin30°
=1/√2.√3/2+1/√2.1/2 =√3+1/2√2
Answer:
c)
10).
BC : AC =√3:2
Answer:
d)
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