SSC CHSL- Practice Aptitude Questions (Mixed Problems) With Solutions Set-48

 Practice Aptitude Questions (Mixed Problems)
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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