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NDA II 2022 Mathematics Test - 3
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NDA II 2022 Mathematics Test - 3
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  • Question 1/10
    2.5 / -0.83

    A tower stands vertically on the grounD. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 600. Find the height of the tower.

    Solutions

    Let AB be the tower standing vertically on the ground and O be the position of the observer we now have:

    OA=20m angle OAB=900 and AOB=600

    Let

    AB=hm

    Now, in the right ∆OAB, we have:

    so, the height of the pole is 34.64m.

     

  • Question 2/10
    2.5 / -0.83

    An observer 1.5m tall is 30m away from a chimney. The angle of elevation of the top of the chimney from his eye is 600. Find the height of the chimney.

    Solutions

    Let CE and AD be the heights of the observer and the chimney, respectively.

    We have,

    BD=CE=1.5m

    BC=DE=30m

    And ACB=600

    In ∆ABC,

    tan 60=AB/BC

    √3=AD-BD/30

    AD-1.5=30√3

    AD=30√3+1.5

    AD=30x1.732+1.5

    AD=51.96+1.5

    AD=53.46m

    So, the height of the chimney is 53.46 m (approx).

     

  • Question 3/10
    2.5 / -0.83

    The angle of elevation of the top of an unfinished tower at a distance of 75m from its base is 300. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 600

    Solutions

    Let AB be the unfinished tower, AC be the raised tower and O be the point of observation

    We have:

    OA =75m

    AOB=300 and

    AOC=600

    Let

    AC=Hm such that

    BC= H-h m

    In AOB we have

    AB/OA=tan30­­­0=1/√3

    h/75=1/√3

    h=75/√3

    in AOC we have

    AC/OA=tan600=√3

    H/75=√3

    H=75x√3

    Required height=(H-h)=75√3-25√3=50√3=86.6m

     

  • Question 4/10
    2.5 / -0.83

    Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide, from a point P between them on the road, the angle of elevation of the top of one pole is 600 and the angle of depression from the top of another pole at P is 300. Find the distance of the point P from the pole having 300 angle..

    Solutions

    Let AB and CD be the equal poles; and BD be the width of the roaD.

    We have,

    AOB=60

    COD=30

    In ∆AOB, tan60=AB/BO

    √3=AB/BO

    BO=AB/√3

    In ∆COD, tan30=CD/DO

    1/√3=CD/DO

    DO=√3CD

    As, BD=80

    BO+DO=80

    AB/√3+√3CD=80

    AB/√3+√3AB=80 (AB=CD)

    AB(1/√3+√3)=80

    AB=80x√3/4

    AB=20√3

    Also, BO=AB/√3=20√3/√3=20m

    DO=80-20=60m

     

  • Question 5/10
    2.5 / -0.83

    Write the principal argument of (1 + i√3)2

    Solutions

    Let, Z = (1+i√3)2

    = (1)2 + (i√3)2 + 2 √3i

    = 1 -3 + 2 √3i

    Z = -2 + 2 √3i

     

  • Question 6/10
    2.5 / -0.83

    If f(x) = x2 - 3x + 4 and f(x)=f(2x+1) find the values of x

    Solutions

     

  • Question 7/10
    2.5 / -0.83

    Evaluate (i57 + i70 + i91 + i101 + i104).

    Solutions

    we have, i57+ i70+ i91+ i101+ i104

    = (i4)14.i + (i4)17.i2 + (i4)22.i3 + (i4)25.i + (i4)26

    We know that, i4 = 1

    ⇒ (i4)14.i + (i4)17.i2 + (i4)22.i3 + (i4)25.i + (i4)26

    = i + i2 + i3 + i +1

    = i -1 –i + i + 1

    =i

     

  • Question 8/10
    2.5 / -0.83

    In a committee, 50 people speak Hindi, 20 speak English and 10 speak both Hindi and English. How many speak at least one of these two languages?

    Solutions

    To Find: People who speak at least one of these two languages

    Let us consider,

    People who speak Hindi = n(H) = 50

    People who speak English = n(E) = 20

    People who speak both Hindi and English = n(H ∩ E) = 10

    People who speak at least one of the two languages = n(H ∪ E)

    Now, we know that,

    n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

    Therefore,

    n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

    = 50 + 20 – 10

    = 60

    Thus, People who speak at least one of the two languages are 60.

     

  • Question 9/10
    2.5 / -0.83

    Find the sum (i + i2 + i3 + i4 +……… up to 400 terms)., where n N.

    Solutions

    we have, i + i2+ i3+ i4+……… up to 400 terms

    We know that given series in GP where a = I, r = I and n = 400

     

  • Question 10/10
    2.5 / -0.83

    Write the coefficient of the middle term in the expansion of 

    Solutions

    To find: that the middle term in the expansion of

    Formula used:

    A General term tr+1 of binomial expansion (x+y)n is given by,

    Tr+1 = nCr Xn-r Yr where

    Total number of terms in the expansion is 11.

    Thus, the middle term of the expansion is T6 and is given by.

     

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