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Mathematics Test - 10

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Mathematics Test - 10
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  • Question 1
    2.5 / -0.83

    The only correct statement about the matrix A is

    Solution



  • Question 2
    2.5 / -0.83

    Find the real numbers x and y such that : (x + iy)(3 + 2i) = 1 + i

    Solution

    (x + iy)(3 + 2i) = (1 + i)
    x + iy = (1 + i)/(3 + 2i)
    x + iy = [(1 + i) * (3 - 2i)] / [(3 + 2i)*(3 - 2i)]
    x + iy = (3 + 3i - 2i + 2) / [(3)2 + (2)2 ]
    x + iy = (5 + i)/[ 9 + 4]
    = (5 + i) / 13
    =>13x + 13iy = 5+i
    13x = 5     13y = 1
    x = 5/13     y = 1/13

  • Question 3
    2.5 / -0.83

    Express the following in standard form : (2 –√3i) (2 + √3i) + 2 –4i

    Solution

    Given: (2 −√3i)(2+√3i) + 2 −4i ​

    (2 −√3i)(2+√3i) = 7

    ⇒7 + 2 - 4i

    ⇒9 - 4i

  • Question 4
    2.5 / -0.83

    Find the reciprocal (or multiplicative inverse) of -2 + 5i  

    Solution

    -2 + 5i
    multiplicative inverse of -2 + 5i is
    1/(-2+5i)
    = 1/(-2+5i) * ((-2-5i)/(-2-5i))
    = -2-5i/(-2)^2 -(5i)^2
    = -2-5i/4-(-25)
    = -2-5i/4+25
    = -2-5i/29
    = -2/29 -5i/29

  • Question 5
    2.5 / -0.83

    The number of ways in which 6 “+ “and 4 “–“signs can be arranged in a line such that no two “–“signs occur together is

    Solution

    ′+′signs can be put in a row in one way creating seven gaps shown as arrows:
    Now 4 ′−′signs must be kept in these gaps. So, no tow ′−′signs should be together.
    Out of these 7 gaps 4 can be chosen in 7C4 ways.

  • Question 6
    2.5 / -0.83

    On a railway track, there are 20 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other is

    Solution

    Number of tickets selected from first station =20
    from second =19
    .... for last station =0
    We have to select 2 consecutive stations
    so total number of possible tickets = P(20,2)

  • Question 7
    2.5 / -0.83

    A class is composed 2 brothers and 6 other boys. In how many ways can all the boys be seated at the round table so that the 2 brothers are not seated besides each other?

    Solution

    Take 1 person from 6 and fix him and 5 others can arranged in -- 5! ways=120

    there are 6 places left in which 2 brothers can sit

    so they can choose any 2 places from 6 - 6C2 ways=15

    2 brothers can arrange themselves in 2! ways=15*2=30

    total ways=120*30=3600

  • Question 8
    2.5 / -0.83

    The minim um value of the s um of re al numb ers a –5 , a–4 , 3a–3 , 1, a8 and a10 with a >0 is

    Solution

    Using AM > GM

    ⇒ Sum = 8

  • Question 9
    2.5 / -0.83

    Let a1 , a2 , a3 , ..., a100 be an arithmetic progression with a1 = 3 and    For any integer n with 1 ≤n ≤20. let m=5n. If Sm /Sn   does not depend on n, then a2 is

    Solution



  • Question 10
    2.5 / -0.83

    Let a1 , a2 , a3 , ..... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which an <0 is 

    Solution




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