CTET Solved Paper - CTET Feburary 2016 Paper2

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141. ctetfeb16 q54

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142. One of the factors of x4 + 4 is

  • Option : C
  • Explanation : x4 + 4 = (x2)2 + (2)2
    = (x2 + 2)2 - 2.(x2).(2) [As a2 + b2 = (a+b)2 - 2ab]
    = (x2 + 2)2 - 4x2
    = (x2 + 2)2 - (2x)2
    = (x2 + 2 + 2x)(x2 + 2 - 2x) [As a2 - b2 = (a+b)(a-b)]
    ∴ (x2 - 2x + 2) is a factor of x4 + 4.
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143. The mean of the five observations x, x + 2, x + 4, x + 6, x + 8 is 11. Then, the mean of the first three observations is

  • Option : D
  • Explanation : Given, the mean of the 5 observations x, x + 2, x + 4, x + 6, x + 8 is 11.
    ctetfeb16 q56_d
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144. A bag has 5 red marbles, 4 green marbles and 3 blue marbles. All marbles are identical in all respects other than colour. A marble is taken out from the bag without looking into it. What is the probability that it is a non-green marble?

  • Option : B
  • Explanation : Given, a bag contains 5 red marbles, 4 green marbles and 3 blue marbles.
    Total number of marbles = 12.
    Total number of non-green marbles =8
    Here, total number of cases =12 and number of favourable cases = 8
    .'. Probability of getting a non-green marble =
    Number of favourable cases/Total number of cases.
    = 8/12. = 2/3.
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145. If AB x BA = BCB, where A, B and C stand for just one digit and A ≠ B ≠ C, then the value of A + B + C is

  • Option : B
  • Explanation : Given, AB x BA = BCB.
    Here, we cannot take AB as 11 because from question A ≠ B ≠ C.
    If we take A = 1 and B = 2, then 12x21 = 252.
    From, here we can say that A = 1, B = 2 and C = 5
    .'. A + B + C = 1 + 2 + 5 = 8.
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