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SSC CGL 2019 Aptitude Test - 12
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SSC CGL 2019 Aptitude Test - 12

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  • Question 1/10
    1 / -0

    From the top of a building of height 150m, the angles of depression of two cars on either sides of the building are 45° and 60°. Then the distance b/w the cars are

    Solutions

    AD is a building

    And B & C are two cars

    In ABD

    tan45=AD/BD

     BD=AD/1=150m

    And

    In ADC

    tan60=AD/DC

    CD=AD/√3=150/√3 m

    BC=BD+DC=150(1+1/√3)

     

  • Question 2/10
    1 / -0

    If x – y = 9, then what is the value of  (x-25)^3-(y-16)^3?

    Solutions

     Given x-y=9

    (x-25-y+16)[(x-25)^2+(y-16)^2+(x-25)(y-16)] = (9-9)[(x-25)^2+(y-16)^2+(x-25)(y-16)] = 0

    (x-25)^3-(y-16)^3=0

     

  • Question 3/10
    1 / -0

    If x-y-√45=1 & x+y=3√5+1 then what is the value of 12xy(x^2-y^2 )?

    Solutions

     ATQ

    x-y=√45-1

    x-y=3√5-1              (i)

    x+y=3√5+1     (ii)

    multiply (i)&(ii)

    we get x^2-y^2=(3√5)^2-1=44     (iii)

    adding (i)&(ii)2x=6√5

    x=3√5

    subtracting (i)&(ii)

    y=1

    12xy(x^2-y^2 )=12×3√5×1×44=1584√5

     

  • Question 4/10
    1 / -0

    Solutions

     

  • Question 5/10
    1 / -0

    O and C are respectively the orthocenter and circumcenter of an acute angled triangle PQR. The points P and O are joined and produced to meet the side QR at S. If ∠PQR = 70° and ∠QCR = 140°, then ∠RPS = ?

    Solutions

    ∠QPR=1/2 ∠QCR=1/2  x 140=70

    And ∠PSQ=90

    In PQS

    ∠QPS=90-70=20

    ∠RPS=∠QPR-∠QPS
    =70-20=50

     

  • Question 6/10
    1 / -0

    In a ∆PQR, PM is the angle bisector of ∠QPR and ∠QPM=60°. What is the length of PM?

    Solutions

     

  • Question 7/10
    1 / -0

    Chords AC and BD of a circle with centre O intersect at right angles E. If ∠OAB = 35°, then the value of ∠EBC is :

    Solutions

    Join OB and OA

    Now in OAB

    OA=OB=radius

    ∠OBA=∠OAB=35

    ∠AOB=180 – (∠OBA+∠OAB)
    =180 – (35+35) =110

    Now angle made by same chord at the center & circumference

    ∠ADB= ½ x 110=55=∠ADE=∠BCE

    ∠EBC=180-(90+55)=35

     

  • Question 8/10
    1 / -0

    The base of a right prism is an equilateral triangle if the lateral surface area and volume is   100   respectively, then the length of the side of base of the prism is

    Solutions

     Lateral surface area = base permeter x h

    150 = 3 x side x h

    Side x h = 50 sq m  (i)

    Volume of prism = area of base x height

    100√3=√3/4×side^2×h

    H x side^2=400 cucm(ii)

    Dividing eq (ii) by (i)

    Side=  400/50=8cm

     

  • Question 9/10
    1 / -0

    Find the value of  (sinA+sinB)/(cosA+cosB)

    Solutions

     

  • Question 10/10
    1 / -0

    Solutions

    We  know  that

     

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