### Quantitative Aptitude Practice Questions (With Solutions) Set-29

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
a)    √3:1
b)    1:√3
c)    √3:√2
d)    √3:2

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
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)
V2=4V1; V =3V1
V1=V2/4 =V3/3
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%
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%
5). Let the length and width be 5x and 2x respectively
Width 2x=40
x=20
Length 5x = 5×20 = 100m
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
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
8). √3tanθ =3sinθ
Tanθ  =3/√3sinθ=√3sinθ
→ sinθ / cosθ=√3sinθ
cos =1/√3
Now, sin2θ-cos2θ =1-2 cos2θ
= 1-2(1/√3)2 =1-2/3 =1/3
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
10).

BC : AC =√3:2