SSC CHSL- Practice Aptitude Questions (Mixed Problems) With Solutions Set-48:
Dear Readers, Important Aptitude Questions for SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.
1).
A hall is 100m long and 80m wide. It has to be designed by squares of maximum
area. Find the number of designed squares.
a)
50
b)
10
c)
20
d)
25
2).
A circle of maximum area is drawn in a rectangular park whose sides are 84m and
98m. Find the area of the park beyond the circle.
a)
2688m2
b)
5544m2
c)
5376m2
d)
2678m2
3).
A copper wire is bent in the form of an equilateral triangle and has arc 484√3
cm². If the same wire is bent into the form of a circle, the area enclosed by
the wire is (Take π=22/7)
a)
1386 cm²
b)
1698 cm²
c)
1586 cm²
d)
1832 cm²
4).
The diameters of a frustum are 36 cm and 24 cm, while the slant height is 10
cm. Find the total surface area of the
frustum. (Approx value)
a)
1206.85 cm²
b)
1536.38 cm²
c)
768.19 cm²
d)
2413.71 cm²
5).
A conical pot having radius 18 cm and height 75 cm is filled by a liquid. The
liquid is dropped into a cylindrical pot having radius 15 cm. What is the
height of the liquid in the cylindrical pot?
a)
36 cm
b)
39 cm
c)
42 cm
d)
30 cm
6).
Find the perimeter of a rhombus diagonals are 18 cm and 24 cm respectively.
a)
56 cm
b)
60 cm
c)
80 cm
d)
72 cm
7).
If sec A=5/3, then find (tan A+2cot A-3sin a)/ (4sin A+ sec A-2tan A)
a)
17/296
b)
16/265
c)
13/266
d)
19/263
8).
Find the least value of 16cosec²a+5sin²a.
a)
19
b)
40
c)
20
d)
41
9).
Find the numerical value of cot36〫[cot54〫.cos32〫+(1/tan54〫.sec² 58〫)]
a)
1
b)
0
c)
4
d)
2
10).
In circular measure, the value of the angle 12〫15’ is:
a)
49π/720
b)
48π/610π
c)
47π/360
d)
47π/720
11).
The sum of two angles is 105〫and their difference is π/2. The value of the
greater angle in radians is:
a) 5π/16 radian
b) 7π radian
c) 5π/12 radian
d) 3π radian
Answers:
1). c) 2).
a) 3).a ) 4). d) 5).a ) 6). b) 7).
c) 8).b ) 9).a ) 10).a ) 11).c)
Solutions:
1).
Length of hall=100m
Width of hall=80m
Side of square design=HCF of 100 and 80.
HCF of 100 and 80 is 20m
∴
Number of designs = (Area of floor)/ (Area of design)
= (100m*80m)/ (20m*20m) =5*4=20
Answer: (c)
2).
Length of rectangular park=98m
Breadth of rectangular park=84m
d= diameter of the circle=breadth of park=84m
∴Radius
of the circle= (Diameter/2) =84/2=42m
Area of the circle=πr²=22/7*42*42=22*42*6=5544m²
Now, area of the rectangle= (98*84)m²=8232m²
Area beyond circle=Area of rectangle-Area of circle= (8232-5544)m²=2688m²
Answer: (a)
3).
Area of the equilateral triangle= √3/4 side²
Then, √3/4 side²= √3*484 cm²
∴ side²=484*4
∴ side=√484*4=22*2=44 cm
Perimeter of the equilateral
triangle=Circumference of the circle
∴ perimeter of equilateral
triangle=3*side=3*44 cm=132 cm
Circumference of the circle=132 cm
We
know, circumference of the circle=2πr
Then,
2*(22/7)*r=132 ∴ r = (132*7)/
(22*2) or, r=21 cm
Area of circle=πr²= [(22/7)*21*21) cm²=1386
cm²
Answer: (a)
4).
Diameters of frustum BC=36 cm
ED=24 cm
∴Radius of frustum=d1/2, d2/2=36/2
cm, 24/2 cm =18 cm, 12 cm
Slant
height=10 cm
R1=18 cm, R2=12
cm
Total surface area of frustum=π [(R₁+R₂)
l+R12+R22] =22/7[
(18+12)10+182+122]
=22/7[30*10+324+144]
=22/7[300+324+144] =22/7[768] = 2413.71 cm2
Answer: (d)
5).
Radius of conical pot=18 cm
Height of conical pot=75 cm
Volume of conical pot=1/3πr²h
=1/3*π*18*18*75 cm3
Radius of cylindrical pot=15 cm
Base area of cylindrical pot=πr² =π*15
cm *15 cm
Height of liquid in the cylindrical
pot
= (Volume of liquid in conical pot)/
(Base area of cylindrical pot)
= (1*π*18*18*75)/ (3*π*15*15) =36 cm
Answer: (a)
6).
Diagonals of the
rhombus are 18 cm and 24 cm respectively. AC=24 cm, BD=18 cm
E is at the center of both diagonals
Then, AE=1/2 AC=1/2*24 cm=12 cm=EC
Again, DE=BE=1/2*18 cm=9 cm
In right-angled triangle DEC
DC²=DE²+EC² or, DC²=9²+12² or, DC²=81+144 or, DC²=225
∴DC=15 cm
∴Perimeter of rhombus=4*side=4*15 cm=60
cm
Answer: (b)
7).
Given sec A=5/3=h/b
p=√h²-b² =√25-9=√16=4
Now, (tan A+2cot A-3sin A)/ (4 sin A+5 sec
A-2 tan A)
= [(4/3) +2*(3/4)-3*(4/5)]/ [4*(4/5)
+5*(5/3)-2*(4/3)]
= [(4/3) + (3/2) – (12/5)]/ [(16/5) +
(25/3) – (8/3)] = [(40+45-72)/30]/ [(48+125-40)/15]
= [(85-72)/30]/ [(173-40)/15] = (13/30)/
(133/15) = (13/30) + (15*133) =13/266
Answer: (c)
8).
Given value=16 cosec²a+25 sin²a = (16/sin² a) +25 sin² a= (4/sin a)² + (5 sin
a)²
[∵ a²+b²= (a-b)²+2ab]
= [(4/sin a)-5sin
a]²+2*(4/sin a)*5 sin a =[(4-5 sin² a)/sin a]²+40
For least value [(4-5sin²a)/sin a] should
be zero.
Hence the least value is 40.
Answer: (b)
9). Cot36°
(cot54°.cos²+1/ (tan 54°.sec58°)
=tan54° (cot54°.cos²+1/ (tan 54°.sec58°)
[∵ tan θ=cot (90°-θ)]
=tan54° [(cos²32°)/ (tan 54°) + (cos²58°)/
(tan 54°)]
=tan54° [(cos²32°+sin²32°)/ (tan54°)]
(∵ cos²θ+sin²θ=1)
=tan 54°*(1/ tan54°) =1
Answer: (a)
10).
Given angle=12°15´
=12° (15/60)° =12°(1/4)° =(49/4)°
= [(49*4)* (π/180°)]c
=49π/720
Answer: (a)
11).
Let α and β be two angles
Given α + β=105° ……………..(ⅰ)
α – β=π/2=90°/2=45° …………..(ⅱ)
By solving equations (ⅰ) and (ⅱ), we get
∴ a=75°,
b=30°
Now, the value of greater angle in
radians=75*(π/180) =5π/12 radian
Answer: (c)
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